Inductance and Transformers
This chapter introduces the inductor and transformer with a discussion of their physical and electrical properties. Circuits containing series and parallel-connected inductors are discussed and analyzed. The property of mutual inductance is introduced.
At the end of this chapter, you should be able to:
1. Define inductance, state the basic unit of inductance, and be able to explain two concepts used to explain the counter EMF of an inductor.
2. Determine the effect of increasing or decreasing the permeability of the core material, the number of turns of conductor, and the cross-sectional area of the core or the length of the core on the value of an inductor.
3. State the relationship between the voltage across an inductor and the current through it.
4. Given a circuit of series-connected inductors or parallel-connected inductors, determine the total inductance of the circuit (no mutual inductance).
5. Given the value of two inductors connected in series or parallel aiding or opposing and their coefficient of coupling, determine the total inductance of their combination.
6. Given the turns ratio of a transformer, the input voltage, and the secondary load resistance, determine the secondary voltage, the primary and secondary current, the secondary frequency, and the primary and secondary power.
7. State the differences between step-up, step-down, and autotransformers, and their advantages and disadvantages.
8. Define inductive reactance and state the units in which it is measured.
9. Given a circuit of series-connected inductors, the applied voltage, and frequency of the applied voltage, determine the inductive reactance of each inductor, the total inductance, the total current, the voltage across each inductor, the reactive power of each inductor, and the total reactive power in the circuit.
10. Given a circuit of parallel-connected inductors, the applied voltage, and frequency of the applied voltage, determine the inductive reactance of each inductor, the total inductive reactance, the total inductance, each inductive branch current, the total current, the reactive power of each inductor, and the total reactive power in the circuit.
INTRODUCTION
In the last two chapters, discussion concerned the capacitor and how to analyze circuits composed of only capacitors or capacitors and resistors. This chapter is about the remaining passive circuit element — the inductor.
Figure 8.1 some types of inductors. Basically, any inductor is a coil of thin wire wrapped on a cylinder called the core. The core may be hollow, of laminated paper — an air core — or made of some type of iron — an iron core. Often an inductor is also called a choke or coil. The turns of wire of the inductor are electrically insulated from each other by a thin, non–conductive coating.
As shown in Figure 8.2 the schematic symbol used to represent the inductor resembles what it is — wire wrapped on a core. The inductor’s letter symbol is a capital L which represents linkages — flux linkages.
An inductor has magnetic properties. Therefore, a brief review of the subject of magnetism should help you understand better the electrical properties of an inductor.
ELECTROMAGNETIC PROPERTIES
Recall that in 1831, Michael Faraday showed that when a conductor connected in a closed circuit is moved through a magnetic field, an electron current flows as a result of a voltage induced in the conductor. (In this chapter, like in all other chapters in this book, current flow refers to electron current flow.) The amount of induced voltage is proportional to the rate of change of the magnetic field — the amount the magnetic flux changes in a specific time period:
As shown in Figures 8.3 and 8.4, it makes no difference if the conductor moves through a stationary magnetic field, or if the conductor is stationary and the magnetic field moves. As long as the lines of flux are caused to cut though the conductor, a voltage is induced across the conductor.
Remember that the direction of the resulting electron current and the polarity of induced voltage can be determined by the left–hand rule for generators, which is shown in Figure 8.5. If the thumb points in the direction of the conductor’s motion and the index finger points in the direction of the magnetic field, then the middle finger points in the direction the electron current is flowing.
If the conductor is not attached to an external circuit, the middle finger points to the negative end of the induced voltage (Figure 8.5).
Oersted’s Discovery
Earlier, in 1819, a Danish physicist, Hans Christian Oersted, had discovered that a magnetic field surrounds any conductor through which a current is flowing. He also found that the greater the current, the stronger the magnetic field, and the larger the area of the magnetic field around the conductor. As illustrated in Figure 8.6, when a small current flows in the conductor, only a small magnetic field is produced. But as current is increased, the magnetic field expands. Conversely, if the current is decreased, the magnetic field decreases in size.
All lines of flux are considered to have a definite direction. The direction of the lines of flux surrounding a conductor can be determined by what is known as the left–hand rule for conductors which is shown in Figure 8.7. It states that if the conductor is grasped with the left–hand with the thumb pointing in the direction of the electron current, the fingers fold in the direction of the lines of flux.
THE PROPERTY OF INDUCTANCE
The principle of inductance should be easier for you to understand if you realize that it is essentially the result of these two properties of electromagnetism: 1) That a voltage can be developed in a conductor which is in a changing magnetic field; 2) That a magnetic field is produced by current flowing in a conductor.
Inductance of a Wire
Consider a single piece of wire, as shown in Figure 8.8a, through which current is flowing in the direction indicated by the arrow. Suppose current is increasing in value from zero to some maximum, such as an alternating current would during a portion of its cycle. According to the left-hand rule, the changing magnetic field’s lines of flux will have a clockwise direction as indicated by the arrows in Figure 8.8b, and will expand as the current increases.
