Electricity
Content
Table of Contents
Introduction ..............................................................................2 Electron Theory .........................................................................4 Conductors, Insulators and Semiconductors ............................5 Electric Charges ........................................................................7 Current......................................................................................9 Voltage .................................................................................... 11 Resistance .............................................................................. 13 Simple Electric Circuit ............................................................. 15 Ohm’s Law ............................................................................. 16 DC Series Circuit .................................................................... 18 DC Parallel Circuit ...................................................................23 Series-Parallel Circuits ............................................................30 Power......................................................................................34 Magnetism .............................................................................37 Electromagnetism ..................................................................39 Introduction to AC ...................................................................42 AC Generators ........................................................................44 Frequency ...............................................................................47 Voltage and Current ................................................................48 Inductance ..............................................................................51 Capacitance ............................................................................56 Inductive and Capacitive Reactance .......................................61 Series R-L-C Circuit .................................................................67 Parallel R-L-C Circuit ................................................................69 Power and Power Factor in an AC Circuit ................................ 71 Transformers ...........................................................................75 Three-Phase Transformers ......................................................80 Review Answers .....................................................................83 Final Exam ..............................................................................84 quickSTEP Online Courses .....................................................88 1
Introduction
Welcome to the first course in the STEP series, Siemens Technical Education Program designed to prepare our distributors to sell Siemens Energy & Automation products more effectively. This course covers Basics of Electricity and is designed to prepare you for subsequent courses on Siemens Energy & Automation products. Upon completion of Basics of Electricity you will be able to:
• • • • • • • • • • • •
Explain the difference between conductors and insulators Use Ohm’s Law to calculate current, voltage, and resistance Calculate equivalent resistance for series, parallel, or series-parallel circuits Calculate voltage drop across a resistor Calculate power given other basic values Identify factors that determine the strength and polarity of a current-carrying coil’s magnetic field Determine peak, instantaneous, and effective values of an AC sine wave Identify factors that effect inductive reactance and capacitive reactance in an AC circuit Calculate total impedance of an AC circuit Explain the difference between real power and apparent power in an AC circuit Calculate primary and secondary voltages of single-phase and three-phase transformers Calculate kVA of a transformer
The objectives listed above may sound strange to you. You may also wonder why you would need to know these things to sell Siemens Energy & Automation products. Developing a basic knowledge of electrical concepts, however, will help you to better understand customer applications. In addition, you will be better able to describe products to customers and determine important differences between products. If you are an employee of a Siemens Energy & Automation authorized distributor, fill out the final exam tear-out card and mail in the card. We will mail you a certificate of completion if you score a passing grade. Good luck with your efforts.
Electron Theory
Elements of an Atom
All matter is composed of molecules which are made up of a combination of atoms. Atoms have a nucleus with electrons orbiting around it. The nucleus is composed of protons and neutrons (not shown). Most atoms have an equal number of electrons and protons. Electrons have a negative charge (-). Protons have a positive charge (+). Neutrons are neutral. The negative charge of the electrons is balanced by the positive charge of the protons. Electrons are bound in their orbit by the attraction of the protons. These are referred to as bound electrons.
Electron Proton Nucleus
Free Electrons
Electrons in the outer band can become free of their orbit by the application of some external force such as movement through a magnetic field, friction, or chemical action. These are referred to as free electrons. A free electron leaves a void which can be filled by an electron forced out of orbit from another atom. As free electrons move from one atom to the next an electron flow is produced. This is the basis of electricity.
Conductors, Insulators and Semiconductors
Conductors
An electric current is produced when free electrons move from one atom to the next. Materials that permit many electrons to move freely are called conductors. Copper, silver, aluminum, zinc, brass, and iron are considered good conductors. Copper is the most common material used for conductors and is relatively inexpensive.
Insulators
Materials that allow few free electrons are called insulators. Materials such as plastic, rubber, glass, mica, and ceramic are good insulators.
An electric cable is one example of how conductors and insulators are used. Electrons flow along a copper conductor to provide energy to an electric device such as a radio, lamp, or a motor. An insulator around the outside of the copper conductor is provided to keep electrons in the conductor.
Rubber Insulator Copper Conductor
Semiconductors
Semiconductor materials, such as silicon, can be used to manufacture devices that have characteristics of both conductors and insulators. Many semiconductor devices will act like a conductor when an external force is applied in one direction. When the external force is applied in the opposite direction, the semiconductor device will act like an insulator. This principle is the basis for transitors, diodes, and other solidstate electronic devices.
Transistor
Diode
Review 1 1. List the three basic elements of an atom and state the charge of each (positive, negative, or neutral). Element Charge ____________ ____________ ____________ ____________ ____________ ____________
2. An electron forced out of orbit by an external force is called a ____________ ____________ . 3. Conductors allow ____________ free electrons to flow when an external electric force is applied. 4. Which of the following materials are good conductors? a. copper b. plastic c. silver d. rubber e. f. g. h. aluminum glass iron mica
5. Semiconductor devices can be manufactured to allow ____________ electrons to flow in one direction and ___ _________ electrons to flow in the opposite direction.
Electric Charges
Neutral State of an Atom
Elements are often identified by the number of electrons in orbit around the nucleus of the atoms making up the element and by the number of protons in the nucleus. A hydrogen atom, for example, has only one electron and one proton. An aluminum atom (illustrated) has 13 electrons and 13 protons. An atom with an equal number of electrons and protons is said to be electrically neutral.
Outer Band
Positive and Negative Charges
Electrons in the outer band of an atom are easily displaced by the application of some external force. Electrons which are forced out of their orbits can result in a lack of electrons where they leave and an excess of electrons where they come to rest. The lack of electrons is called a positive charge because there are more protons than electrons. The excess of electrons has a negative charge. A positive or negative charge is caused by an absence or excess of electrons. The number of protons remains constant.
Neutral Charge
Negative Charge
Positive Charge
Attraction and Repulsion of Electric Charges
The old saying, “opposites attract, ” is true when dealing with electric charges. Charged bodies have an invisible electric field around them. When two like-charged bodies are brought together, their electric fields repel one body from the other. When two unlike-charged bodies are brought together, their electric fields attract one body to the other. The electric field around a charged body forms invisible lines of force. These invisible lines of force cause the attraction or repulsion. Lines of force are shown leaving a body with a positive charge and entering a body with a negative charge.
Unlike Charges Attract Like Charges Repel
Coulomb’s Law
During the 18th century a French scientist, Charles A. Coulomb, studied fields of force that surround charged bodies. Coulomb discovered that charged bodies attract or repel each other with a force that is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them. Today we call this Coulomb’s Law of Charges. Simply put, the force of attraction or repulsion depends on the strength of the charges and the distance between them.
Current
Electricity is the flow of free electrons in a conductor from one atom to the next atom in the same general direction. This flow of electrons is referred to as current and is designated by the symbol “I” . Electrons move through a conductor at different rates and electric current has different values. Current is determined by the number of electrons that pass through a cross-section of a conductor in one second. We must remember that atoms are very small. It takes about 1,000,000,000,000,000,000,000,000 atoms to fill one cubic centimeter of a copper conductor. This number can be simplified using mathematical exponents. Instead of writing 24 zeros after the number 1, write 1024. Trying to measure even small values of current would result in unimaginably large numbers. For this reason current is measured in amperes which is abbreviated “amps” . The letter “A” is the symbol for amps. A current of one amp means that in one second about 6.24 x 1018 electrons move through a cross-section of conductor. These numbers are given for information only and you do not need to be concerned with them. It is important, however, to understand the concept of current flow.
Units of Measurement
The following chart reflects special prefixes that are used when dealing with very small or large values of current: Prefix Symbol Decimal 1 kiloampere 1 kA 1000 A 1 milliampere 1 mA 1/1000 A 1 microampere 1 mA 1/1,000,000 A
Direction of Current Flow
Some authorities distinguish between electron flow and current flow. Conventional current flow theory ignores the flow of electrons and states that current flows from positive to negative. To avoid confusion, this book will use the electron flow concept which states that electrons flow from negative to positive.
_
+
Electron Flow
_
+
Conventional Current Flow
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Voltage
Electricity can be compared with water flowing through a pipe. A force is required to get water to flow through a pipe. This force comes from either a water pump or gravity. Voltage is the force that is applied to a conductor that causes electric current to flow.
Water Flow Through a Pipe
Current Flow Through a Conductor
Electrons are negative and are attracted by positive charges. They will always be attracted from a source having an excess of electrons, thus having a negative charge, to a source having a deficiency of electrons, giving it a positive charge. The force required to make electricity flow through a conductor is called a difference in potential, electromotive force (emf), or voltage. Voltage is designated by the letter “E” , or the letter “V” . The unit of measurement for voltage is volts which is also designated by the letter “V” .
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Voltage Sources
An electrical voltage can be generated in various ways. A battery uses an electrochemical process. A car’s alternator and a power plant generator utilize a magnetic induction process. All voltage sources share the characteristic of an excess of electrons at one terminal and a shortage at the other terminal. This results in a difference of potential between the two terminals.
Shortage of Electrons Excess of Electrons
+
_
Batter y
Voltage Circuit Symbol
The terminals of a battery are indicated symbolically on an electrical drawing by two lines. The longer line indicates the positive terminal. The shorter line indicates the negative terminal.
+ _
Units of Measurement
The following chart reflects special prefixes that are used when dealing with very small or large values of voltage: Prefix Symbol Decimal 1 kilovolt 1 kV 1000 V 1 millivolt 1 mV 1/1000 V 1 microvolt 1 mV 1/1,000,000 V
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Resistance
A third factor that plays a role in an electrical circuit is resistance. All material impedes the flow of electrical current to some extent. The amount of resistance depends upon composition, length, cross-section and temperature of the resistive material. As a rule of thumb, resistance of a conductor increases with an increase of length or a decrease of crosssection. Resistance is designated by the symbol “R” . The unit of measurement for resistance is ohms (W). Resistance Circuit Symbols Resistance is usually indicated symbolically on an electrical drawing by one of two ways. An unfilled rectangle is commonly used. A zigzag line may also be used.
Resistance can be in the form of various components. A resistor may be placed in the circuit, or the circuit might contain other devices that have resistance. Units of Measurement The following chart reflects special prefixes that are commonly used when dealing with values of resistance: Prefix Symbol Decimal 1 kilohm 1 kW 1000 W 1 megohm 1 MW 1,000,000 W
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Review 2 1. Elements are identified by the number of ____________ in orbit around the nucleus.
2. A material that has an excess of electrons is said to have a ____________ charge. 3. A material that has a deficiency of electrons is said to have a ____________ charge. 4. Like charges ____________ and unlike charges ____________ . 5. The force that is applied to a conductor to cause current flow is ____________ . 6. Electrons move from ____________ . 7 . a. positive to negative b. negative to positive With an increase of length or a decrease of crosssection of a conductor, resistance will ____________ . a. increase b. decrease
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Simple Electric Circuit
An Electric Circuit
A fundamental relationship exists between current, voltage, and resistance. A simple electric circuit consists of a voltage source, some type of load, and a conductor to allow electrons to flow between the voltage source and the load. In the following circuit a battery provides the voltage source, electrical wire is used for the conductor, and a light provides the resistance. An additional component has been added to this circuit, a switch. There must be a complete path for current to flow. If the switch is open, the path is incomplete and the light will not illuminate. Closing the switch completes the path, allowing electrons to leave the negative terminal and flow through the light to the positive terminal.
+
Switch
_
+
_
An Electrical Circuit Schematic The following schematic is a representation of an electrical circuit, consisting of a battery, a resistor, a voltmeter and an ammeter. The ammeter, connected in series with the circuit, will show how much current flows in the circuit. The voltmeter, connected across the voltage source, will show the value of voltage supplied from the battery. Before an analysis can be made of a circuit, we need to understand Ohm’s Law.
+ + V _ R _ A
+ _
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Ohm’s Law
George Simon Ohm and Ohm’s Law
The relationship between current, voltage and resistance was studied by the 19th century German mathematician, George Simon Ohm. Ohm formulated a law which states that current varies directly with voltage and inversely with resistance. From this law the following formula is derived:
I= E R or Current = Voltage Resistance
Ohm’s Law is the basic formula used in all electrical circuits. Electrical designers must decide how much voltage is needed for a given load, such as computers, clocks, lamps and motors. Decisions must be made concerning the relationship of current, voltage and resistance. All electrical design and analysis begins with Ohm’s Law. There are three mathematical ways to express Ohm’s Law. Which of the formulas is used depends on what facts are known before starting and what facts need to be known.
I= E R E=IxR R= E I
Ohm’s Law Triangle
There is an easy way to remember which formula to use. By arranging current, voltage and resistance in a triangle, one can quickly determine the correct formula.
16
Using the Triangle
To use the triangle, cover the value you want to calculate. The remaining letters make up the formula.
I=
E R
E=IxR
R=
E I
Ohm’s Law can only give the correct answer when the correct values are used. Remember the following three rules: • • • Examples of Solving Ohm’s Law Current is always expressed in amperes or amps Voltage is always expressed in volts Resistance is always expressed in ohms
Using the simple circuit below, assume that the voltage supplied by the battery is 10 volts, and the resistance is 5 W.
+ + V _ R _ A
+ _
To find how much current is flowing through the circuit, cover the “I” in the triangle and use the resulting equation.
I= E R I= 10 Volts 5Ω I = 2 Amps
Using the same circuit, assume the ammeter reads 200 mA and the resistance is known to be 10 W. To solve for voltage, cover the “E” in the triangle and use the resulting equation.
E=IxR E = 0.2 x 10 E = 2 Volts
Remember to use the correct decimal equivalent when dealing with numbers that are preceded with milli (m), micro (m) or kilo (k). In this example, had 200 been used instead of converting the value to 0.2, the wrong answer of 2000 volts would have been calculated.
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DC Series Circuit
Resistance in a Series Circuit
A series circuit is formed when any number of resistors are connected end-to-end so that there is only one path for current to flow. The resistors can be actual resistors or other devices that have resistance. The following illustration shows four resistors connected end-to-end. There is one path of current flow from the negative terminal of the battery through R4, R3, R2, R1 returning to the positive terminal.
R1 + _ R2 R3 R4
Formula for Series Resistance
The values of resistance add in a series circuit. If a 4 W resistor is placed in series with a 6 W resistor, the total value will be 10 W. This is true when other types of resistive devices are placed in series. The mathematical formula for resistance in series is:
Rt = R1 + R2 + R3 + R4 + R5 ... + Rn
In this example, the circuit includes five series resistors.
11 KΩ R1 2 KΩ R2 2 KΩ R3 100 Ω R4 1 KΩ R5
+ _
Rt = R1 + R2 + R3 + R4 + R5 Rt = 11,000 + 2,000 + 2,000 + 100 + 1,000 Rt = 16,100 Ω
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Current in a Series Circuit
The equation for total resistance in a series circuit allows us to simplify a circuit. Using Ohm’s Law, the value of current can be calculated. Current is the same anywhere it is measured in a series circuit.
E R 12 I= 10 I= I = 1.2 Amps 5Ω R1
+ _
1Ω R2
2Ω R3
2Ω R4
+ _
10 Ω Rt
12 Volts
12 Volts
Original Circuit
Equivalent Circuit
Voltage in a Series Circuit
Voltage can be measured across each of the resistors in a circuit. The voltage across a resistor is referred to as a voltage drop. A German physicist, Gustav Kirchhoff, formulated a law which states the sum of the voltage drops across the resistances of a closed circuit equals the total voltage applied to the circuit. In the following illustration, four equal value resistors of 1.5 W each have been placed in series with a 12 volt battery. Ohm’s Law can be applied to show that each resistor will “drop” an equal amount of voltage.
12 V
3V 1.5 Ω R1
3V 1.5 Ω R2
+ _
3V 1.5 Ω R3
3V 1.5 Ω R4
12 Volt Battery
19
First, solve for total resistance:
Rt = R1 + R2 + R3 + R4 Rt = 1.5 + 1.5 + 1.5 + 1.5 Rt = 6 Ω
Second, solve for current:
I= E R I = 12 6 I = 2 Amps
Third, solve for voltage across any resistor:
E=IxR E = 2 x 1.5 E = 3 Volts
If voltage were measured across any single resistor, the meter would read three volts. If voltage were read across a combination of R3 and R4 the meter would read six volts. If voltage were read across a combination of R2, R3, and R4 the meter would read nine volts. If the voltage drops of all four resistors were added together the sum would be 12 volts, the original supply voltage of the battery. Voltage Division in a Series Circuit It is often desirable to use a voltage potential that is lower than the supply voltage. To do this, a voltage divider, similar to the one illustrated, can be used. The battery represents Ein which in this case is 50 volts. The desired voltage is represented by Eout, which mathematically works out to be 40 volts. To calculate this voltage, first solve for total resistance.
Rt = R1 + R2 Rt = 5 + 20 Rt = 25 Ω
20
Second, solve for current:
Ein Rt 50 I= 25 I= I = 2 Amps
Finally, solve for voltage:
Eout = I x R2 Eout = 2 x 20 Eout = 40 Volts
+ _
R1
Ein
5Ω
R2
20 Ω
Eout 40 Volts
21
Review 3 1. The basic Ohm’s Law formula is ____________ . 2. When solving circuit problems; current must always be expressed in ____________ , voltage must always be expressed in ____________ , and resistance must always be expressed in ____________ . 3. The total current of a simple circuit with a voltage supply of 12 volts and a resistance of 24 W is ____________ amps. 4. What is the total resistance of a series circuit with the following values: R1=10 W, R2=15 W, and R3=20 W? ____________ W. 5. What is total current of a series circuit that has a 120 volt supply and 60 W resistance? 6. In the following circuit, the voltage dropped across R1 is ____________ volts and R2 is ____________ volts.
1.5 Ω
R1
1.5 Ω
R2
+ _
12 Volts
7 . In the following circuit, voltage dropped across R1 is ____________ volts and across R2 is ____________ volts.
5Ω
R1
20 Ω
R2
+ _
100 Volts
22
DC Parallel Circuit
Resistance in a Parallel Circuit
A parallel circuit is formed when two or more resistances are placed in a circuit side-by-side so that current can flow through more than one path. The illustration shows two resistors placed side-by-side. There are two paths of current flow. One path is from the negative terminal of the battery through R1 returning to the positive terminal. The second path is from the negative terminal of the battery through R2 returning to the positive terminal of the battery.
+ _
R1
R2
Formula for Equal Value Resistors in a Parallel Circuit
To determine the total resistance when resistors are of equal value in a parallel circuit, use the following formula:
Rt = Value of any one Resistor Number of Resistors
In the following illustration there are three 15 W resistors. The total resistance is:
Rt = Value of any one Resistor Number of Resistor
15 Rt = 3 Rt = 5 Ω
+ _
R1 15 Ω
R2 15 Ω
R3 15 Ω
23
Formula for Unequal There are two formulas to determine total resistance for Resistors in a Parallel Circuit resistors of any value in a parallel circuit. The first formula is used when there are any number of resistors.
1 = 1 + 1 + 1 ... + 1 Rt R1 R2 R3 Rn
In the following illustration, there are three resistors, each of different value. Solve for the total resistance as follows:
1 = 1 + 1 + 1 Rt R1 R2 R3 1 = Rt 1 + 1 + 1 5 10 20 Insert Values for the Resistors Find the Lowest Common Denominator Add the Numerators Invert Both Sides of the Equation Divide
1 = 4 + 2 + 1 Rt 20 20 20 1 = 7 Rt 20 Rt = 20 1 7 Rt = 2.86 Ω
+ _
R1
R2
R3
5Ω
10 Ω
20 Ω
24
The second formula is used when there are only two resistors.
R1 x R2 Rt = R1 + R2
In the following illustration there are two resistors, each of different value. The total resistance is:
R1 x R2 Rt = R1 + R2 Rt = 5 x 10 5 + 10
50 Rt = 15 Rt =
3.33 Ω
+ _
R1 5Ω
R2 10 Ω
Voltage in a Parallel Circuit
When resistors are placed in parallel across a voltage source, the voltage is the same across each resistor. In the following illustration three resistors are placed in parallel across a 12 volt battery. Each resistor has 12 volts available to it.
12 Volt Battery + _ R1 12 V R2 12 V R3 12 V
25
Current in a Parallel Circuit
Current flowing through a parallel circuit divides and flows through each branch of the circuit.
It
+ _
R1 I1 It
R2 I2
R3 I3
Total current in a parallel circuit is equal to the sum of the current in each branch. The following formula applies to current in a parallel circuit.
It = I1 + I2 + I3 ... + In
Current Flow with Equal Value Resistors in a Parallel Circuit
When equal resistances are placed in a parallel circuit, opposition to current flow is the same in each branch. In the following circuit R1 and R2 are of equal value. If total current (It) is 10 amps, then 5 amps would flow through R1 and 5 amps would flow through R2.
It = 10 Amps
+ _ I1 = 5 Amps
R1 I2 = 5 Amps
R2
It = 10 Amps It = I 1 + I 2 It = 5 Amps + 5 Amps It = 10 Amps
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Current Flow with Unequal Value Resistors in a Parallel Circuit
When unequal value resistors are placed in a parallel circuit, opposition to current flow is not the same in every circuit branch. Current is greater through the path of least resistance. In the following circuit R1 is 40 W and R2 is 20 W. Small values of resistance means less opposition to current flow. More current will flow through R2 than R1.
12 Volts + _
R1 40 Ω I1 = 0.3 Amps It = 0.9 Amps
R2 20 Ω I2 = 0.6 Amps
Using Ohm’s Law, the total current for each circuit can be calculated.
I1 = I1 = E R1 12 Volts 40 Ω
I1 = 0.3 Amps
I2 = I2 =
E R2 12 Volts 20 Ω
I2 = 0.6 Amps
It = I 1 + I 2 It = 0.3 Amps + 0.6 Amps It = 0.9 Amps
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Total current can also be calculated by first calculating total resistance, then applying the formula for Ohm’s Law.
R1 x R2 Rt = R1 + R2 40 Ω x 20 Ω Rt = 40 Ω + 20 Ω 800 Ω Rt = 60 Ω Rt = 13.333 Ω
It = E Rt It = 12 Volts 13.333 Ω It = 0.9 Amps
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Review 4 1. The total resistance of a parallel circuit that has four 20 W resistors is ____________ W.
2. Rt for the following circuit is ____________ W.
+ _
R1 10 Ω
R2 20 Ω
R3 30 Ω
3. Rt for the following circuit is ____________ W.
+ _
R1 5Ω
R2 10 Ω
4. Voltage available at R2 in the following circuit is ____________ volts.
12 Volts
+ _
R1 5Ω
R2 10 Ω
5. In a parallel circuit with two resistors of equal value and a total current flow of 12 amps, the value of current through each resistor is ____________ amps. 6. In the following circuit, current flow through R1 is _____ _______ amps, and through R2 is ____________ amps.
24 Volts + _
R1 10 Ω
R2 15 Ω
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Series-Parallel Circuits
Series-parallel circuits are also known as compound circuits. At least three resistors are required to form a series-parallel circuit. The following illustrations show the two simplest ways a series-parallel combination can be represented.
Parallel Branches + _
Parallel Branches Devices in Series + _
Simplifying a Series-Parallel The formulas required for solving current, voltage and resistance problems have already been defined. To solve a series-parallel circuit, reduce the compound circuits to equivalent simple circuits. In the following illustration, R1 and R2 are parallel with each other. R3 is in series with the parallel circuit of R1 and R2.
R1 10 Ω R3 10 Ω + _
R2 10 Ω
30
First, use the formula to determine total resistance of a parallel circuit to find the total resistance of R1 and R2. When the resistors in a parallel circuit are equal, the following formula is used:
R = Value of any One Resistor Number of Resistors
R = 10 Ω 2 R = 5Ω
Second, redraw the circuit showing the equivalent values. The result is a simple series circuit which uses already learned equations and methods of problem solving.
R3 10Ω + _ R3 5Ω
Simplifying a Series-Parallel Circuit to a Parallel Circuit
In the following illustration R1 and R2 are in series with each other. R3 is in parallel with the series circuit of R1 and R2.
R1 10Ω + _ R2 10Ω R3 20Ω
First, use the formula to determine total resistance of a series circuit to find the total resistance of R1 and R2. The following formula is used:
R = R1 + R2 R = 10 Ω + 10 Ω R = 20 Ω
31
Second, redraw the circuit showing the equivalent values. The result is a simple parallel circuit which uses already learned equations and methods of problem solving.
+ _
R = 20 Ω
R3 = 20 Ω
+ _
Rt = 10 Ω
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Review 5 1.
R3 10 Ω
Calculate equivalent resistance for R1 and R2 and total resistance for the entire circuit.
R1 20 Ω
+ _
R2 30 Ω
R1,R2 equivalent resistance = ____________ W Total resistance = ____________ W
2. Calculate equivalent resistance for R1 and R2 and total resistance for the entire circuit.
R1 30 Ω + _ R2 10 Ω R3 20 Ω
R1,R2 equivalent resistance = ____________ W Total resistance = ____________ W
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Power
Work
Whenever a force of any kind causes motion, work is accomplished. In the illustration below work is done when a mechanical force is used to lift a weight. If a force were exerted without causing motion, then no work is done.
Electric Power
In an electrical circuit, voltage applied to a conductor will cause electrons to flow. Voltage is the force and electron flow is the motion. The rate at which work is done is called power and is represented by the symbol “P” . Power is measured in watts, represented by the symbol “W” . In a direct current circuit, one watt is the rate work is done in a circuit when 1 amp flows with 1 volt applied. In a DC circuit, power is the product of voltage times current. Later in this course, you will learn a slightly different version of this relationship for an alternating current (AC) circuit.
P=ExI or P = EI
Power Formulas
Two other power equations can be derived from this formula by substituting other components of Ohm’s Law.
P = I2R and P= E2 R
34
DC Circuit Example
In the following illustration, power can be calculated using any of the power formulas.
I = 2 Amps + _ R=6Ω
12 Volts
P = EI P = 12 Volts x 2 Amps P = 24 Watts P = I2 R P = (2 Amps)2 x 6 Ω P = 24 Watts P= P= P= E2 R (12 Volts)2 6Ω 144 6
P = 24 Watts
Additional Calculations
Electrical equipment often has a power rating expressed in watts. This rating is an indication of the rate at which electrical equipment converts electrical energy into some other form of energy, such as heat or mechanical energy. If the power associated with a device and its operating voltage are known, other quantities can be easily calculated. For example, a common household lamp may be rated for 120 volts and 100 watts. Using Ohm’s Law, the rated value of resistance of the lamp can be calculated.
P= R= E2 R
2 which can be transposed to R = E P
(120 Volts)2 100 Watts
R = 144 Ω
35
Using the basic Ohm’s Law formula, the amount of current flow for the 120 volt, 100 watt lamp can be calculated.
I= I= E R 120 Volts 144 Ω
I = 0.833 Amps
By comparison, a lamp rated for 120 volts and 75 watts has a resistance of 192 W and a current of 0.625 amps would flow if the lamp had the rated voltage applied to it.
R= R= E2 P (120 Volts)2 75 Watts
R = 192 Ω I= I= E2 P 120 Volts 192 Ω
I = 0.625 Amps
36
Magnetism
The principles of magnetism are an integral part of electricity. In fact, magnetism can be used to produce electric current and vice versa. Types of Magnets When we think of a permanent magnet, we often envision a horse-shoe or bar magnet or a compass needle, but permanent magnets come in many shapes.
However, all magnets have two characteristics. They attract iron and, if free to move (like the compass needle), a magnet will assume a north-south orientation. Magnetic Lines of Flux Every magnet has two poles, one north pole and one south pole. Invisible magnetic lines of flux leave the north pole and enter the south pole. While the lines of flux are invisible, the effects of magnetic fields can be made visible. When a sheet of paper is placed on a magnet and iron filings loosely scattered over it, the filings will arrange themselves along the invisible lines of flux.
37
By drawing lines the way the iron filings have arranged themselves, the following picture is obtained. Broken lines indicate the paths of magnetic flux lines. The field lines exist outside and inside the magnet. The magnetic lines of flux always form closed loops. Magnetic lines of flux leave the north pole and enter the south pole, returning to the north pole through the magnet.
Interaction between Two Magnets
When two magnets are brought together, the magnetic flux field around the magnets causes some form of interaction. Two unlike poles brought together cause the magnets to attract each other. Two like poles brought together cause the magnets to repel each other.
38
Electromagnetism
Left-Hand Rule for Conductors
An electromagnetic field is a magnetic field generated by current flow in a conductor. Every electric current generates a magnetic field and a relationship exists between the direction of current flow and the direction of the magnetic field. The lefthand rule for conductors demonstrates this relationship. If a current-carrying conductor is grasped with the left hand with the thumb pointing in the direction of electron flow, the fingers will point in the direction of the magnetic lines of flux.
Current-Carrying Coil
A coil of wire carrying a current, acts like a magnet. Individual loops of wire act as small magnets. The individual fields add together to form one magnet. The strength of the field can be increased by adding more turns to the coil, increasing the amount of current, or winding the coil around a material such as iron that conducts magnetic flux more easily than air.
39
Left-Hand Rule for Coils
A left-hand rule exists for coils to determine the direction of the magnetic field. The fingers of the left hand are wrapped around the coil in the direction of electron flow. The thumb points to the north pole of the coil.
Electromagnets
An electromagnet is composed of a coil of wire wound around a core. The core is made of soft iron or some other material that easily conducts magnetic lines of force. When current is passed through the coil, the core becomes magnetized. The ability to control the strength and direction of the magnetic force makes electromagnets useful. As with permanent magnets, opposite poles attract. An electromagnet can be made to control the strength of its field which controls the strength of the magnetic poles. A large variety of electrical devices such as motors, circuit breakers, contactors, relays and motor starters use electromagnetic principles.
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Review 6 1. The rate at which work is done is called ___________ . 2. The basic formulas for power in a DC circuit are: P = I x __ , P = I2 x __ , and P = _ R 3. In a circuit with a 12 volt supply and 4 W resistance, the power consumed is ____________ watts. 4. The two characteristics of all magnets are: they attract and hold ____________ , and, if free to move, they will assume roughly a ____________ position. 5. Lines of flux always leave the ____________ pole and enter the ____________ pole. 6. The left-hand rule for conductors states that, when the ___________ hand is placed on a current-carrying conductor with the ____________ pointing in the direction of electron flow, the fingers will point in the direction of ____________ .
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Introduction to AC
The supply of current for electrical devices may come from a direct current (DC) source, or an alternating current (AC) source. In a direct current circuit, electrons flow continuously in one direction from the source of power through a conductor to a load and back to the source of power. Voltage in direct current remains constant. DC power sources include batteries and DC generators. By contrast, an AC generator makes electrons flow first in one direction then in another. In fact, an AC generator reverses its terminal polarities many times a second, causing current to change direction with each reversal.
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AC Sine Wave
Alternating voltage and current vary continuously. The graphic representation for AC is a sine wave. A sine wave can represent current or voltage. There are two axes. The vertical axis represents the direction and magnitude of current or voltage. The horizontal axis represents time.
+ Direction
0
Time
- Direction
When the waveform is above the time axis, current is flowing in one direction. This is referred to as the positive direction. When the waveform is below the time axis, current is flowing in the opposite direction. This is referred to as the negative direction. A sine wave moves through a complete rotation of 360 degrees, which is referred to as one cycle. Alternating current goes through many of these cycles each second. Single-Phase and Three-Phase AC Power Alternating current is divided into single-phase and threephase types. Single-phase power is used for small electrical demands such as found in the home. Three-phase power is used where large blocks of power are required, such as found in commercial applications and industrial plants. Single-phase power is shown in the above illustration. Three-phase power, as shown in the following illustration, is a continuous series of three overlapping AC cycles. Each wave represents a phase, and is offset by 120 electrical degrees.
Phase 1 Phase 2 Phase 3 +
0
-
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AC Generators
Basic Generator
A basic generator consists of a magnetic field, an armature, slip rings, brushes and a resistive load. In a commercial generator, the magnetic field is created by an electromagnet, but, for this simple generator, permanent magnets are used. An armature is any number of conductive wires wound in loops which rotates through the magnetic field. For simplicity, one loop is shown. When a conductor is moved through a magnetic field, a voltage is induced in the conductor. As the armature rotates through the magnetic field, a voltage is generated in the armature which causes current to flow. Slip rings are attached to the armature and rotate with it. Carbon brushes ride against the slip rings to conduct current from the armature to a resistive load.
Pole Piece
Magnetic Field Armature
Brush Slip Ring
R1
Basic Generator Operation
An armature rotates through the magnetic field. At an initial position of zero degrees, the armature conductors are moving parallel to the magnetic field and not cutting through any magnetic lines of flux. No voltage is induced.
R1
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Generator Operation from Zero to 90 Degrees
As the armature rotates from zero to 90 degrees, the conductors cut through more and more lines of flux, building up to a maximum induced voltage in the positive direction.
90 Degrees
R1
Generator Operation from 90 to 180 Degrees
The armature continues to rotate from 90 to 180 degrees, cutting fewer lines of flux. The induced voltage decreases from a maximum positive value to zero.
180 Degrees
S
R1
Generator Operation from 180 to 270 Degrees
As the armature continues to rotate from 180 degrees to 270 degrees, the conductors cut more lines of flux, but in the opposite direction, and voltage is induced in the negative direction, building up to a maximum at 270 degrees.
270 Degrees
R1
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Generator Operation from 270 to 360 Degrees
As the armature continues to rotate from 270 to 360 degrees, induced voltage decreases from a maximum negative value to zero. This completes one cycle. The armature continues to rotate at a constant speed causing the cycle to repeat as long as the armature rotates.
360 Degrees
S
One Revolution R1
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Frequency
The number of cycles per second of voltage induced in the armature is the frequency of the generator. If the armature rotates at a speed of 60 revolutions per second, the generated voltage will be 60 cycles per second. The recognized unit for frequency is hertz, abbreviated “Hz” . 1 Hz is equal to 1 cycle per second. Power companies generate and distribute electricity at very low frequencies. The standard power line frequency in the United States and many other countries is 60 Hz. 50 Hz is also a common power line frequency used throughout the world. The following illustration shows 15 cycles in 1/4 second which is equivalent to 60 Hz.
1/4 Second
Four-Pole AC Generator
The frequency is the same as the number of rotations per second if the magnetic field is produced by only two poles. An increase in the number of poles, would cause an increase in the number of cycles completed in a revolution. A two-pole generator would complete one cycle per revolution and a fourpole generator would complete two cycles per revolution. An AC generator produces one cycle per revolution for each pair of poles.
One Revolution R1
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Voltage and Current
Peak Value
Voltage and current in an AC circuit rise and fall over time in a pattern referred to as a sine wave. The peak value of a sine wave occurs twice each cycle, once at the positive maximum value and once at the negative maximum value.
+ Peak Value
0
Time
-
Peak Value
Peak-to-Peak Value
The value of the voltage or current between the peak positive and peak negative values is called the peak-to-peak value.
+
0
Time
Peak-to-Peak Value
-
Instantaneous Value
The instantaneous value is the value at any one point in the sine wave.
+ Instantaneous Value
0
Time
-
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Calculating Instantaneous Voltage
The voltage waveform produced as the armature rotates through 360 degrees rotation is called a sine wave because the instantaneous voltage (e) is related to the sine trigonometric function. The sine of an angle is represented symbolically as sin q, where the Greek letter theta (q) represents the angle. The sine curve is a graph of the following equation for values of q from 0 to 360 degrees:
e = Epeak x sin θ
Instantaneous voltage is equal to the peak voltage times the sine of the angle of the generator armature. The sine value is obtained from trigonometric tables. The following table shows selected instantaneous values.
Angle 30 Degrees 60 Degrees 90 Degrees 120 Degrees 150 Degrees 180 Degrees Sin 0.5 0.866 1 0.866 0.5 0
θ
Angle 210 Degrees 240 Degrees 270 Degrees 300 Degrees 330 Degrees 360 Degrees
Sin
θ
-0.5 -0.866 -1 -0.866 -0.5 0
The following example illustrates instantaneous values at 90, 150, and 240 degrees. The peak voltage is equal to 100 volts. By substituting the sine at the instantaneous angle value, the instantaneous voltage can be calculated.
+ 90° = +100 Volts 150° = +50 Volts 0 240° = -86.6 Volts -
Any instantaneous value can be calculated. For example:
240° e = 100 x -0.866 e = -86.6 volts
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Effective Value of an AC Sine Wave
The instantaneous value of an alternating voltage and current are constantly changing values. However, there is a method for translating the varying values into an equivalent constant value, referred to at the effective value of voltage or current. This is also known as the RMS value. RMS is an abbreviation of the mathematical term root-mean-square. For example, a common voltage used in many applications is 120 volts, this is an RMS value, which is equal to the peak value times 0.707 .
+ Peak Value 169.7 Volts 0
-
169.7 Vpeak x 0.707 = 120 Vrms
The effective value of AC is defined in terms of an equivalent heating effect when compared to DC. One RMS ampere of current flowing through a resistance will produce heat at the same rate as one DC ampere. It is sometimes necessary to know the peak value of an AC voltage or current when the RMS value is known. To calculate the peak value, multiply the effective value by 1.41. For example, if the effective value is 100 volts, the peak value is 141 volts. Review 7 1. A ____________ is the graphic representation of AC voltage or current values over time.
2. Each phase of three phase AC power is offset by ____________ degrees. 3. An AC generator produces ____________ cycle(s) per revolution for each pair of poles. 4. What is the instantaneous voltage at 240 degrees for a sine wave with a peak voltage of 150 volts? 5. What is the effective voltage for a sine wave with a peak voltage of 150 volts?
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Inductance
The circuits studied to this point have been resistive. Resistance and voltage are not the only circuit properties that effect current flow, however. Inductance is the property of an electric circuit that opposes any change in electric current. Resistance opposes current flow, inductance opposes change in current flow. Inductance is designated by the letter “L ” . . The unit of measurement for inductance is the henry (h). Current Flow and Field Strength Current flow produces a magnetic field in a conductor. The amount of current determines the strength of the magnetic field. As current flow increases, field strength increases, and as current flow decreases, field strength decreases.
0 Degrees No Current 30 Degrees Increasing Current 90 Degrees Maximum Current
Any change in current causes a corresponding change in the magnetic field surrounding the conductor. Current is constant for a regulated DC source, except when the circuit is turned on and off, or when there is a load change. However, alternating current is constantly changing, and inductance is continually opposing the change. A change in the magnetic field surrounding the conductor induces a voltage in the conductor. This self-induced voltage opposes the change in current. This is known as counter emf. This opposition causes a delay in the time it takes current to attain its new steady value. If current increases, inductance tries to hold it down. If current decreases, inductance tries to hold it up. Inductance is somewhat like mechanical inertia which must be overcome to get a mechanical object moving or to stop a mechanical object from moving. A vehicle, for example, takes a few moments to accelerate to a desired speed, or decelerate to a stop. 51
Inductors
All conductors have inductance, but inductors are coils of wire wound for a specific inductance. For some applications, inductors are wound around a metal core to further concentrate the inductance. The inductance of a coil is determined by the number of turns in the coil, the coil diameter and length, and the core material. An inductor is usually indicated symbolically on an electrical drawing as a curled line or a filled rectangle.
Simple Inductive Circuit
In a resistive circuit, current change is considered instantaneous. If an inductor is used, the current does not change as quickly. For the purpose of explanation, a DC circuit is used here to describe the operation of an inductor. There will always be some amount of resistance and inductance in any circuit. The electrical wire used in the circuit has some resistance and inductance. In addition, inductors also have resistance. However, to simplify examples in this book, the resistance and inductance of the wiring and the resistance of inductors are not considered. In the following circuit, initially the switch is in position 2, and there is no current flowing through the ammeter (A). When the switch is moved to position 1, current will rise rapidly at first, then more slowly as the maximum value is approached.
1 2 A
+ _
L1
R1
Inductive Time Constant
52
The time required for the current to rise to its maximum value is determined by the ratio of inductance (in henrys) to resistance (in ohms). This ratio is called the time constant of the inductive circuit. A time constant is the time (in seconds) required for the circuit current to rise to 63.2% of its maximum value. When the switch is closed in the previous circuit, current will begin to flow. During the first time constant current rises to 63.2% of its maximum value. During the second time constant, current rises to 63.2% of the remaining 36.8%, or a total of 86.4%. It takes about five time constants for current to reach its maximum value.
100.0% 98.1% 94.9% 86.4%
63.2%
First Time Constant
Second Time Constant
Third Time Constant
Fourth Time Constant
Fifth Time Constant
Similarly, when the switch in the previous circuit is returned to position 2, the magnetic field around the inductor will begin to collapse, returning stored energy to the circuit, and it will take about five time constants for current to reach zero.
100.0% First Time Constant Second Time Constant Third Time Constant Fourth Time Constant Fifth Time Constant
36.8% 13.6% 5.1% 1.9% 0%
Calculating the Time Constant of an Inductive Circuit
The time constant is designated by the symbol “t” . To determine the time constant of an inductive circuit use one of the following formulas:
τ (in seconds) = L (henrys) R (ohms) L (millihenrys) R (ohms) L (microhenrys) R (ohms)
τ (in milliseconds) =
τ (in microseconds) =
53
In the following illustration, L1 is equal to 15 millihenrys and R1 is equal to 5 W. When the switch is closed, it will take 3 milliseconds for current to rise from zero to 63.2% of its maximum value and approximately 15 milliseconds for full current to be reached.
+ _
L1 15 mh
R1 5 Ω
τ=
15 mh 5Ω
τ = 3 milliseconds
Formula for Series Inductors The same rules for calculating total resistance can be applied to calculating total inductance. In the following circuit, an AC generator is used to supply electrical power to four inductors. Total inductance of series inductors is calculated using the following formula:
Lt = L1 + L2 + L3 ... + Ln 2 mh L1 AC Generator 2 mh L2 1 mh L3 1 mh L4
Lt = L1 + L2 + L3 + L4 Lt = 2 mh + 2 mh + 1 mh + 1 mh Lt = 6 mh
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Formula for Parallel Inductors
In the following circuit, an AC generator is used to supply electrical power to three inductors. Total inductance is calculated using the following formula:
1 = 1 + 1 + 1 ... + 1 Lt L1 L2 L3 Ln
L1 5 mh
L2 10 mh
L3 20 mh
1 = Lt
1 + 1 + 1 5 10 20
1 = 7 Lt 20 Lt = 20 7 Lt = 2.86 mh
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Capacitance
Capacitance and Capacitors Capacitance is a measure of a circuit’s ability to store an electrical charge. A device manufactured to have a specific amount of capacitance is called a capacitor. A capacitor is made up of a pair of conductive plates separated by a thin layer of insulating material. Another name for the insulating material is dielectric material. When a voltage is applied to the plates, electrons are forced onto one plate. That plate has an excess of electrons while the other plate has a deficiency of electrons. The plate with an excess of electrons is negatively charged. The plate with a deficiency of electrons is positively charged.
Negative Plate Dielectric Material Positive Plate
Direct current cannot flow through the dielectric material because it is an insulator; however, the electric field created when the capacitor is charged is felt through the dielectric. Capacitors are rated for the amount of charge they can hold. The capacitance of a capacitor depends on the area of the plates, the distance between the plates, and type of dielectric material used. The unit of measurement for capacitance is farads (F). However, the farad is a large unit and capacitors are often rated in microfarads (mF) or picofarads (pF). Capacitor Circuit Symbols Capacitance is usually indicated symbolically on an electrical drawing by a combination of a straight line with a curved line, or two straight lines.
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Simple Capacitive Circuit
For the purpose of explanation, a DC circuit is used here to describe the operation of an capacitor. The resistance of wiring and stray values of capacitance found in any circuit are not considered in this explanation. In a resistive circuit, voltage change is instantaneous. In a circuit with a resistor and capacitor in series, the voltage across the capacitor does not change as quickly. In the following circuit, initially the switch is in position 2 and no voltage is measured by the voltmeter (V). When the switch is moved to position 1, voltage across the capacitor will rise rapidly at first, then more slowly as the maximum value is approached.
1 2
+ _
3
R1 C1
V
Capacitive Time Constant
The time required for voltage across the capacitor in this simple circuit to rise to its maximum value is determined by the product of capacitance, in farads, times resistance, in ohms. This product is the time constant of a capacitive circuit. The time constant gives the time in seconds required for voltage across the capacitor to reach 63.2% of its maximum value. When the switch in the previous circuit is moved to position 1, the voltage measured by the voltmeter will begin to rise. During the first time constant, voltage will rise to 63.2% of its maximum value. During the second time constant, voltage will rise to 63.2% of the remaining 36.8%, or a total of 86.4%. It takes about five time constants for voltage across the capacitor to reach its maximum value.
100.0% 98.1% 94.9% 86.4%
63.2%
First Time Constant
Second Time Constant
Third Time Constant
Fourth Time Constant
Fifth Time Constant
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The voltage across the capacitor will reach its maximum value when it is equal to the supply voltage. At that point, current flow will reach zero. When the switch in the previous circuit is returned to position 2, the capacitor will retain its charge because there is no path for current flow. When the switch is moved to position 3, the capacitor will begin to discharge, and it will take about five time constants for the voltage across the capacitor and the current through the resistor to reach zero.
100.0% First Time Constant Second Time Constant Third Time Constant Fourth Time Constant Fifth Time Constant
36.8% 13.6% 5.1% 1.9% 0%
Calculating the Time Constant of a Capacitive Circuit
To determine the time constant of a capacitive circuit, use one of the following formulas:
τ (in seconds) = R (megohms) x C (microfarads) τ (in microseconds) = R (megohms) x C (picofarads) τ (in microseconds) = R (ohms) x C (microfarads)
In the following illustration, C1 is equal to 2 mF , and R1 is equal to 10 W. When the switch is closed, it will take 20 microseconds for voltage across the capacitor to rise from zero to 63.2% of its maximum value. It will take about five time constants, 100 microseconds, for this voltage to rise to its maximum value.
+ _
C1 2µF
R1 10 Ω
V
τ = RC τ = 2µF x 10 Ω τ = 20 microseconds
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Formula for Series Capacitors
Connecting capacitors in series decreases total capacitance. The formula for series capacitors is similar to the formula for parallel resistors. In the following circuit, an AC generator supplies electrical power to three capacitors. Total capacitance is calculated using the following formula:
1 = 1 + 1 + 1 ... + 1 Cn Ct C1 C2 C3 5 µF 10 µF 20 µF
C1
C2
C3
Formula for Parallel Capacitors
Adding capacitors in parallel increases circuit capacitance. In the following circuit, an AC generator is used to supply electrical power to three capacitors. Total capacitance is calculated using the following formula:
Ct = C1 + C2 + C3 ... + Cn
1 = 1 + 1 + 1 Ct 5 10 20 1 = 7 Ct 20 20 Ct = 7 Ct = 2.86 µF
C1 5 µF
C2 10 µF
C3 20 µF
Ct = 5 µF + 10 µF + 20 µF Ct = 35 µF
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Review 8 1. The total inductance for this circuit is __________ mh.
4 mh L1 2 mh L2 3 mh L3 1 mh L4
2. The total inductance for this circuit is ___________ mh.
L1 5 mh
L2 10 mh
L3 10 mh
3. The total capacitance for this circuit is __________ mF .
5 µF 10 µF 10 µF
C1
C2
C3
4. The total capacitance for this circuit is __________ mF .
C1 5 µF
C2 10 µF
C3 10 µF
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Inductive and Capacitive Reactance
In a purely resistive AC circuit, resistance is the only opposition to current flow. In an AC circuit with only inductance, capacitance, or both inductance and capacitance, but no resistance, opposition to current flow is called reactance, designated by the symbol “X” . Total opposition to current flow in an AC circuit that contains both reactance and resistance is called impedance, designated by the symbol “Z” . Just like resistance, reactance and impedance are expressed in ohms. Inductive Reactance Inductance only affects current flow when the current is changing. Inductance produces a self-induced voltage (counter emf) that opposes changes in current. In an AC circuit, current is changing constantly. Inductance in an AC circuit, therefore, causes a continual opposition. This opposition to current flow is called inductive reactance and is designated by the symbol “XL. ” Inductive reactance is proportional to both the inductance and the frequency applied. The formula for inductive reactance is:
XL = 2πfL XL = 2 x 3.14 x frequency x inductance
For a 60 hertz circuit containing a 10 mh inductor, the inductive reactance is:
XL = 2πfL XL = 2 x 3.14 x 60 x 0.010 XL = 3.768 Ω
For this example, the resistance is zero so the impedance is equal to the reactance. If the voltage is known, Ohm’s Law can be used to calculate the current. If, for example, the voltage is 10 volts, the current is calculated as follows;
I= I= E Z 10 3.768
I = 2.65 Amps
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Phase Relationship between Current and Voltage in an Inductive Circuit
In a purely resistive circuit, current and voltage rise and fall at the same time. They are said to be “in phase. ” In this circuit, there is no inductance. Resistance and impedance are the same.
+ Voltage Current
0
_
In a purely inductive circuit, current lags behind voltage by 90 degrees. Current and voltage are said to be “out of phase” . In this circuit, impedance and inductive reactance are the same.
90 Degrees + Voltage Current
0
_
In a circuit with both resistance and inductive reactance, AC current will lag the voltage by more than 0 degrees and less than 90 degrees. The exact amount of lag will depend upon the relative amounts of resistance and inductive reactance. The more resistive a circuit is, the closer it is to being in phase. The more reactive a circuit is, the more out of phase it is. In the following illustration, resistance and inductive reactance are equal. Current lags voltage by 45 degrees.
45 Degrees + Voltage Current XL = 10 Ω
0 R = 10 Ω _
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Calculating Impedance in an Inductive Circuit
When working with a circuit containing elements of inductance, capacitance, and resistance, impedance must be calculated. Because electrical concepts deal with trigonometric functions, this is not a simple matter of subtraction and addition. The following formula is used to calculate impedance in a circuit with resistance and inductive reactance:
Z = R2 + XL2
In the previous example, resistance and inductive reactance are each 10 ohms. Impedance for this circuit can be calculated as follows:
Z = 102 + 102 Z= 200
Z = 14.1421 Ω
Vectors
A common way to represent AC circuit values is with a vector. A vector is a graphic representation of a quantity that has direction and magnitude. A vector on a map might indicate that one city is 50 miles southwest from another. The magnitude is 50 miles and the direction is southwest. Vectors are also used to show electrical relationships. The following vector illustrates the relationship between resistance and inductive reactance for a circuit containing 10 ohms of each. The angle between the vectors is the phase angle represented by the symbol q. When inductive reactance is equal to resistance the resultant angle is 45 degrees. This angle represents how much current lags voltage for this circuit.
14 .14 21 Z = Ω
XL = 10 Ω
θ
R = 10 Ω
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Capacitive Reactance
Capacitance also opposes AC current flow. Capacitive reactance is designated by the symbol XC. The larger the capacitor, the smaller the capacitive reactance. Current flow in a capacitive AC circuit is also dependent on frequency. The following formula is used to calculate capacitive reactance:
1 XC = 2πfC
The capacitive reactance for a 60 hertz circuit with a 10 microfarad capacitor is calculated as follows:
1 XC = 2πfC 1 XC = 2 x 3.14 x 60 x 0.000010 XC = 265.39 Ω
For this example, the resistance is zero so the impedance is equal to the reactance. If the voltage is known, Ohm’s Law can be used to calculate the current. If, for example, the voltage is 10 volts, the current is calculated as follows:
I= I= E Z 10 265.39
I = 0.0376 Amps
Phase Relationship between The phase relationship between current and voltage are Current and Voltage opposite to the phase relationship of an inductive circuit. In a purely capacitive circuit, current leads voltage by 90 degrees.
90 Degrees + Voltage Current 0
_
64
In a circuit with both resistance and capacitive reactance, AC current will lead the voltage by more than 0 degrees and less than 90 degrees. The exact amount of lead will depend upon the relative amounts of resistance and capacitive reactance. The more resistive a circuit is, the closer it is to being in phase. The more reactive a circuit is, the more out of phase it is. In the following illustration, resistance and capacitive reactance are equal. Current leads voltage by 45 degrees.
45 Degrees + Voltage Current 0 R = 10 Ω _ XC = 10 Ω
Calculating Impedance in a Capacitive Circuit
The following formula is used to calculate impedance in a circuit with resistance and capacitive reactance:
Z = R2 + XC2
In the previous example, resistance and capacitive reactance are each 10 ohms. Impedance for this circuit can be calculated as follows:
Z = 102 + 102 Z= 200
Z = 14.1421 Ω
The following vector illustrates the relationship between resistance and capacitive reactance for a circuit containing 10 ohms of each. The angle between the vectors is the phase angle represented by the symbol q. When capacitive reactance is equal to resistance, the resultant angle is 45 degrees. This angle represents how much current leads voltage for this circuit.
R = 10 Ω
θ
Z = 21 .14 14 Ω
XC = 10 Ω
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Review 9 1. Opposition to current flow in an AC circuit caused by inductors and capacitors is called ____________ .
2. Total opposition to current flow in an AC circuit with both resistance and reactance is called ____________ . 3. In a 50 hertz circuit, containing a 10 mh inductor, the inductive reactance is ____________ ohms. 4. In a purely inductive circuit, ____________ a. current and voltage are in phase b. current leads voltage by 90 degrees c. current lags voltage by 90 degrees
5. In a purely capacitive circuit, ____________ a. current and voltage are in phase b. current leads voltage by 90 degrees c. current lags voltage by 90 degrees
6. In a 50 hertz circuit, containing a 10 microfarad capacitor, the capacitive reactance is ____________ ohms. 7 . In a circuit with 5 W resistance, and 10 W inductive reactance, impedance is ____________ ohms.
8. In a circuit with 5 W resistance, and 4 W capacitive reactance, impedance is ____________ ohms.
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Series R-L-C Circuit
Circuits often contain elements of resistance, inductance, and capacitance. In an inductive AC circuit, current lags voltage by 90 degrees. In a capacitive AC circuit, current leads voltage by 90 degrees. When represented in vector form, inductive and capacitive reactance are 180 degrees apart. As a result, the net reactance is determined by taking the difference between the two quantities.
XL R
XC
An AC circuit is: • • • Calculating Total Impedance in a Series R-L-C Circuit Resistive if XL and XC are equal Inductive if XL is greater than XC Capacitive if XC is greater than XL
The following formula is used to calculate total impedance of a circuit containing resistance, capacitance, and inductance:
Z = R2 + (XL - XC)2
In the case where inductive reactance is greater than capacitive reactance, subtracting XC from XL results in a positive number. The positive phase angle is an indicator that the net circuit reactance is inductive, and current lags voltage. In the case where capacitive reactance is greater than inductive reactance, subtracting XC from XL results in a negative number. The negative phase angle is an indicator that the net circuit reactance is capacitive and current leads voltage. In either case, the value squared will result in a positive number.
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Calculating Reactance and Impedance in a Series R-L-C Circuit
In the following 120 volt, 60 hertz circuit, resistance is 1000 W, inductance is 5 mh, and capacitance is 2 mF . The following example shows the method for calculating impedance for this circuit.
R = 1000 Ω
L = 5 mh
C = 2 µF
XL = 2πfL XL = 6.28 x 60 x 0.005 XL = 1.884 Ω 1 XC = 2πfC 1 XC = 6.28 x 60 x 0.000002 XC = 1,327 Ω
Z = R2 + (XL - XC)2 Z = 10002 + (1.884 - 1,327)2 Z = 1,000,000 + ( - 1,325.116)2 Z = 1,000,000 + 1,755,932.41 Z = 2,755,932.41 Z = 1,660.1 Ω
Given that the applied voltage is 120 volts, current can be calculated as follows:
I= E Z I= 120 1,660.1
I = 0.0723 Amps
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Parallel R-L-C Circuit
Calculating Impedance in a Parallel R-L-C Circuit
Total impedance (Zt) can be calculated in a parallel R-L-C circuit if values of resistance and reactance are known. One method of calculating impedance involves first calculating total current, then using the following formula:
E Zt = t It
Total current is the vector sum of current flowing through the resistance plus, the difference between inductive current and capacitive current. This is expressed in the following formula:
It = IR2 + (IC - IL)2
In the following 120 volt, 60 hertz circuit, capacitive reactance is 25 W, inductive reactance is 50 W, and resistance is 1000 W. A simple application of Ohm’s Law will find the branch currents. Remember, voltage is constant throughout a parallel circuit.
120 V
R = 1000 Ω
XL = 50 Ω
XC = 25 Ω
IR = IR =
E R 120 1000
E IL = XL 120 IL = 50 IL = 2.4 Amps
IC = IC =
E XC 120 25
IR = 0.12 Amps
IC = 4.8 Amps
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Once the branch currents are known, total current can be calculated.
It = It = It = It = IR2 + (IC - IL)2 0.122 + (4.8 - 2.4)2 0.0144 + 5.76 5.7744
It = 2.403 Amps
Impedance can then be calculated as follows:
E Zt = t It Zt = 120 2.403
Zt = 49.94 Ω
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Power and Power Factor in an AC Circuit
Power consumed by a resistor is dissipated in heat and not returned to the source. This is called true power because it is the rate at which energy is used. Current in an AC circuit rises to peak values and diminishes to zero many times a second. The energy stored in the magnetic field of an inductor, or plates of a capacitor, is returned to the source when current changes direction. Although reactive components do not consume energy, they do increase the amount of energy that must be generated to do the same amount of work. The rate at which this non-working energy must be generated is called reactive power. If voltage and current are 90 degrees out of phase, as would be the case in a purely capacitive or purely inductive circuit, the average value of true power is equal to zero. There are high positive and negative peak values of power, but when added together the result is zero. Power in an AC circuit is the vector sum of true power and reactive power. This is called apparent power. True power is equal to apparent power in a purely resistive circuit because voltage and current are in phase. Voltage and current are also in phase in a circuit containing equal values of inductive reactance and capacitive reactance. In most circuits, however, apparent power is composed of both true power and reactive power. True Power and Apparent Power Formulas The formula for apparent power is:
P = EI
Apparent power is measured in volt-amps (VA). True power is calculated from another trigonometric function, the cosine of the phase angle (cos θ). The formula for true power is:
P = EI cos θ
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In a purely resistive circuit, current and voltage are in phase. There is a zero degree angle displacement between current and voltage. The cosine of zero is one. Multiplying a value by one does not change the value. Therefore, in a purely resistive circuit, the cosine of the angle is ignored. In a purely reactive circuit, either inductive or capacitive, current and voltage are 90 degrees out of phase. The cosine of 90 degrees is zero. Multiplying a value times zero results in a zero product. Therefore, no power is consumed in a purely reactive circuit. Calculating Apparent Power In the following 120 volt circuit, current is equal to 84.9 mA. in a simple R-L-C Circuit Inductive reactance is 100 W and capacitive reactance is 1100 W. The phase angle is -45 degrees. By referring to a trigonometric table, the cosine of -45 degrees is found to be 0.7071.
R = 1000 Ω XL = 100 Ω XC = 1100 Ω 120 V
The apparent power consumed by the circuit is:
P = EI P = 120 x 0.0849 P = 10.2 VA
The true power consumed by the circuit is:
P = EI cos θ P = 120 x 0.0849 x 0.7071 P=7 .2 Watts
Another formula for true power is:
P = I2R P = 0.08492 x 1000 P=7 .2 Watts
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Power Factor
Power factor is the ratio of true power to apparent power in an AC circuit. Power factor is expressed in the following formula:
PF = True Power Apparent Power
Power factor can also be expressed using the formulas for true power and apparent power. The value of EI cancels out because it is the same in the numerator and denominator. Power factor is the cosine of the angle.
PF = EI cos θ EI
PF = cos θ
In a purely resistive circuit, where current and voltage are in phase, there is no angle of displacement between current and voltage. The cosine of a zero degree angle is one. The power factor is one. This means that all energy delivered by the source is consumed by the circuit and dissipated in the form of heat. In a purely reactive circuit, voltage and current are 90 degrees apart. The cosine of a 90 degree angle is zero. The power factor is zero. This means the circuit returns all energy it receives from the source to the source. In a circuit where reactance and resistance are equal, voltage and current are displaced by 45 degrees. The cosine of a 45 degree angle is .7071. The power factor is .7071. This means the circuit uses approximately 70% of the energy supplied by the source and returns approximately 30%.
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Review 10 1. An AC circuit is ____________ if inductive reactance and capacitive reactance are equal.
2. A series AC circuit is ____________ if there is more inductive reactance than capacitive reactance. 3. A series AC circuit is ____________ if there is more capacitive reactance than inductive reactance. 4. In a 120 VAC, 60 hertz series circuit, with 1000 W of resistance, 10 mh of inductance and 4 mF of capacitance, impedance is ____________ W and current is ____________ amps. 5. For the following circuit, calculate impedance and current.
R = 1000 Ω XL = 200 Ω XC = 1200 Ω 120 V 60 Hz
Impedance is __________ W , and It is __________ amps.
6. For a circuit with a 120 volt AC source and a current of 10 amps, the apparent power is __________ VA. 7 . For a circuit with an apparent power of 3000 VA and a power factor of 0.8, the true power is __________ watts.
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Transformers
Mutual Induction
Transformers are electromagnetic devices that transfer electrical energy from one circuit to another by mutual induction. A single-phase transformer has two coils, a primary and a secondary. Mutual induction is the transfer of electrical energy from the primary to the secondary through magnetic fields . The following circuit illustrates mutual induction. The AC generator provides electrical power to the primary coil. The magnetic field produced by the primary induces a voltage into the secondary coil, which supplies power to a load.
Primary Coil
Secondary Coil
Transformers are used to step a voltage up to a higher level, or down to a lower level. To understand the need to stepping up or down voltages, consider how electrical power is generated and distributed. Generators used by power companies typically generate voltages of 30 KV or less. While this is a relatively high voltage compared to the voltages used by power customers, it is more efficient for utilities to transmit this power at still higher voltages, up to as high at 765 KV. The electrical power is received at substation transformers many miles away where it is stepped down and distributed locally. When it arrives at the customer’s location, it is further stepped down to the level needed for the type of customer. Even within a customer’s facility, voltage may need to be stepped down further to meet requirements of some equipment.
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This process of stepping up or down the voltage throughout a power distribution system is most often accomplished using transformers. The size and ratings of the transformers vary, but the basic operation of these devices is the same. Coefficient of Coupling Mutual inductance between two coils depends on their flux linkage. Maximum coupling occurs when all the lines of flux from the primary coil cut through the secondary winding. The amount of coupling which takes place is referred to as coefficient of coupling. To maximize coefficient of coupling, both coils are often wound on an iron core which is used to provide a path for the lines of flux. The following discussion of step-up and step-down transformers applies to transformers with an iron core.
Lines of Flux Confined to Iron Core
Lines of Flux that don’t Couple
Voltage, Current, and the Number of Turns in a Coil
There is a direct relationship between voltage, impedance, current, and the number of coil turns in a transformer. This relationship can be used to find either primary or secondary voltage, current, and the number of turns in each coil. It is the number of turns which determine if a transformer is a step up or step down transformer. The following “rules-of-thumb” apply to transformers: • • If the primary coil has fewer turns than the secondary coil, the transformer is a step-up transformer. If the primary coil has more turns than the secondary coil, the transformer is a step-down transformer.
When the number of turns on the primary and secondary coils of a transformer are equal, input voltage, impedance, and current are equal to output voltage, impedance, and current.
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Step-Up Transformer
A step-up transformer is used when it is desirable to step voltage up in value. The following circuit illustrates a stepup transformer. The primary coil has fewer turns than the secondary coil. When the primary has fewer turns than the secondary, voltage and impedance are stepped up. In the circuit illustrated, The transformer secondary has twice as many turns as the primary and voltage is stepped up from 120 VAC to 240 VAC. Because impedance is also stepped up, current is stepped down from 10 amps to 5 amps.
1:2
Primary Coil 900 Turns
Secondary Coil 1800 Turns
120 VAC Supply 10 Amps
240 VAC 5 Amps
Step-Down Transformer
A step-down transformer is used when it is desirable to step voltage down in value. The following circuit illustrates a step-down transformer. The primary coil has more turns than the secondary coil. The step-down ratio is 2:1. voltage and impedance are stepped down, current is stepped up.
2:1
Primary Coil 1800 Turns
Seconday Coil 900 Turns
240 VAC Supply 5 Amps
120 VAC Out 10 Amps
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Single-Phase Transformer 120 or 240 VAC single-phase transformers are used to supply lighting, receptacle, and small appliance loads. A transformer with a 240 VAC secondary can be used to supply 240 VAC to larger appliances such as stoves, air conditioners and heaters. A 240 VAC secondary can be tapped in the center to provide two sources of 120 VAC power.
Primary Primary
120 VAC
120 VAC
120 VAC Secondary Ground
240 VAC Secondary
Formulas for Calculating the Number of Primary and Secondary Turns of a Transformer
There are a number of useful formulas for calculating, voltage, current, and the number of turns between the primary and secondary of a transformer. These formulas can be used with either step-up or step-down transformers. The following legend applies to the transformer formulas: ES EP IS IP NS NP = = = = = = secondary voltage primary voltage secondary current primary current turns in the secondary coil turns in the primary coil
To find voltage:
ES = EP x IP IS EP = ES x IS IP
To find current:
IS = EP x IP ES IP = ES x IS EP
To find the number of coil turns:
NS = ES x NP EP NP = EP x NS ES
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Using the values for the step-down transformer in the example of the previous page, the secondary voltage can be verified.
ES = ES = ES = EP x IP IS 240 Volts x 5 Amps 10 Amps 1200 10
ES = 120 Volts
Transformer Ratings
Transformers are rated for the amount of apparent power they can provide. Because values of apparent power are often large, the transformer rating is frequently given in kVA (kilovoltamps). The kVA rating determines the current and voltage a transformer can deliver to its load without overheating. Given kVA and volts, amps can be calculated.
kVA = Volts x Amps 1000 kVA x 1000 Volts
Amps =
Using the illustrated step-down transformer, the kVA rating can be calculated. The kVA rating of a transformer is the same for both the primary and the secondary.
Primary kVA = 240 x 5 1000
Primary kVA = 1.2 kVA Secondary kVA = 120 x 10 1000
Secondary kVA = 1.2 kVA
Transformer Losses
Most of the electrical energy provided to the primary of a transformer is transferred to the secondary. Some energy, however, is lost in heat in the wiring or the core. Some losses in the core can be reduced by building the core of a number of flat sections called laminations.
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Three-Phase Transformers
Delta Connections
Three-phase transformers are used when three-phase power is required for larger loads such as industrial motors. There are two basic three-phase transformer connections, delta and wye. A delta is like three single-phase transformers connected together. The secondary of a delta transformer is illustrated below. For simplicity, the primary is not shown in this example. The voltages shown on the illustration are examples. Just as with a single-phase transformer, the secondary voltage depends on both the primary voltage and the turns ratio. Delta transformers are schematically drawn in a triangle. The voltages across each winding of the delta triangle represents one phase of a three phase system. The voltage is always the same between any two wires. A single phase (such as L1 to L2) can be used to supply single phase loads. All three phases are used to supply three phase loads.
L1
480 Volts
480 Volts
480 Volts
L2 480 Volts 480 Volts
L3
L1 to L2 = 480 volts L2 to L3 = 480 volts L1 to L3 = 480 volts
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Balanced Delta Current
When current is the same in all three coils, it is said to be balanced. In each phase, current has two paths to follow. For example, current flowing from L1 to the connection point at the top of the delta can flow down through one coil to L2, and down through another coil to L3. When current is balanced, the current in each line is equal to the square root of 3 times the current in each coil. The following example shows the line current calculation with the current in each coil equal to 29 amps.
IL = 3
x 29 = 1.732 x 29 = 50.2 Amps
50.2 Amps L1
29 Amps
29 Amps
50.2 Amps 29 Amps
L2
50.2 Amps
L3
Unbalanced Delta Current
When current is not the same in all three coils, it is unbalanced. The following diagram depicts an example of an unbalanced system.
43.6 Amps Coil A 30 Amps Coil B 20 Amps L1
26.5 Amps Coil C 10 Amps
L2
36.1 Amps
L3
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Wye Connections
The wye connection is also known as a star connection. Three coils are connected to form a “Y” shape. The wye transformer secondary has four leads, three phase leads and one neutral lead. The voltage across any phase (line-to-neutral) will always be less than the line-to-line voltage. The line-to-line voltage is the square root of 3 times the line-to-neutral voltage. The following example shows a wye transformer secondary with a line-to-neutral voltage is 277 volts and a line-to-line voltage of 480 volts.
ELine-to-Line = 3
x 277 = 1.732 x 277 = 480 Volts
L1
480 Volts 277 Volts L2
277 Volts
480 Volts
N
480 Volts
277 Volts
L3
Review 11 1. If the primary of a transformer has more turns than the secondary, it is a ____________ transformer.
2. If the primary of a transformer has fewer turns than the secondary, it is a ____________ transformer. 3. The secondary voltage of an iron-core transformer with 240 volts on the primary, 40 amps on the primary, and 20 amps on the secondary is ____________ volts. 4. A transformer with a 480 volt, 10 amp primary, and a 240 volt, 20 amp secondary will be rated for ____________ kVA. 5. A wye connected, three-phase transformer secondary, with 240 volts line-to-line will have ____________ volts line-to-neutral.
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Review Answers
Review 1 1) electron (-), proton (+), neutron (neutral); 2) free electron; 3) many; 4) a, c, e, g; 5) many, few. Review 2 1) electrons; 2) negative; 3) positive; 4) repel, attract; 5) voltage; 6) b; 7) a. Review 3 1) I = R ; 2) amps, volts, ohms; 3) 0.5; 4) 45; 5) 2 amps; 6) 6, 6; 7) 20, 80. Review 4 1) 5; 2) 5.45; 3) 3.33; 4) 12; 5) 6; 6) 2.4, 1.6. Review 5 1) 12, 22; 2) 40, 13.33. Review 6 1) power; 2) E, R, E2; 3) 36; 4) iron, north-south; 5) north, south; 6) left, thumb, lines of flux. Review 7 1) sine wave; 2) 120; 3) one; 4) -129.9 volts; 5) 106.05 volts RMS. Review 8 1) 10; 2) 2.5; 3) 2.5; 4) 25. Review 9 1) reactance; 2) impedance; 3) 3.14; 4) c; 5) b; 6) 318.5; 7) 11.18; 8) 6.4. Review 10 1) resistive; 2) inductive; 3) capacitive; 4) 1198, 0.1; 5) 1414.2, 0.0849; 6) 1200; 7) 2400. Review 11 1) step-down; 2) step-up; 3) 480; 4) 4.8; 5) 138.56.
E
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Final Exam
The final exam is intended to be a learning tool. The book may be used during the exam. A tear-out answer sheet is provided. After completing the test, mail the answer sheet in for grading. A grade of 70% or better is passing. Upon successful completion of the test a certificate will be issued. Questions 1. 2. 3. ________ is a good electrical insulator. a. b. copper aluminum c. d. silver rubber
A material with more protons than electrons has a ________. a. b. negative charge positive charge c. d. neutral charge no charge
In a simple electric circuit with a 12 volt supply, and a 24 W resistor, current is ________.
a. 2 amps c. 0.2 amps b. 5 amps d. 0.5 amps 4. The total resistance in a series circuit containing three, 10 W, resistors is ________. 5. a. 10 W b. 30 W c. 3.33 W d. 100 W
In a 12 volt series circuit where R1=10 W, R2=20 W, and R3=10 W, current flow is ________. a. b. 0.3 amps 0.5 amps c. d. 0.25 amps 3.33 amps
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6. 7 . 8. 9. 10. 11. 12. 13.
In a circuit containing three 30 W resistors in parallel, the total resistance is ________. a. b. 30 W 90 W c. 10 W d. 0.1 W
The rate at which work is done is called ________. a. b. energy power c. d. efficiency power factor
Power in a simple DC 12 volt, 4 amp series circuit is ________. a. b. 3 amps 3 watts c. d. 48 amps 48 watts
The instantaneous voltage at 150 degrees of an AC sine wave whose peak voltage is 480 volts is ________. a. b. 415.7 volts 480 volts c. d. 240 volts 0 volts
The effective voltage of an AC sine wave whose peak voltage is 480 volts is ________. a. b. 415.7 volts 339.4 volts c. d. 480 volts 679 volts
The time constant of a series circuit with a 10 mh inductor, and a 5 W resistor is ________. a. b. 2 milliseconds 2 seconds c. d. 2 microseconds 0.5 seconds
The total inductance of a series circuit containing three inductors with values of 10mh, 20 mh, and 40 mh is ________. a. b. 5.7 pF 5.7 mh c. d. 70 W 70 mh
The time constant for a series circuit with a 20 W resistor and a 4 mF capacitor is ________. a. b. 80 microseconds 80 milliseconds c. d. 5 microseconds 5 milliseconds
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14. 15. 16. 17 . 18. 19.
Total capacitance for a series circuit containing a 2 mF , 4 mF , and 8 mF capacitors is ________. a. 14 mF b. 0.875 mF c. 1.14 mF d. 4 mF
Total opposition to current flow in an AC circuit that contains both reactance and resistance is called ________. a. b. resistance reactance c. d. impedance capacitance
In a 60 hertz circuit containing 5 millihenrys of inductance, inductive reactance is ________. a. 1.884 W b. 1884 W c. d. 0.0005 W 0.05 W
In a purely inductive circuit ________. a. b. c. d. current leads voltage by 90 degrees current lags voltage by 90 degrees current and voltage are in phase current leads voltage by 30 degrees
In a series AC circuit with a 20 W resistor and a 10 W inductive reactance, impedance is ________. a. 30 W b. 10 W c. 14.1 W d. 22.4 W
A series AC circuit containing more capacitive reactance than inductive reactance is ________.
a. inductive c. capacitive b. resistive d. in phase 20. An iron-core transformer with a primary voltage of 120 volts, primary current of 10 amps, and secondary current of 5 amps has a secondary voltage of _______. a. b. 60 volts 240 volts c. d. 480 volts 30 volts
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quickSTEP Online Courses
quickSTEP online courses are available at http://www.sea.siemens.com/step. The quickSTEP training site is divided into three sections: Courses, Downloads, and a Glossary. Online courses include reviews, a final exam, the ability to print a certificate of completion, and the opportunity to register in the Sales & Distributor training database to maintain a record of your accomplishments. From this site the complete text of all STEP courses can be downloaded in PDF format. These files contain the most recent changes and updates to the STEP courses. A unique feature of the quickSTEP site is our pictorial glossary. The pictorial glossary can be accessed from anywhere within a quickSTEP course. This enables the student to look up an unfamiliar word without leaving the current work area.
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Introduction ..............................................................................2 Electron Theory .........................................................................4 Conductors, Insulators and Semiconductors ............................5 Electric Charges ........................................................................7 Current......................................................................................9 Voltage .................................................................................... 11 Resistance .............................................................................. 13 Simple Electric Circuit ............................................................. 15 Ohm’s Law ............................................................................. 16 DC Series Circuit .................................................................... 18 DC Parallel Circuit ...................................................................23 Series-Parallel Circuits ............................................................30 Power......................................................................................34 Magnetism .............................................................................37 Electromagnetism ..................................................................39 Introduction to AC ...................................................................42 AC Generators ........................................................................44 Frequency ...............................................................................47 Voltage and Current ................................................................48 Inductance ..............................................................................51 Capacitance ............................................................................56 Inductive and Capacitive Reactance .......................................61 Series R-L-C Circuit .................................................................67 Parallel R-L-C Circuit ................................................................69 Power and Power Factor in an AC Circuit ................................ 71 Transformers ...........................................................................75 Three-Phase Transformers ......................................................80 Review Answers .....................................................................83 Final Exam ..............................................................................84 quickSTEP Online Courses .....................................................88 1
Introduction
Welcome to the first course in the STEP series, Siemens Technical Education Program designed to prepare our distributors to sell Siemens Energy & Automation products more effectively. This course covers Basics of Electricity and is designed to prepare you for subsequent courses on Siemens Energy & Automation products. Upon completion of Basics of Electricity you will be able to:
• • • • • • • • • • • •
Explain the difference between conductors and insulators Use Ohm’s Law to calculate current, voltage, and resistance Calculate equivalent resistance for series, parallel, or series-parallel circuits Calculate voltage drop across a resistor Calculate power given other basic values Identify factors that determine the strength and polarity of a current-carrying coil’s magnetic field Determine peak, instantaneous, and effective values of an AC sine wave Identify factors that effect inductive reactance and capacitive reactance in an AC circuit Calculate total impedance of an AC circuit Explain the difference between real power and apparent power in an AC circuit Calculate primary and secondary voltages of single-phase and three-phase transformers Calculate kVA of a transformer
The objectives listed above may sound strange to you. You may also wonder why you would need to know these things to sell Siemens Energy & Automation products. Developing a basic knowledge of electrical concepts, however, will help you to better understand customer applications. In addition, you will be better able to describe products to customers and determine important differences between products. If you are an employee of a Siemens Energy & Automation authorized distributor, fill out the final exam tear-out card and mail in the card. We will mail you a certificate of completion if you score a passing grade. Good luck with your efforts.
Electron Theory
Elements of an Atom
All matter is composed of molecules which are made up of a combination of atoms. Atoms have a nucleus with electrons orbiting around it. The nucleus is composed of protons and neutrons (not shown). Most atoms have an equal number of electrons and protons. Electrons have a negative charge (-). Protons have a positive charge (+). Neutrons are neutral. The negative charge of the electrons is balanced by the positive charge of the protons. Electrons are bound in their orbit by the attraction of the protons. These are referred to as bound electrons.
Electron Proton Nucleus
Free Electrons
Electrons in the outer band can become free of their orbit by the application of some external force such as movement through a magnetic field, friction, or chemical action. These are referred to as free electrons. A free electron leaves a void which can be filled by an electron forced out of orbit from another atom. As free electrons move from one atom to the next an electron flow is produced. This is the basis of electricity.
Conductors, Insulators and Semiconductors
Conductors
An electric current is produced when free electrons move from one atom to the next. Materials that permit many electrons to move freely are called conductors. Copper, silver, aluminum, zinc, brass, and iron are considered good conductors. Copper is the most common material used for conductors and is relatively inexpensive.
Insulators
Materials that allow few free electrons are called insulators. Materials such as plastic, rubber, glass, mica, and ceramic are good insulators.
An electric cable is one example of how conductors and insulators are used. Electrons flow along a copper conductor to provide energy to an electric device such as a radio, lamp, or a motor. An insulator around the outside of the copper conductor is provided to keep electrons in the conductor.
Rubber Insulator Copper Conductor
Semiconductors
Semiconductor materials, such as silicon, can be used to manufacture devices that have characteristics of both conductors and insulators. Many semiconductor devices will act like a conductor when an external force is applied in one direction. When the external force is applied in the opposite direction, the semiconductor device will act like an insulator. This principle is the basis for transitors, diodes, and other solidstate electronic devices.
Transistor
Diode
Review 1 1. List the three basic elements of an atom and state the charge of each (positive, negative, or neutral). Element Charge ____________ ____________ ____________ ____________ ____________ ____________
2. An electron forced out of orbit by an external force is called a ____________ ____________ . 3. Conductors allow ____________ free electrons to flow when an external electric force is applied. 4. Which of the following materials are good conductors? a. copper b. plastic c. silver d. rubber e. f. g. h. aluminum glass iron mica
5. Semiconductor devices can be manufactured to allow ____________ electrons to flow in one direction and ___ _________ electrons to flow in the opposite direction.
Electric Charges
Neutral State of an Atom
Elements are often identified by the number of electrons in orbit around the nucleus of the atoms making up the element and by the number of protons in the nucleus. A hydrogen atom, for example, has only one electron and one proton. An aluminum atom (illustrated) has 13 electrons and 13 protons. An atom with an equal number of electrons and protons is said to be electrically neutral.
Outer Band
Positive and Negative Charges
Electrons in the outer band of an atom are easily displaced by the application of some external force. Electrons which are forced out of their orbits can result in a lack of electrons where they leave and an excess of electrons where they come to rest. The lack of electrons is called a positive charge because there are more protons than electrons. The excess of electrons has a negative charge. A positive or negative charge is caused by an absence or excess of electrons. The number of protons remains constant.
Neutral Charge
Negative Charge
Positive Charge
Attraction and Repulsion of Electric Charges
The old saying, “opposites attract, ” is true when dealing with electric charges. Charged bodies have an invisible electric field around them. When two like-charged bodies are brought together, their electric fields repel one body from the other. When two unlike-charged bodies are brought together, their electric fields attract one body to the other. The electric field around a charged body forms invisible lines of force. These invisible lines of force cause the attraction or repulsion. Lines of force are shown leaving a body with a positive charge and entering a body with a negative charge.
Unlike Charges Attract Like Charges Repel
Coulomb’s Law
During the 18th century a French scientist, Charles A. Coulomb, studied fields of force that surround charged bodies. Coulomb discovered that charged bodies attract or repel each other with a force that is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them. Today we call this Coulomb’s Law of Charges. Simply put, the force of attraction or repulsion depends on the strength of the charges and the distance between them.
Current
Electricity is the flow of free electrons in a conductor from one atom to the next atom in the same general direction. This flow of electrons is referred to as current and is designated by the symbol “I” . Electrons move through a conductor at different rates and electric current has different values. Current is determined by the number of electrons that pass through a cross-section of a conductor in one second. We must remember that atoms are very small. It takes about 1,000,000,000,000,000,000,000,000 atoms to fill one cubic centimeter of a copper conductor. This number can be simplified using mathematical exponents. Instead of writing 24 zeros after the number 1, write 1024. Trying to measure even small values of current would result in unimaginably large numbers. For this reason current is measured in amperes which is abbreviated “amps” . The letter “A” is the symbol for amps. A current of one amp means that in one second about 6.24 x 1018 electrons move through a cross-section of conductor. These numbers are given for information only and you do not need to be concerned with them. It is important, however, to understand the concept of current flow.
Units of Measurement
The following chart reflects special prefixes that are used when dealing with very small or large values of current: Prefix Symbol Decimal 1 kiloampere 1 kA 1000 A 1 milliampere 1 mA 1/1000 A 1 microampere 1 mA 1/1,000,000 A
Direction of Current Flow
Some authorities distinguish between electron flow and current flow. Conventional current flow theory ignores the flow of electrons and states that current flows from positive to negative. To avoid confusion, this book will use the electron flow concept which states that electrons flow from negative to positive.
_
+
Electron Flow
_
+
Conventional Current Flow
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Voltage
Electricity can be compared with water flowing through a pipe. A force is required to get water to flow through a pipe. This force comes from either a water pump or gravity. Voltage is the force that is applied to a conductor that causes electric current to flow.
Water Flow Through a Pipe
Current Flow Through a Conductor
Electrons are negative and are attracted by positive charges. They will always be attracted from a source having an excess of electrons, thus having a negative charge, to a source having a deficiency of electrons, giving it a positive charge. The force required to make electricity flow through a conductor is called a difference in potential, electromotive force (emf), or voltage. Voltage is designated by the letter “E” , or the letter “V” . The unit of measurement for voltage is volts which is also designated by the letter “V” .
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Voltage Sources
An electrical voltage can be generated in various ways. A battery uses an electrochemical process. A car’s alternator and a power plant generator utilize a magnetic induction process. All voltage sources share the characteristic of an excess of electrons at one terminal and a shortage at the other terminal. This results in a difference of potential between the two terminals.
Shortage of Electrons Excess of Electrons
+
_
Batter y
Voltage Circuit Symbol
The terminals of a battery are indicated symbolically on an electrical drawing by two lines. The longer line indicates the positive terminal. The shorter line indicates the negative terminal.
+ _
Units of Measurement
The following chart reflects special prefixes that are used when dealing with very small or large values of voltage: Prefix Symbol Decimal 1 kilovolt 1 kV 1000 V 1 millivolt 1 mV 1/1000 V 1 microvolt 1 mV 1/1,000,000 V
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Resistance
A third factor that plays a role in an electrical circuit is resistance. All material impedes the flow of electrical current to some extent. The amount of resistance depends upon composition, length, cross-section and temperature of the resistive material. As a rule of thumb, resistance of a conductor increases with an increase of length or a decrease of crosssection. Resistance is designated by the symbol “R” . The unit of measurement for resistance is ohms (W). Resistance Circuit Symbols Resistance is usually indicated symbolically on an electrical drawing by one of two ways. An unfilled rectangle is commonly used. A zigzag line may also be used.
Resistance can be in the form of various components. A resistor may be placed in the circuit, or the circuit might contain other devices that have resistance. Units of Measurement The following chart reflects special prefixes that are commonly used when dealing with values of resistance: Prefix Symbol Decimal 1 kilohm 1 kW 1000 W 1 megohm 1 MW 1,000,000 W
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Review 2 1. Elements are identified by the number of ____________ in orbit around the nucleus.
2. A material that has an excess of electrons is said to have a ____________ charge. 3. A material that has a deficiency of electrons is said to have a ____________ charge. 4. Like charges ____________ and unlike charges ____________ . 5. The force that is applied to a conductor to cause current flow is ____________ . 6. Electrons move from ____________ . 7 . a. positive to negative b. negative to positive With an increase of length or a decrease of crosssection of a conductor, resistance will ____________ . a. increase b. decrease
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Simple Electric Circuit
An Electric Circuit
A fundamental relationship exists between current, voltage, and resistance. A simple electric circuit consists of a voltage source, some type of load, and a conductor to allow electrons to flow between the voltage source and the load. In the following circuit a battery provides the voltage source, electrical wire is used for the conductor, and a light provides the resistance. An additional component has been added to this circuit, a switch. There must be a complete path for current to flow. If the switch is open, the path is incomplete and the light will not illuminate. Closing the switch completes the path, allowing electrons to leave the negative terminal and flow through the light to the positive terminal.
+
Switch
_
+
_
An Electrical Circuit Schematic The following schematic is a representation of an electrical circuit, consisting of a battery, a resistor, a voltmeter and an ammeter. The ammeter, connected in series with the circuit, will show how much current flows in the circuit. The voltmeter, connected across the voltage source, will show the value of voltage supplied from the battery. Before an analysis can be made of a circuit, we need to understand Ohm’s Law.
+ + V _ R _ A
+ _
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Ohm’s Law
George Simon Ohm and Ohm’s Law
The relationship between current, voltage and resistance was studied by the 19th century German mathematician, George Simon Ohm. Ohm formulated a law which states that current varies directly with voltage and inversely with resistance. From this law the following formula is derived:
I= E R or Current = Voltage Resistance
Ohm’s Law is the basic formula used in all electrical circuits. Electrical designers must decide how much voltage is needed for a given load, such as computers, clocks, lamps and motors. Decisions must be made concerning the relationship of current, voltage and resistance. All electrical design and analysis begins with Ohm’s Law. There are three mathematical ways to express Ohm’s Law. Which of the formulas is used depends on what facts are known before starting and what facts need to be known.
I= E R E=IxR R= E I
Ohm’s Law Triangle
There is an easy way to remember which formula to use. By arranging current, voltage and resistance in a triangle, one can quickly determine the correct formula.
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Using the Triangle
To use the triangle, cover the value you want to calculate. The remaining letters make up the formula.
I=
E R
E=IxR
R=
E I
Ohm’s Law can only give the correct answer when the correct values are used. Remember the following three rules: • • • Examples of Solving Ohm’s Law Current is always expressed in amperes or amps Voltage is always expressed in volts Resistance is always expressed in ohms
Using the simple circuit below, assume that the voltage supplied by the battery is 10 volts, and the resistance is 5 W.
+ + V _ R _ A
+ _
To find how much current is flowing through the circuit, cover the “I” in the triangle and use the resulting equation.
I= E R I= 10 Volts 5Ω I = 2 Amps
Using the same circuit, assume the ammeter reads 200 mA and the resistance is known to be 10 W. To solve for voltage, cover the “E” in the triangle and use the resulting equation.
E=IxR E = 0.2 x 10 E = 2 Volts
Remember to use the correct decimal equivalent when dealing with numbers that are preceded with milli (m), micro (m) or kilo (k). In this example, had 200 been used instead of converting the value to 0.2, the wrong answer of 2000 volts would have been calculated.
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DC Series Circuit
Resistance in a Series Circuit
A series circuit is formed when any number of resistors are connected end-to-end so that there is only one path for current to flow. The resistors can be actual resistors or other devices that have resistance. The following illustration shows four resistors connected end-to-end. There is one path of current flow from the negative terminal of the battery through R4, R3, R2, R1 returning to the positive terminal.
R1 + _ R2 R3 R4
Formula for Series Resistance
The values of resistance add in a series circuit. If a 4 W resistor is placed in series with a 6 W resistor, the total value will be 10 W. This is true when other types of resistive devices are placed in series. The mathematical formula for resistance in series is:
Rt = R1 + R2 + R3 + R4 + R5 ... + Rn
In this example, the circuit includes five series resistors.
11 KΩ R1 2 KΩ R2 2 KΩ R3 100 Ω R4 1 KΩ R5
+ _
Rt = R1 + R2 + R3 + R4 + R5 Rt = 11,000 + 2,000 + 2,000 + 100 + 1,000 Rt = 16,100 Ω
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Current in a Series Circuit
The equation for total resistance in a series circuit allows us to simplify a circuit. Using Ohm’s Law, the value of current can be calculated. Current is the same anywhere it is measured in a series circuit.
E R 12 I= 10 I= I = 1.2 Amps 5Ω R1
+ _
1Ω R2
2Ω R3
2Ω R4
+ _
10 Ω Rt
12 Volts
12 Volts
Original Circuit
Equivalent Circuit
Voltage in a Series Circuit
Voltage can be measured across each of the resistors in a circuit. The voltage across a resistor is referred to as a voltage drop. A German physicist, Gustav Kirchhoff, formulated a law which states the sum of the voltage drops across the resistances of a closed circuit equals the total voltage applied to the circuit. In the following illustration, four equal value resistors of 1.5 W each have been placed in series with a 12 volt battery. Ohm’s Law can be applied to show that each resistor will “drop” an equal amount of voltage.
12 V
3V 1.5 Ω R1
3V 1.5 Ω R2
+ _
3V 1.5 Ω R3
3V 1.5 Ω R4
12 Volt Battery
19
First, solve for total resistance:
Rt = R1 + R2 + R3 + R4 Rt = 1.5 + 1.5 + 1.5 + 1.5 Rt = 6 Ω
Second, solve for current:
I= E R I = 12 6 I = 2 Amps
Third, solve for voltage across any resistor:
E=IxR E = 2 x 1.5 E = 3 Volts
If voltage were measured across any single resistor, the meter would read three volts. If voltage were read across a combination of R3 and R4 the meter would read six volts. If voltage were read across a combination of R2, R3, and R4 the meter would read nine volts. If the voltage drops of all four resistors were added together the sum would be 12 volts, the original supply voltage of the battery. Voltage Division in a Series Circuit It is often desirable to use a voltage potential that is lower than the supply voltage. To do this, a voltage divider, similar to the one illustrated, can be used. The battery represents Ein which in this case is 50 volts. The desired voltage is represented by Eout, which mathematically works out to be 40 volts. To calculate this voltage, first solve for total resistance.
Rt = R1 + R2 Rt = 5 + 20 Rt = 25 Ω
20
Second, solve for current:
Ein Rt 50 I= 25 I= I = 2 Amps
Finally, solve for voltage:
Eout = I x R2 Eout = 2 x 20 Eout = 40 Volts
+ _
R1
Ein
5Ω
R2
20 Ω
Eout 40 Volts
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Review 3 1. The basic Ohm’s Law formula is ____________ . 2. When solving circuit problems; current must always be expressed in ____________ , voltage must always be expressed in ____________ , and resistance must always be expressed in ____________ . 3. The total current of a simple circuit with a voltage supply of 12 volts and a resistance of 24 W is ____________ amps. 4. What is the total resistance of a series circuit with the following values: R1=10 W, R2=15 W, and R3=20 W? ____________ W. 5. What is total current of a series circuit that has a 120 volt supply and 60 W resistance? 6. In the following circuit, the voltage dropped across R1 is ____________ volts and R2 is ____________ volts.
1.5 Ω
R1
1.5 Ω
R2
+ _
12 Volts
7 . In the following circuit, voltage dropped across R1 is ____________ volts and across R2 is ____________ volts.
5Ω
R1
20 Ω
R2
+ _
100 Volts
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DC Parallel Circuit
Resistance in a Parallel Circuit
A parallel circuit is formed when two or more resistances are placed in a circuit side-by-side so that current can flow through more than one path. The illustration shows two resistors placed side-by-side. There are two paths of current flow. One path is from the negative terminal of the battery through R1 returning to the positive terminal. The second path is from the negative terminal of the battery through R2 returning to the positive terminal of the battery.
+ _
R1
R2
Formula for Equal Value Resistors in a Parallel Circuit
To determine the total resistance when resistors are of equal value in a parallel circuit, use the following formula:
Rt = Value of any one Resistor Number of Resistors
In the following illustration there are three 15 W resistors. The total resistance is:
Rt = Value of any one Resistor Number of Resistor
15 Rt = 3 Rt = 5 Ω
+ _
R1 15 Ω
R2 15 Ω
R3 15 Ω
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Formula for Unequal There are two formulas to determine total resistance for Resistors in a Parallel Circuit resistors of any value in a parallel circuit. The first formula is used when there are any number of resistors.
1 = 1 + 1 + 1 ... + 1 Rt R1 R2 R3 Rn
In the following illustration, there are three resistors, each of different value. Solve for the total resistance as follows:
1 = 1 + 1 + 1 Rt R1 R2 R3 1 = Rt 1 + 1 + 1 5 10 20 Insert Values for the Resistors Find the Lowest Common Denominator Add the Numerators Invert Both Sides of the Equation Divide
1 = 4 + 2 + 1 Rt 20 20 20 1 = 7 Rt 20 Rt = 20 1 7 Rt = 2.86 Ω
+ _
R1
R2
R3
5Ω
10 Ω
20 Ω
24
The second formula is used when there are only two resistors.
R1 x R2 Rt = R1 + R2
In the following illustration there are two resistors, each of different value. The total resistance is:
R1 x R2 Rt = R1 + R2 Rt = 5 x 10 5 + 10
50 Rt = 15 Rt =
3.33 Ω
+ _
R1 5Ω
R2 10 Ω
Voltage in a Parallel Circuit
When resistors are placed in parallel across a voltage source, the voltage is the same across each resistor. In the following illustration three resistors are placed in parallel across a 12 volt battery. Each resistor has 12 volts available to it.
12 Volt Battery + _ R1 12 V R2 12 V R3 12 V
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Current in a Parallel Circuit
Current flowing through a parallel circuit divides and flows through each branch of the circuit.
It
+ _
R1 I1 It
R2 I2
R3 I3
Total current in a parallel circuit is equal to the sum of the current in each branch. The following formula applies to current in a parallel circuit.
It = I1 + I2 + I3 ... + In
Current Flow with Equal Value Resistors in a Parallel Circuit
When equal resistances are placed in a parallel circuit, opposition to current flow is the same in each branch. In the following circuit R1 and R2 are of equal value. If total current (It) is 10 amps, then 5 amps would flow through R1 and 5 amps would flow through R2.
It = 10 Amps
+ _ I1 = 5 Amps
R1 I2 = 5 Amps
R2
It = 10 Amps It = I 1 + I 2 It = 5 Amps + 5 Amps It = 10 Amps
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Current Flow with Unequal Value Resistors in a Parallel Circuit
When unequal value resistors are placed in a parallel circuit, opposition to current flow is not the same in every circuit branch. Current is greater through the path of least resistance. In the following circuit R1 is 40 W and R2 is 20 W. Small values of resistance means less opposition to current flow. More current will flow through R2 than R1.
12 Volts + _
R1 40 Ω I1 = 0.3 Amps It = 0.9 Amps
R2 20 Ω I2 = 0.6 Amps
Using Ohm’s Law, the total current for each circuit can be calculated.
I1 = I1 = E R1 12 Volts 40 Ω
I1 = 0.3 Amps
I2 = I2 =
E R2 12 Volts 20 Ω
I2 = 0.6 Amps
It = I 1 + I 2 It = 0.3 Amps + 0.6 Amps It = 0.9 Amps
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Total current can also be calculated by first calculating total resistance, then applying the formula for Ohm’s Law.
R1 x R2 Rt = R1 + R2 40 Ω x 20 Ω Rt = 40 Ω + 20 Ω 800 Ω Rt = 60 Ω Rt = 13.333 Ω
It = E Rt It = 12 Volts 13.333 Ω It = 0.9 Amps
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Review 4 1. The total resistance of a parallel circuit that has four 20 W resistors is ____________ W.
2. Rt for the following circuit is ____________ W.
+ _
R1 10 Ω
R2 20 Ω
R3 30 Ω
3. Rt for the following circuit is ____________ W.
+ _
R1 5Ω
R2 10 Ω
4. Voltage available at R2 in the following circuit is ____________ volts.
12 Volts
+ _
R1 5Ω
R2 10 Ω
5. In a parallel circuit with two resistors of equal value and a total current flow of 12 amps, the value of current through each resistor is ____________ amps. 6. In the following circuit, current flow through R1 is _____ _______ amps, and through R2 is ____________ amps.
24 Volts + _
R1 10 Ω
R2 15 Ω
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Series-Parallel Circuits
Series-parallel circuits are also known as compound circuits. At least three resistors are required to form a series-parallel circuit. The following illustrations show the two simplest ways a series-parallel combination can be represented.
Parallel Branches + _
Parallel Branches Devices in Series + _
Simplifying a Series-Parallel The formulas required for solving current, voltage and resistance problems have already been defined. To solve a series-parallel circuit, reduce the compound circuits to equivalent simple circuits. In the following illustration, R1 and R2 are parallel with each other. R3 is in series with the parallel circuit of R1 and R2.
R1 10 Ω R3 10 Ω + _
R2 10 Ω
30
First, use the formula to determine total resistance of a parallel circuit to find the total resistance of R1 and R2. When the resistors in a parallel circuit are equal, the following formula is used:
R = Value of any One Resistor Number of Resistors
R = 10 Ω 2 R = 5Ω
Second, redraw the circuit showing the equivalent values. The result is a simple series circuit which uses already learned equations and methods of problem solving.
R3 10Ω + _ R3 5Ω
Simplifying a Series-Parallel Circuit to a Parallel Circuit
In the following illustration R1 and R2 are in series with each other. R3 is in parallel with the series circuit of R1 and R2.
R1 10Ω + _ R2 10Ω R3 20Ω
First, use the formula to determine total resistance of a series circuit to find the total resistance of R1 and R2. The following formula is used:
R = R1 + R2 R = 10 Ω + 10 Ω R = 20 Ω
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Second, redraw the circuit showing the equivalent values. The result is a simple parallel circuit which uses already learned equations and methods of problem solving.
+ _
R = 20 Ω
R3 = 20 Ω
+ _
Rt = 10 Ω
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Review 5 1.
R3 10 Ω
Calculate equivalent resistance for R1 and R2 and total resistance for the entire circuit.
R1 20 Ω
+ _
R2 30 Ω
R1,R2 equivalent resistance = ____________ W Total resistance = ____________ W
2. Calculate equivalent resistance for R1 and R2 and total resistance for the entire circuit.
R1 30 Ω + _ R2 10 Ω R3 20 Ω
R1,R2 equivalent resistance = ____________ W Total resistance = ____________ W
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Power
Work
Whenever a force of any kind causes motion, work is accomplished. In the illustration below work is done when a mechanical force is used to lift a weight. If a force were exerted without causing motion, then no work is done.
Electric Power
In an electrical circuit, voltage applied to a conductor will cause electrons to flow. Voltage is the force and electron flow is the motion. The rate at which work is done is called power and is represented by the symbol “P” . Power is measured in watts, represented by the symbol “W” . In a direct current circuit, one watt is the rate work is done in a circuit when 1 amp flows with 1 volt applied. In a DC circuit, power is the product of voltage times current. Later in this course, you will learn a slightly different version of this relationship for an alternating current (AC) circuit.
P=ExI or P = EI
Power Formulas
Two other power equations can be derived from this formula by substituting other components of Ohm’s Law.
P = I2R and P= E2 R
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DC Circuit Example
In the following illustration, power can be calculated using any of the power formulas.
I = 2 Amps + _ R=6Ω
12 Volts
P = EI P = 12 Volts x 2 Amps P = 24 Watts P = I2 R P = (2 Amps)2 x 6 Ω P = 24 Watts P= P= P= E2 R (12 Volts)2 6Ω 144 6
P = 24 Watts
Additional Calculations
Electrical equipment often has a power rating expressed in watts. This rating is an indication of the rate at which electrical equipment converts electrical energy into some other form of energy, such as heat or mechanical energy. If the power associated with a device and its operating voltage are known, other quantities can be easily calculated. For example, a common household lamp may be rated for 120 volts and 100 watts. Using Ohm’s Law, the rated value of resistance of the lamp can be calculated.
P= R= E2 R
2 which can be transposed to R = E P
(120 Volts)2 100 Watts
R = 144 Ω
35
Using the basic Ohm’s Law formula, the amount of current flow for the 120 volt, 100 watt lamp can be calculated.
I= I= E R 120 Volts 144 Ω
I = 0.833 Amps
By comparison, a lamp rated for 120 volts and 75 watts has a resistance of 192 W and a current of 0.625 amps would flow if the lamp had the rated voltage applied to it.
R= R= E2 P (120 Volts)2 75 Watts
R = 192 Ω I= I= E2 P 120 Volts 192 Ω
I = 0.625 Amps
36
Magnetism
The principles of magnetism are an integral part of electricity. In fact, magnetism can be used to produce electric current and vice versa. Types of Magnets When we think of a permanent magnet, we often envision a horse-shoe or bar magnet or a compass needle, but permanent magnets come in many shapes.
However, all magnets have two characteristics. They attract iron and, if free to move (like the compass needle), a magnet will assume a north-south orientation. Magnetic Lines of Flux Every magnet has two poles, one north pole and one south pole. Invisible magnetic lines of flux leave the north pole and enter the south pole. While the lines of flux are invisible, the effects of magnetic fields can be made visible. When a sheet of paper is placed on a magnet and iron filings loosely scattered over it, the filings will arrange themselves along the invisible lines of flux.
37
By drawing lines the way the iron filings have arranged themselves, the following picture is obtained. Broken lines indicate the paths of magnetic flux lines. The field lines exist outside and inside the magnet. The magnetic lines of flux always form closed loops. Magnetic lines of flux leave the north pole and enter the south pole, returning to the north pole through the magnet.
Interaction between Two Magnets
When two magnets are brought together, the magnetic flux field around the magnets causes some form of interaction. Two unlike poles brought together cause the magnets to attract each other. Two like poles brought together cause the magnets to repel each other.
38
Electromagnetism
Left-Hand Rule for Conductors
An electromagnetic field is a magnetic field generated by current flow in a conductor. Every electric current generates a magnetic field and a relationship exists between the direction of current flow and the direction of the magnetic field. The lefthand rule for conductors demonstrates this relationship. If a current-carrying conductor is grasped with the left hand with the thumb pointing in the direction of electron flow, the fingers will point in the direction of the magnetic lines of flux.
Current-Carrying Coil
A coil of wire carrying a current, acts like a magnet. Individual loops of wire act as small magnets. The individual fields add together to form one magnet. The strength of the field can be increased by adding more turns to the coil, increasing the amount of current, or winding the coil around a material such as iron that conducts magnetic flux more easily than air.
39
Left-Hand Rule for Coils
A left-hand rule exists for coils to determine the direction of the magnetic field. The fingers of the left hand are wrapped around the coil in the direction of electron flow. The thumb points to the north pole of the coil.
Electromagnets
An electromagnet is composed of a coil of wire wound around a core. The core is made of soft iron or some other material that easily conducts magnetic lines of force. When current is passed through the coil, the core becomes magnetized. The ability to control the strength and direction of the magnetic force makes electromagnets useful. As with permanent magnets, opposite poles attract. An electromagnet can be made to control the strength of its field which controls the strength of the magnetic poles. A large variety of electrical devices such as motors, circuit breakers, contactors, relays and motor starters use electromagnetic principles.
40
Review 6 1. The rate at which work is done is called ___________ . 2. The basic formulas for power in a DC circuit are: P = I x __ , P = I2 x __ , and P = _ R 3. In a circuit with a 12 volt supply and 4 W resistance, the power consumed is ____________ watts. 4. The two characteristics of all magnets are: they attract and hold ____________ , and, if free to move, they will assume roughly a ____________ position. 5. Lines of flux always leave the ____________ pole and enter the ____________ pole. 6. The left-hand rule for conductors states that, when the ___________ hand is placed on a current-carrying conductor with the ____________ pointing in the direction of electron flow, the fingers will point in the direction of ____________ .
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Introduction to AC
The supply of current for electrical devices may come from a direct current (DC) source, or an alternating current (AC) source. In a direct current circuit, electrons flow continuously in one direction from the source of power through a conductor to a load and back to the source of power. Voltage in direct current remains constant. DC power sources include batteries and DC generators. By contrast, an AC generator makes electrons flow first in one direction then in another. In fact, an AC generator reverses its terminal polarities many times a second, causing current to change direction with each reversal.
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AC Sine Wave
Alternating voltage and current vary continuously. The graphic representation for AC is a sine wave. A sine wave can represent current or voltage. There are two axes. The vertical axis represents the direction and magnitude of current or voltage. The horizontal axis represents time.
+ Direction
0
Time
- Direction
When the waveform is above the time axis, current is flowing in one direction. This is referred to as the positive direction. When the waveform is below the time axis, current is flowing in the opposite direction. This is referred to as the negative direction. A sine wave moves through a complete rotation of 360 degrees, which is referred to as one cycle. Alternating current goes through many of these cycles each second. Single-Phase and Three-Phase AC Power Alternating current is divided into single-phase and threephase types. Single-phase power is used for small electrical demands such as found in the home. Three-phase power is used where large blocks of power are required, such as found in commercial applications and industrial plants. Single-phase power is shown in the above illustration. Three-phase power, as shown in the following illustration, is a continuous series of three overlapping AC cycles. Each wave represents a phase, and is offset by 120 electrical degrees.
Phase 1 Phase 2 Phase 3 +
0
-
43
AC Generators
Basic Generator
A basic generator consists of a magnetic field, an armature, slip rings, brushes and a resistive load. In a commercial generator, the magnetic field is created by an electromagnet, but, for this simple generator, permanent magnets are used. An armature is any number of conductive wires wound in loops which rotates through the magnetic field. For simplicity, one loop is shown. When a conductor is moved through a magnetic field, a voltage is induced in the conductor. As the armature rotates through the magnetic field, a voltage is generated in the armature which causes current to flow. Slip rings are attached to the armature and rotate with it. Carbon brushes ride against the slip rings to conduct current from the armature to a resistive load.
Pole Piece
Magnetic Field Armature
Brush Slip Ring
R1
Basic Generator Operation
An armature rotates through the magnetic field. At an initial position of zero degrees, the armature conductors are moving parallel to the magnetic field and not cutting through any magnetic lines of flux. No voltage is induced.
R1
44
Generator Operation from Zero to 90 Degrees
As the armature rotates from zero to 90 degrees, the conductors cut through more and more lines of flux, building up to a maximum induced voltage in the positive direction.
90 Degrees
R1
Generator Operation from 90 to 180 Degrees
The armature continues to rotate from 90 to 180 degrees, cutting fewer lines of flux. The induced voltage decreases from a maximum positive value to zero.
180 Degrees
S
R1
Generator Operation from 180 to 270 Degrees
As the armature continues to rotate from 180 degrees to 270 degrees, the conductors cut more lines of flux, but in the opposite direction, and voltage is induced in the negative direction, building up to a maximum at 270 degrees.
270 Degrees
R1
45
Generator Operation from 270 to 360 Degrees
As the armature continues to rotate from 270 to 360 degrees, induced voltage decreases from a maximum negative value to zero. This completes one cycle. The armature continues to rotate at a constant speed causing the cycle to repeat as long as the armature rotates.
360 Degrees
S
One Revolution R1
46
Frequency
The number of cycles per second of voltage induced in the armature is the frequency of the generator. If the armature rotates at a speed of 60 revolutions per second, the generated voltage will be 60 cycles per second. The recognized unit for frequency is hertz, abbreviated “Hz” . 1 Hz is equal to 1 cycle per second. Power companies generate and distribute electricity at very low frequencies. The standard power line frequency in the United States and many other countries is 60 Hz. 50 Hz is also a common power line frequency used throughout the world. The following illustration shows 15 cycles in 1/4 second which is equivalent to 60 Hz.
1/4 Second
Four-Pole AC Generator
The frequency is the same as the number of rotations per second if the magnetic field is produced by only two poles. An increase in the number of poles, would cause an increase in the number of cycles completed in a revolution. A two-pole generator would complete one cycle per revolution and a fourpole generator would complete two cycles per revolution. An AC generator produces one cycle per revolution for each pair of poles.
One Revolution R1
47
Voltage and Current
Peak Value
Voltage and current in an AC circuit rise and fall over time in a pattern referred to as a sine wave. The peak value of a sine wave occurs twice each cycle, once at the positive maximum value and once at the negative maximum value.
+ Peak Value
0
Time
-
Peak Value
Peak-to-Peak Value
The value of the voltage or current between the peak positive and peak negative values is called the peak-to-peak value.
+
0
Time
Peak-to-Peak Value
-
Instantaneous Value
The instantaneous value is the value at any one point in the sine wave.
+ Instantaneous Value
0
Time
-
48
Calculating Instantaneous Voltage
The voltage waveform produced as the armature rotates through 360 degrees rotation is called a sine wave because the instantaneous voltage (e) is related to the sine trigonometric function. The sine of an angle is represented symbolically as sin q, where the Greek letter theta (q) represents the angle. The sine curve is a graph of the following equation for values of q from 0 to 360 degrees:
e = Epeak x sin θ
Instantaneous voltage is equal to the peak voltage times the sine of the angle of the generator armature. The sine value is obtained from trigonometric tables. The following table shows selected instantaneous values.
Angle 30 Degrees 60 Degrees 90 Degrees 120 Degrees 150 Degrees 180 Degrees Sin 0.5 0.866 1 0.866 0.5 0
θ
Angle 210 Degrees 240 Degrees 270 Degrees 300 Degrees 330 Degrees 360 Degrees
Sin
θ
-0.5 -0.866 -1 -0.866 -0.5 0
The following example illustrates instantaneous values at 90, 150, and 240 degrees. The peak voltage is equal to 100 volts. By substituting the sine at the instantaneous angle value, the instantaneous voltage can be calculated.
+ 90° = +100 Volts 150° = +50 Volts 0 240° = -86.6 Volts -
Any instantaneous value can be calculated. For example:
240° e = 100 x -0.866 e = -86.6 volts
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Effective Value of an AC Sine Wave
The instantaneous value of an alternating voltage and current are constantly changing values. However, there is a method for translating the varying values into an equivalent constant value, referred to at the effective value of voltage or current. This is also known as the RMS value. RMS is an abbreviation of the mathematical term root-mean-square. For example, a common voltage used in many applications is 120 volts, this is an RMS value, which is equal to the peak value times 0.707 .
+ Peak Value 169.7 Volts 0
-
169.7 Vpeak x 0.707 = 120 Vrms
The effective value of AC is defined in terms of an equivalent heating effect when compared to DC. One RMS ampere of current flowing through a resistance will produce heat at the same rate as one DC ampere. It is sometimes necessary to know the peak value of an AC voltage or current when the RMS value is known. To calculate the peak value, multiply the effective value by 1.41. For example, if the effective value is 100 volts, the peak value is 141 volts. Review 7 1. A ____________ is the graphic representation of AC voltage or current values over time.
2. Each phase of three phase AC power is offset by ____________ degrees. 3. An AC generator produces ____________ cycle(s) per revolution for each pair of poles. 4. What is the instantaneous voltage at 240 degrees for a sine wave with a peak voltage of 150 volts? 5. What is the effective voltage for a sine wave with a peak voltage of 150 volts?
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Inductance
The circuits studied to this point have been resistive. Resistance and voltage are not the only circuit properties that effect current flow, however. Inductance is the property of an electric circuit that opposes any change in electric current. Resistance opposes current flow, inductance opposes change in current flow. Inductance is designated by the letter “L ” . . The unit of measurement for inductance is the henry (h). Current Flow and Field Strength Current flow produces a magnetic field in a conductor. The amount of current determines the strength of the magnetic field. As current flow increases, field strength increases, and as current flow decreases, field strength decreases.
0 Degrees No Current 30 Degrees Increasing Current 90 Degrees Maximum Current
Any change in current causes a corresponding change in the magnetic field surrounding the conductor. Current is constant for a regulated DC source, except when the circuit is turned on and off, or when there is a load change. However, alternating current is constantly changing, and inductance is continually opposing the change. A change in the magnetic field surrounding the conductor induces a voltage in the conductor. This self-induced voltage opposes the change in current. This is known as counter emf. This opposition causes a delay in the time it takes current to attain its new steady value. If current increases, inductance tries to hold it down. If current decreases, inductance tries to hold it up. Inductance is somewhat like mechanical inertia which must be overcome to get a mechanical object moving or to stop a mechanical object from moving. A vehicle, for example, takes a few moments to accelerate to a desired speed, or decelerate to a stop. 51
Inductors
All conductors have inductance, but inductors are coils of wire wound for a specific inductance. For some applications, inductors are wound around a metal core to further concentrate the inductance. The inductance of a coil is determined by the number of turns in the coil, the coil diameter and length, and the core material. An inductor is usually indicated symbolically on an electrical drawing as a curled line or a filled rectangle.
Simple Inductive Circuit
In a resistive circuit, current change is considered instantaneous. If an inductor is used, the current does not change as quickly. For the purpose of explanation, a DC circuit is used here to describe the operation of an inductor. There will always be some amount of resistance and inductance in any circuit. The electrical wire used in the circuit has some resistance and inductance. In addition, inductors also have resistance. However, to simplify examples in this book, the resistance and inductance of the wiring and the resistance of inductors are not considered. In the following circuit, initially the switch is in position 2, and there is no current flowing through the ammeter (A). When the switch is moved to position 1, current will rise rapidly at first, then more slowly as the maximum value is approached.
1 2 A
+ _
L1
R1
Inductive Time Constant
52
The time required for the current to rise to its maximum value is determined by the ratio of inductance (in henrys) to resistance (in ohms). This ratio is called the time constant of the inductive circuit. A time constant is the time (in seconds) required for the circuit current to rise to 63.2% of its maximum value. When the switch is closed in the previous circuit, current will begin to flow. During the first time constant current rises to 63.2% of its maximum value. During the second time constant, current rises to 63.2% of the remaining 36.8%, or a total of 86.4%. It takes about five time constants for current to reach its maximum value.
100.0% 98.1% 94.9% 86.4%
63.2%
First Time Constant
Second Time Constant
Third Time Constant
Fourth Time Constant
Fifth Time Constant
Similarly, when the switch in the previous circuit is returned to position 2, the magnetic field around the inductor will begin to collapse, returning stored energy to the circuit, and it will take about five time constants for current to reach zero.
100.0% First Time Constant Second Time Constant Third Time Constant Fourth Time Constant Fifth Time Constant
36.8% 13.6% 5.1% 1.9% 0%
Calculating the Time Constant of an Inductive Circuit
The time constant is designated by the symbol “t” . To determine the time constant of an inductive circuit use one of the following formulas:
τ (in seconds) = L (henrys) R (ohms) L (millihenrys) R (ohms) L (microhenrys) R (ohms)
τ (in milliseconds) =
τ (in microseconds) =
53
In the following illustration, L1 is equal to 15 millihenrys and R1 is equal to 5 W. When the switch is closed, it will take 3 milliseconds for current to rise from zero to 63.2% of its maximum value and approximately 15 milliseconds for full current to be reached.
+ _
L1 15 mh
R1 5 Ω
τ=
15 mh 5Ω
τ = 3 milliseconds
Formula for Series Inductors The same rules for calculating total resistance can be applied to calculating total inductance. In the following circuit, an AC generator is used to supply electrical power to four inductors. Total inductance of series inductors is calculated using the following formula:
Lt = L1 + L2 + L3 ... + Ln 2 mh L1 AC Generator 2 mh L2 1 mh L3 1 mh L4
Lt = L1 + L2 + L3 + L4 Lt = 2 mh + 2 mh + 1 mh + 1 mh Lt = 6 mh
54
Formula for Parallel Inductors
In the following circuit, an AC generator is used to supply electrical power to three inductors. Total inductance is calculated using the following formula:
1 = 1 + 1 + 1 ... + 1 Lt L1 L2 L3 Ln
L1 5 mh
L2 10 mh
L3 20 mh
1 = Lt
1 + 1 + 1 5 10 20
1 = 7 Lt 20 Lt = 20 7 Lt = 2.86 mh
55
Capacitance
Capacitance and Capacitors Capacitance is a measure of a circuit’s ability to store an electrical charge. A device manufactured to have a specific amount of capacitance is called a capacitor. A capacitor is made up of a pair of conductive plates separated by a thin layer of insulating material. Another name for the insulating material is dielectric material. When a voltage is applied to the plates, electrons are forced onto one plate. That plate has an excess of electrons while the other plate has a deficiency of electrons. The plate with an excess of electrons is negatively charged. The plate with a deficiency of electrons is positively charged.
Negative Plate Dielectric Material Positive Plate
Direct current cannot flow through the dielectric material because it is an insulator; however, the electric field created when the capacitor is charged is felt through the dielectric. Capacitors are rated for the amount of charge they can hold. The capacitance of a capacitor depends on the area of the plates, the distance between the plates, and type of dielectric material used. The unit of measurement for capacitance is farads (F). However, the farad is a large unit and capacitors are often rated in microfarads (mF) or picofarads (pF). Capacitor Circuit Symbols Capacitance is usually indicated symbolically on an electrical drawing by a combination of a straight line with a curved line, or two straight lines.
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Simple Capacitive Circuit
For the purpose of explanation, a DC circuit is used here to describe the operation of an capacitor. The resistance of wiring and stray values of capacitance found in any circuit are not considered in this explanation. In a resistive circuit, voltage change is instantaneous. In a circuit with a resistor and capacitor in series, the voltage across the capacitor does not change as quickly. In the following circuit, initially the switch is in position 2 and no voltage is measured by the voltmeter (V). When the switch is moved to position 1, voltage across the capacitor will rise rapidly at first, then more slowly as the maximum value is approached.
1 2
+ _
3
R1 C1
V
Capacitive Time Constant
The time required for voltage across the capacitor in this simple circuit to rise to its maximum value is determined by the product of capacitance, in farads, times resistance, in ohms. This product is the time constant of a capacitive circuit. The time constant gives the time in seconds required for voltage across the capacitor to reach 63.2% of its maximum value. When the switch in the previous circuit is moved to position 1, the voltage measured by the voltmeter will begin to rise. During the first time constant, voltage will rise to 63.2% of its maximum value. During the second time constant, voltage will rise to 63.2% of the remaining 36.8%, or a total of 86.4%. It takes about five time constants for voltage across the capacitor to reach its maximum value.
100.0% 98.1% 94.9% 86.4%
63.2%
First Time Constant
Second Time Constant
Third Time Constant
Fourth Time Constant
Fifth Time Constant
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The voltage across the capacitor will reach its maximum value when it is equal to the supply voltage. At that point, current flow will reach zero. When the switch in the previous circuit is returned to position 2, the capacitor will retain its charge because there is no path for current flow. When the switch is moved to position 3, the capacitor will begin to discharge, and it will take about five time constants for the voltage across the capacitor and the current through the resistor to reach zero.
100.0% First Time Constant Second Time Constant Third Time Constant Fourth Time Constant Fifth Time Constant
36.8% 13.6% 5.1% 1.9% 0%
Calculating the Time Constant of a Capacitive Circuit
To determine the time constant of a capacitive circuit, use one of the following formulas:
τ (in seconds) = R (megohms) x C (microfarads) τ (in microseconds) = R (megohms) x C (picofarads) τ (in microseconds) = R (ohms) x C (microfarads)
In the following illustration, C1 is equal to 2 mF , and R1 is equal to 10 W. When the switch is closed, it will take 20 microseconds for voltage across the capacitor to rise from zero to 63.2% of its maximum value. It will take about five time constants, 100 microseconds, for this voltage to rise to its maximum value.
+ _
C1 2µF
R1 10 Ω
V
τ = RC τ = 2µF x 10 Ω τ = 20 microseconds
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Formula for Series Capacitors
Connecting capacitors in series decreases total capacitance. The formula for series capacitors is similar to the formula for parallel resistors. In the following circuit, an AC generator supplies electrical power to three capacitors. Total capacitance is calculated using the following formula:
1 = 1 + 1 + 1 ... + 1 Cn Ct C1 C2 C3 5 µF 10 µF 20 µF
C1
C2
C3
Formula for Parallel Capacitors
Adding capacitors in parallel increases circuit capacitance. In the following circuit, an AC generator is used to supply electrical power to three capacitors. Total capacitance is calculated using the following formula:
Ct = C1 + C2 + C3 ... + Cn
1 = 1 + 1 + 1 Ct 5 10 20 1 = 7 Ct 20 20 Ct = 7 Ct = 2.86 µF
C1 5 µF
C2 10 µF
C3 20 µF
Ct = 5 µF + 10 µF + 20 µF Ct = 35 µF
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Review 8 1. The total inductance for this circuit is __________ mh.
4 mh L1 2 mh L2 3 mh L3 1 mh L4
2. The total inductance for this circuit is ___________ mh.
L1 5 mh
L2 10 mh
L3 10 mh
3. The total capacitance for this circuit is __________ mF .
5 µF 10 µF 10 µF
C1
C2
C3
4. The total capacitance for this circuit is __________ mF .
C1 5 µF
C2 10 µF
C3 10 µF
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Inductive and Capacitive Reactance
In a purely resistive AC circuit, resistance is the only opposition to current flow. In an AC circuit with only inductance, capacitance, or both inductance and capacitance, but no resistance, opposition to current flow is called reactance, designated by the symbol “X” . Total opposition to current flow in an AC circuit that contains both reactance and resistance is called impedance, designated by the symbol “Z” . Just like resistance, reactance and impedance are expressed in ohms. Inductive Reactance Inductance only affects current flow when the current is changing. Inductance produces a self-induced voltage (counter emf) that opposes changes in current. In an AC circuit, current is changing constantly. Inductance in an AC circuit, therefore, causes a continual opposition. This opposition to current flow is called inductive reactance and is designated by the symbol “XL. ” Inductive reactance is proportional to both the inductance and the frequency applied. The formula for inductive reactance is:
XL = 2πfL XL = 2 x 3.14 x frequency x inductance
For a 60 hertz circuit containing a 10 mh inductor, the inductive reactance is:
XL = 2πfL XL = 2 x 3.14 x 60 x 0.010 XL = 3.768 Ω
For this example, the resistance is zero so the impedance is equal to the reactance. If the voltage is known, Ohm’s Law can be used to calculate the current. If, for example, the voltage is 10 volts, the current is calculated as follows;
I= I= E Z 10 3.768
I = 2.65 Amps
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Phase Relationship between Current and Voltage in an Inductive Circuit
In a purely resistive circuit, current and voltage rise and fall at the same time. They are said to be “in phase. ” In this circuit, there is no inductance. Resistance and impedance are the same.
+ Voltage Current
0
_
In a purely inductive circuit, current lags behind voltage by 90 degrees. Current and voltage are said to be “out of phase” . In this circuit, impedance and inductive reactance are the same.
90 Degrees + Voltage Current
0
_
In a circuit with both resistance and inductive reactance, AC current will lag the voltage by more than 0 degrees and less than 90 degrees. The exact amount of lag will depend upon the relative amounts of resistance and inductive reactance. The more resistive a circuit is, the closer it is to being in phase. The more reactive a circuit is, the more out of phase it is. In the following illustration, resistance and inductive reactance are equal. Current lags voltage by 45 degrees.
45 Degrees + Voltage Current XL = 10 Ω
0 R = 10 Ω _
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Calculating Impedance in an Inductive Circuit
When working with a circuit containing elements of inductance, capacitance, and resistance, impedance must be calculated. Because electrical concepts deal with trigonometric functions, this is not a simple matter of subtraction and addition. The following formula is used to calculate impedance in a circuit with resistance and inductive reactance:
Z = R2 + XL2
In the previous example, resistance and inductive reactance are each 10 ohms. Impedance for this circuit can be calculated as follows:
Z = 102 + 102 Z= 200
Z = 14.1421 Ω
Vectors
A common way to represent AC circuit values is with a vector. A vector is a graphic representation of a quantity that has direction and magnitude. A vector on a map might indicate that one city is 50 miles southwest from another. The magnitude is 50 miles and the direction is southwest. Vectors are also used to show electrical relationships. The following vector illustrates the relationship between resistance and inductive reactance for a circuit containing 10 ohms of each. The angle between the vectors is the phase angle represented by the symbol q. When inductive reactance is equal to resistance the resultant angle is 45 degrees. This angle represents how much current lags voltage for this circuit.
14 .14 21 Z = Ω
XL = 10 Ω
θ
R = 10 Ω
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Capacitive Reactance
Capacitance also opposes AC current flow. Capacitive reactance is designated by the symbol XC. The larger the capacitor, the smaller the capacitive reactance. Current flow in a capacitive AC circuit is also dependent on frequency. The following formula is used to calculate capacitive reactance:
1 XC = 2πfC
The capacitive reactance for a 60 hertz circuit with a 10 microfarad capacitor is calculated as follows:
1 XC = 2πfC 1 XC = 2 x 3.14 x 60 x 0.000010 XC = 265.39 Ω
For this example, the resistance is zero so the impedance is equal to the reactance. If the voltage is known, Ohm’s Law can be used to calculate the current. If, for example, the voltage is 10 volts, the current is calculated as follows:
I= I= E Z 10 265.39
I = 0.0376 Amps
Phase Relationship between The phase relationship between current and voltage are Current and Voltage opposite to the phase relationship of an inductive circuit. In a purely capacitive circuit, current leads voltage by 90 degrees.
90 Degrees + Voltage Current 0
_
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In a circuit with both resistance and capacitive reactance, AC current will lead the voltage by more than 0 degrees and less than 90 degrees. The exact amount of lead will depend upon the relative amounts of resistance and capacitive reactance. The more resistive a circuit is, the closer it is to being in phase. The more reactive a circuit is, the more out of phase it is. In the following illustration, resistance and capacitive reactance are equal. Current leads voltage by 45 degrees.
45 Degrees + Voltage Current 0 R = 10 Ω _ XC = 10 Ω
Calculating Impedance in a Capacitive Circuit
The following formula is used to calculate impedance in a circuit with resistance and capacitive reactance:
Z = R2 + XC2
In the previous example, resistance and capacitive reactance are each 10 ohms. Impedance for this circuit can be calculated as follows:
Z = 102 + 102 Z= 200
Z = 14.1421 Ω
The following vector illustrates the relationship between resistance and capacitive reactance for a circuit containing 10 ohms of each. The angle between the vectors is the phase angle represented by the symbol q. When capacitive reactance is equal to resistance, the resultant angle is 45 degrees. This angle represents how much current leads voltage for this circuit.
R = 10 Ω
θ
Z = 21 .14 14 Ω
XC = 10 Ω
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Review 9 1. Opposition to current flow in an AC circuit caused by inductors and capacitors is called ____________ .
2. Total opposition to current flow in an AC circuit with both resistance and reactance is called ____________ . 3. In a 50 hertz circuit, containing a 10 mh inductor, the inductive reactance is ____________ ohms. 4. In a purely inductive circuit, ____________ a. current and voltage are in phase b. current leads voltage by 90 degrees c. current lags voltage by 90 degrees
5. In a purely capacitive circuit, ____________ a. current and voltage are in phase b. current leads voltage by 90 degrees c. current lags voltage by 90 degrees
6. In a 50 hertz circuit, containing a 10 microfarad capacitor, the capacitive reactance is ____________ ohms. 7 . In a circuit with 5 W resistance, and 10 W inductive reactance, impedance is ____________ ohms.
8. In a circuit with 5 W resistance, and 4 W capacitive reactance, impedance is ____________ ohms.
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Series R-L-C Circuit
Circuits often contain elements of resistance, inductance, and capacitance. In an inductive AC circuit, current lags voltage by 90 degrees. In a capacitive AC circuit, current leads voltage by 90 degrees. When represented in vector form, inductive and capacitive reactance are 180 degrees apart. As a result, the net reactance is determined by taking the difference between the two quantities.
XL R
XC
An AC circuit is: • • • Calculating Total Impedance in a Series R-L-C Circuit Resistive if XL and XC are equal Inductive if XL is greater than XC Capacitive if XC is greater than XL
The following formula is used to calculate total impedance of a circuit containing resistance, capacitance, and inductance:
Z = R2 + (XL - XC)2
In the case where inductive reactance is greater than capacitive reactance, subtracting XC from XL results in a positive number. The positive phase angle is an indicator that the net circuit reactance is inductive, and current lags voltage. In the case where capacitive reactance is greater than inductive reactance, subtracting XC from XL results in a negative number. The negative phase angle is an indicator that the net circuit reactance is capacitive and current leads voltage. In either case, the value squared will result in a positive number.
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Calculating Reactance and Impedance in a Series R-L-C Circuit
In the following 120 volt, 60 hertz circuit, resistance is 1000 W, inductance is 5 mh, and capacitance is 2 mF . The following example shows the method for calculating impedance for this circuit.
R = 1000 Ω
L = 5 mh
C = 2 µF
XL = 2πfL XL = 6.28 x 60 x 0.005 XL = 1.884 Ω 1 XC = 2πfC 1 XC = 6.28 x 60 x 0.000002 XC = 1,327 Ω
Z = R2 + (XL - XC)2 Z = 10002 + (1.884 - 1,327)2 Z = 1,000,000 + ( - 1,325.116)2 Z = 1,000,000 + 1,755,932.41 Z = 2,755,932.41 Z = 1,660.1 Ω
Given that the applied voltage is 120 volts, current can be calculated as follows:
I= E Z I= 120 1,660.1
I = 0.0723 Amps
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Parallel R-L-C Circuit
Calculating Impedance in a Parallel R-L-C Circuit
Total impedance (Zt) can be calculated in a parallel R-L-C circuit if values of resistance and reactance are known. One method of calculating impedance involves first calculating total current, then using the following formula:
E Zt = t It
Total current is the vector sum of current flowing through the resistance plus, the difference between inductive current and capacitive current. This is expressed in the following formula:
It = IR2 + (IC - IL)2
In the following 120 volt, 60 hertz circuit, capacitive reactance is 25 W, inductive reactance is 50 W, and resistance is 1000 W. A simple application of Ohm’s Law will find the branch currents. Remember, voltage is constant throughout a parallel circuit.
120 V
R = 1000 Ω
XL = 50 Ω
XC = 25 Ω
IR = IR =
E R 120 1000
E IL = XL 120 IL = 50 IL = 2.4 Amps
IC = IC =
E XC 120 25
IR = 0.12 Amps
IC = 4.8 Amps
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Once the branch currents are known, total current can be calculated.
It = It = It = It = IR2 + (IC - IL)2 0.122 + (4.8 - 2.4)2 0.0144 + 5.76 5.7744
It = 2.403 Amps
Impedance can then be calculated as follows:
E Zt = t It Zt = 120 2.403
Zt = 49.94 Ω
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Power and Power Factor in an AC Circuit
Power consumed by a resistor is dissipated in heat and not returned to the source. This is called true power because it is the rate at which energy is used. Current in an AC circuit rises to peak values and diminishes to zero many times a second. The energy stored in the magnetic field of an inductor, or plates of a capacitor, is returned to the source when current changes direction. Although reactive components do not consume energy, they do increase the amount of energy that must be generated to do the same amount of work. The rate at which this non-working energy must be generated is called reactive power. If voltage and current are 90 degrees out of phase, as would be the case in a purely capacitive or purely inductive circuit, the average value of true power is equal to zero. There are high positive and negative peak values of power, but when added together the result is zero. Power in an AC circuit is the vector sum of true power and reactive power. This is called apparent power. True power is equal to apparent power in a purely resistive circuit because voltage and current are in phase. Voltage and current are also in phase in a circuit containing equal values of inductive reactance and capacitive reactance. In most circuits, however, apparent power is composed of both true power and reactive power. True Power and Apparent Power Formulas The formula for apparent power is:
P = EI
Apparent power is measured in volt-amps (VA). True power is calculated from another trigonometric function, the cosine of the phase angle (cos θ). The formula for true power is:
P = EI cos θ
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In a purely resistive circuit, current and voltage are in phase. There is a zero degree angle displacement between current and voltage. The cosine of zero is one. Multiplying a value by one does not change the value. Therefore, in a purely resistive circuit, the cosine of the angle is ignored. In a purely reactive circuit, either inductive or capacitive, current and voltage are 90 degrees out of phase. The cosine of 90 degrees is zero. Multiplying a value times zero results in a zero product. Therefore, no power is consumed in a purely reactive circuit. Calculating Apparent Power In the following 120 volt circuit, current is equal to 84.9 mA. in a simple R-L-C Circuit Inductive reactance is 100 W and capacitive reactance is 1100 W. The phase angle is -45 degrees. By referring to a trigonometric table, the cosine of -45 degrees is found to be 0.7071.
R = 1000 Ω XL = 100 Ω XC = 1100 Ω 120 V
The apparent power consumed by the circuit is:
P = EI P = 120 x 0.0849 P = 10.2 VA
The true power consumed by the circuit is:
P = EI cos θ P = 120 x 0.0849 x 0.7071 P=7 .2 Watts
Another formula for true power is:
P = I2R P = 0.08492 x 1000 P=7 .2 Watts
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Power Factor
Power factor is the ratio of true power to apparent power in an AC circuit. Power factor is expressed in the following formula:
PF = True Power Apparent Power
Power factor can also be expressed using the formulas for true power and apparent power. The value of EI cancels out because it is the same in the numerator and denominator. Power factor is the cosine of the angle.
PF = EI cos θ EI
PF = cos θ
In a purely resistive circuit, where current and voltage are in phase, there is no angle of displacement between current and voltage. The cosine of a zero degree angle is one. The power factor is one. This means that all energy delivered by the source is consumed by the circuit and dissipated in the form of heat. In a purely reactive circuit, voltage and current are 90 degrees apart. The cosine of a 90 degree angle is zero. The power factor is zero. This means the circuit returns all energy it receives from the source to the source. In a circuit where reactance and resistance are equal, voltage and current are displaced by 45 degrees. The cosine of a 45 degree angle is .7071. The power factor is .7071. This means the circuit uses approximately 70% of the energy supplied by the source and returns approximately 30%.
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Review 10 1. An AC circuit is ____________ if inductive reactance and capacitive reactance are equal.
2. A series AC circuit is ____________ if there is more inductive reactance than capacitive reactance. 3. A series AC circuit is ____________ if there is more capacitive reactance than inductive reactance. 4. In a 120 VAC, 60 hertz series circuit, with 1000 W of resistance, 10 mh of inductance and 4 mF of capacitance, impedance is ____________ W and current is ____________ amps. 5. For the following circuit, calculate impedance and current.
R = 1000 Ω XL = 200 Ω XC = 1200 Ω 120 V 60 Hz
Impedance is __________ W , and It is __________ amps.
6. For a circuit with a 120 volt AC source and a current of 10 amps, the apparent power is __________ VA. 7 . For a circuit with an apparent power of 3000 VA and a power factor of 0.8, the true power is __________ watts.
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Transformers
Mutual Induction
Transformers are electromagnetic devices that transfer electrical energy from one circuit to another by mutual induction. A single-phase transformer has two coils, a primary and a secondary. Mutual induction is the transfer of electrical energy from the primary to the secondary through magnetic fields . The following circuit illustrates mutual induction. The AC generator provides electrical power to the primary coil. The magnetic field produced by the primary induces a voltage into the secondary coil, which supplies power to a load.
Primary Coil
Secondary Coil
Transformers are used to step a voltage up to a higher level, or down to a lower level. To understand the need to stepping up or down voltages, consider how electrical power is generated and distributed. Generators used by power companies typically generate voltages of 30 KV or less. While this is a relatively high voltage compared to the voltages used by power customers, it is more efficient for utilities to transmit this power at still higher voltages, up to as high at 765 KV. The electrical power is received at substation transformers many miles away where it is stepped down and distributed locally. When it arrives at the customer’s location, it is further stepped down to the level needed for the type of customer. Even within a customer’s facility, voltage may need to be stepped down further to meet requirements of some equipment.
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This process of stepping up or down the voltage throughout a power distribution system is most often accomplished using transformers. The size and ratings of the transformers vary, but the basic operation of these devices is the same. Coefficient of Coupling Mutual inductance between two coils depends on their flux linkage. Maximum coupling occurs when all the lines of flux from the primary coil cut through the secondary winding. The amount of coupling which takes place is referred to as coefficient of coupling. To maximize coefficient of coupling, both coils are often wound on an iron core which is used to provide a path for the lines of flux. The following discussion of step-up and step-down transformers applies to transformers with an iron core.
Lines of Flux Confined to Iron Core
Lines of Flux that don’t Couple
Voltage, Current, and the Number of Turns in a Coil
There is a direct relationship between voltage, impedance, current, and the number of coil turns in a transformer. This relationship can be used to find either primary or secondary voltage, current, and the number of turns in each coil. It is the number of turns which determine if a transformer is a step up or step down transformer. The following “rules-of-thumb” apply to transformers: • • If the primary coil has fewer turns than the secondary coil, the transformer is a step-up transformer. If the primary coil has more turns than the secondary coil, the transformer is a step-down transformer.
When the number of turns on the primary and secondary coils of a transformer are equal, input voltage, impedance, and current are equal to output voltage, impedance, and current.
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Step-Up Transformer
A step-up transformer is used when it is desirable to step voltage up in value. The following circuit illustrates a stepup transformer. The primary coil has fewer turns than the secondary coil. When the primary has fewer turns than the secondary, voltage and impedance are stepped up. In the circuit illustrated, The transformer secondary has twice as many turns as the primary and voltage is stepped up from 120 VAC to 240 VAC. Because impedance is also stepped up, current is stepped down from 10 amps to 5 amps.
1:2
Primary Coil 900 Turns
Secondary Coil 1800 Turns
120 VAC Supply 10 Amps
240 VAC 5 Amps
Step-Down Transformer
A step-down transformer is used when it is desirable to step voltage down in value. The following circuit illustrates a step-down transformer. The primary coil has more turns than the secondary coil. The step-down ratio is 2:1. voltage and impedance are stepped down, current is stepped up.
2:1
Primary Coil 1800 Turns
Seconday Coil 900 Turns
240 VAC Supply 5 Amps
120 VAC Out 10 Amps
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Single-Phase Transformer 120 or 240 VAC single-phase transformers are used to supply lighting, receptacle, and small appliance loads. A transformer with a 240 VAC secondary can be used to supply 240 VAC to larger appliances such as stoves, air conditioners and heaters. A 240 VAC secondary can be tapped in the center to provide two sources of 120 VAC power.
Primary Primary
120 VAC
120 VAC
120 VAC Secondary Ground
240 VAC Secondary
Formulas for Calculating the Number of Primary and Secondary Turns of a Transformer
There are a number of useful formulas for calculating, voltage, current, and the number of turns between the primary and secondary of a transformer. These formulas can be used with either step-up or step-down transformers. The following legend applies to the transformer formulas: ES EP IS IP NS NP = = = = = = secondary voltage primary voltage secondary current primary current turns in the secondary coil turns in the primary coil
To find voltage:
ES = EP x IP IS EP = ES x IS IP
To find current:
IS = EP x IP ES IP = ES x IS EP
To find the number of coil turns:
NS = ES x NP EP NP = EP x NS ES
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Using the values for the step-down transformer in the example of the previous page, the secondary voltage can be verified.
ES = ES = ES = EP x IP IS 240 Volts x 5 Amps 10 Amps 1200 10
ES = 120 Volts
Transformer Ratings
Transformers are rated for the amount of apparent power they can provide. Because values of apparent power are often large, the transformer rating is frequently given in kVA (kilovoltamps). The kVA rating determines the current and voltage a transformer can deliver to its load without overheating. Given kVA and volts, amps can be calculated.
kVA = Volts x Amps 1000 kVA x 1000 Volts
Amps =
Using the illustrated step-down transformer, the kVA rating can be calculated. The kVA rating of a transformer is the same for both the primary and the secondary.
Primary kVA = 240 x 5 1000
Primary kVA = 1.2 kVA Secondary kVA = 120 x 10 1000
Secondary kVA = 1.2 kVA
Transformer Losses
Most of the electrical energy provided to the primary of a transformer is transferred to the secondary. Some energy, however, is lost in heat in the wiring or the core. Some losses in the core can be reduced by building the core of a number of flat sections called laminations.
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Three-Phase Transformers
Delta Connections
Three-phase transformers are used when three-phase power is required for larger loads such as industrial motors. There are two basic three-phase transformer connections, delta and wye. A delta is like three single-phase transformers connected together. The secondary of a delta transformer is illustrated below. For simplicity, the primary is not shown in this example. The voltages shown on the illustration are examples. Just as with a single-phase transformer, the secondary voltage depends on both the primary voltage and the turns ratio. Delta transformers are schematically drawn in a triangle. The voltages across each winding of the delta triangle represents one phase of a three phase system. The voltage is always the same between any two wires. A single phase (such as L1 to L2) can be used to supply single phase loads. All three phases are used to supply three phase loads.
L1
480 Volts
480 Volts
480 Volts
L2 480 Volts 480 Volts
L3
L1 to L2 = 480 volts L2 to L3 = 480 volts L1 to L3 = 480 volts
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Balanced Delta Current
When current is the same in all three coils, it is said to be balanced. In each phase, current has two paths to follow. For example, current flowing from L1 to the connection point at the top of the delta can flow down through one coil to L2, and down through another coil to L3. When current is balanced, the current in each line is equal to the square root of 3 times the current in each coil. The following example shows the line current calculation with the current in each coil equal to 29 amps.
IL = 3
x 29 = 1.732 x 29 = 50.2 Amps
50.2 Amps L1
29 Amps
29 Amps
50.2 Amps 29 Amps
L2
50.2 Amps
L3
Unbalanced Delta Current
When current is not the same in all three coils, it is unbalanced. The following diagram depicts an example of an unbalanced system.
43.6 Amps Coil A 30 Amps Coil B 20 Amps L1
26.5 Amps Coil C 10 Amps
L2
36.1 Amps
L3
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Wye Connections
The wye connection is also known as a star connection. Three coils are connected to form a “Y” shape. The wye transformer secondary has four leads, three phase leads and one neutral lead. The voltage across any phase (line-to-neutral) will always be less than the line-to-line voltage. The line-to-line voltage is the square root of 3 times the line-to-neutral voltage. The following example shows a wye transformer secondary with a line-to-neutral voltage is 277 volts and a line-to-line voltage of 480 volts.
ELine-to-Line = 3
x 277 = 1.732 x 277 = 480 Volts
L1
480 Volts 277 Volts L2
277 Volts
480 Volts
N
480 Volts
277 Volts
L3
Review 11 1. If the primary of a transformer has more turns than the secondary, it is a ____________ transformer.
2. If the primary of a transformer has fewer turns than the secondary, it is a ____________ transformer. 3. The secondary voltage of an iron-core transformer with 240 volts on the primary, 40 amps on the primary, and 20 amps on the secondary is ____________ volts. 4. A transformer with a 480 volt, 10 amp primary, and a 240 volt, 20 amp secondary will be rated for ____________ kVA. 5. A wye connected, three-phase transformer secondary, with 240 volts line-to-line will have ____________ volts line-to-neutral.
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Review Answers
Review 1 1) electron (-), proton (+), neutron (neutral); 2) free electron; 3) many; 4) a, c, e, g; 5) many, few. Review 2 1) electrons; 2) negative; 3) positive; 4) repel, attract; 5) voltage; 6) b; 7) a. Review 3 1) I = R ; 2) amps, volts, ohms; 3) 0.5; 4) 45; 5) 2 amps; 6) 6, 6; 7) 20, 80. Review 4 1) 5; 2) 5.45; 3) 3.33; 4) 12; 5) 6; 6) 2.4, 1.6. Review 5 1) 12, 22; 2) 40, 13.33. Review 6 1) power; 2) E, R, E2; 3) 36; 4) iron, north-south; 5) north, south; 6) left, thumb, lines of flux. Review 7 1) sine wave; 2) 120; 3) one; 4) -129.9 volts; 5) 106.05 volts RMS. Review 8 1) 10; 2) 2.5; 3) 2.5; 4) 25. Review 9 1) reactance; 2) impedance; 3) 3.14; 4) c; 5) b; 6) 318.5; 7) 11.18; 8) 6.4. Review 10 1) resistive; 2) inductive; 3) capacitive; 4) 1198, 0.1; 5) 1414.2, 0.0849; 6) 1200; 7) 2400. Review 11 1) step-down; 2) step-up; 3) 480; 4) 4.8; 5) 138.56.
E
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Final Exam
The final exam is intended to be a learning tool. The book may be used during the exam. A tear-out answer sheet is provided. After completing the test, mail the answer sheet in for grading. A grade of 70% or better is passing. Upon successful completion of the test a certificate will be issued. Questions 1. 2. 3. ________ is a good electrical insulator. a. b. copper aluminum c. d. silver rubber
A material with more protons than electrons has a ________. a. b. negative charge positive charge c. d. neutral charge no charge
In a simple electric circuit with a 12 volt supply, and a 24 W resistor, current is ________.
a. 2 amps c. 0.2 amps b. 5 amps d. 0.5 amps 4. The total resistance in a series circuit containing three, 10 W, resistors is ________. 5. a. 10 W b. 30 W c. 3.33 W d. 100 W
In a 12 volt series circuit where R1=10 W, R2=20 W, and R3=10 W, current flow is ________. a. b. 0.3 amps 0.5 amps c. d. 0.25 amps 3.33 amps
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6. 7 . 8. 9. 10. 11. 12. 13.
In a circuit containing three 30 W resistors in parallel, the total resistance is ________. a. b. 30 W 90 W c. 10 W d. 0.1 W
The rate at which work is done is called ________. a. b. energy power c. d. efficiency power factor
Power in a simple DC 12 volt, 4 amp series circuit is ________. a. b. 3 amps 3 watts c. d. 48 amps 48 watts
The instantaneous voltage at 150 degrees of an AC sine wave whose peak voltage is 480 volts is ________. a. b. 415.7 volts 480 volts c. d. 240 volts 0 volts
The effective voltage of an AC sine wave whose peak voltage is 480 volts is ________. a. b. 415.7 volts 339.4 volts c. d. 480 volts 679 volts
The time constant of a series circuit with a 10 mh inductor, and a 5 W resistor is ________. a. b. 2 milliseconds 2 seconds c. d. 2 microseconds 0.5 seconds
The total inductance of a series circuit containing three inductors with values of 10mh, 20 mh, and 40 mh is ________. a. b. 5.7 pF 5.7 mh c. d. 70 W 70 mh
The time constant for a series circuit with a 20 W resistor and a 4 mF capacitor is ________. a. b. 80 microseconds 80 milliseconds c. d. 5 microseconds 5 milliseconds
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14. 15. 16. 17 . 18. 19.
Total capacitance for a series circuit containing a 2 mF , 4 mF , and 8 mF capacitors is ________. a. 14 mF b. 0.875 mF c. 1.14 mF d. 4 mF
Total opposition to current flow in an AC circuit that contains both reactance and resistance is called ________. a. b. resistance reactance c. d. impedance capacitance
In a 60 hertz circuit containing 5 millihenrys of inductance, inductive reactance is ________. a. 1.884 W b. 1884 W c. d. 0.0005 W 0.05 W
In a purely inductive circuit ________. a. b. c. d. current leads voltage by 90 degrees current lags voltage by 90 degrees current and voltage are in phase current leads voltage by 30 degrees
In a series AC circuit with a 20 W resistor and a 10 W inductive reactance, impedance is ________. a. 30 W b. 10 W c. 14.1 W d. 22.4 W
A series AC circuit containing more capacitive reactance than inductive reactance is ________.
a. inductive c. capacitive b. resistive d. in phase 20. An iron-core transformer with a primary voltage of 120 volts, primary current of 10 amps, and secondary current of 5 amps has a secondary voltage of _______. a. b. 60 volts 240 volts c. d. 480 volts 30 volts
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