Imagine that the magnetic field starts at the center of the conductor and expands out as the current increases as shown in Figure 8.8c. As the magnetic field first develops and expands, it cuts through the body of the conductor itself and induces a voltage that opposes the voltage producing the initial current. In other words, the induced voltage produces a circuit current opposite in direction to the initial current.
This opposing current is called a counter-induced current. This inductive action occurs in any piece of wire through which an electrical current flows.
So there are two properties working together. The initial current produces a changing magnetic field which cuts a conductor and develops a voltage to produce a current as shown in Figure 8.9. This current flows in the circuit in a direction opposite to the initial current, and it opposes the changes of the initial current.
Figure 8.10 illustrates this action in a circuit. When a current that changes in value passes through a conductor, it will induce a voltage that opposes the current change as shown in Figure 8.10a. The voltage induced is opposite in polarity to the voltage that produced the intial voltage. If the initial voltage is increasing to produce an increasing initial current, then the induced voltage is of a polarity to produce a current that opposes the increase of the initial current. If the initial circuit voltage now decreases to cause the initial current to decrease as shown in Figure 8.10b, the induced voltage is of a polarity to oppose the decrease of the initial current. An aiding current is produced which keeps the initial current flowing and opposes its change.
North–South Magnetic Field of a Coil
What happens if this wire is wound into a coil as shown in Figure 8.11? In a coil the lines of magnetic flux emanating from one turn of the coil not only cut through the wire from which they originate, but they also cut through other turns of the coil adjacent to it. The more turns that are present, the more the induced voltage and the more the opposition to any changes in current within the coil and within the circuit containing the coil. When a current passes through a coil of wire, the individual magnetic fields set up by each turn of wire in the coil add to form a magnetic field. Therefore, the inductance of a coil is greater than the inductance of an identical length of wire not coiled. The magnetic field formed by the coil is similar to that of a bar magnet.
The orientation of the magnetic field around the coil can be determined by using the left–hand rule for coils. The rule states that if the fingers of the left hand are wrapped around the coil in the direction of current flow as shown in Figure 8.12, thumb points toward the north pole end of the coil.
Inductance of a Coil
In a coil, the inductive action is multiplied because the lines of magnetic flux emanating from one section of the coil not only cut through the wire from which they originate, but they also cut through the sections of the coil adjacent to it. This is illustrated in Figure 8.13a and 8.13b The more wire that is present, the more the opposition will be to any changes in current within that wire. Thus, the number of turns of wire determines a coil’s inductance value. The magnetic field of a coil is shown in Figure 8.13c. It is like the magnetic field of a bar magnet (Figure 8.13d).
Counter EMF
If an inductor is placed in a circuit as shown in Figure 8.14 which has a changing voltage, EA, applied to it, a current which changes in magnitude will attempt to flow in the circuit. The property of inductance that the inductor exhibits, however, will oppose the change of current through it. Thus, a voltage will be set up across the inductor due to the inductive action. This voltage has been given a special name It is called the counter EMF, abbreviated CEMF. This is the voltage that appears across inductors in ac circuits. It is due to the property called self–inductance.
Recall that earlier it was stated that a coil must be composed of turns of wire in which all turns are insulated one from the other. It should be clear now why this is true. If the turns of wire are not electrically separate, then the coil would not act as an inductor, but as a conductor.
Units of Inductance
The amount of inductance in a coil of wire is measured in henrys or millihenrys (thousandths of a henry). Millihenrys is abbreviated mH.
Physical Properties that Determine Inductance of a Coil
The inductance of a coil can be determined by this equation:
This equation is valid for a single–layer coil only, but it can be used to help you understand the physical properties of any type of coil.
In the equation, L represents the magnetic flux linkages that connect one part of a coil to the next part, causing the property of inductance. The permeability of the core material is μ, the ability of a material to conduct magnetic lines of force, also called magnetic flux. Together, μ and μorepresent a constant.
From this equation and, as shown in Figure 8.15, you can see that if the number of turns of wire is increased, the inductance of the coil will increase by the square of the number of turns. As shown in Figure 8.16, a coil with a large cross-sectional area will have a greater value of inductance than one of a smaller cross–sectional area. As shown in Figure 8.17, a long
coil will have a smaller value of inductance than a shorter one. Remember that when determining the effect of varying one specific factor of a coil, that all other factors must be held constant. From equation 8–2, the value of a manufactured inductor depends basically on these four factors: 1) the number of turns of wire used, 2) the cross–sectional area of the coil, 3) the overall length of the coil, and 4) the permeability of the core material.
Working an example using equation 8–2 should help you learn to use it to determine the inductance of a coil. In this example, as shown in Figure 8.18, a coil has 400 turns of wire wound on a round core 0.013 meter (0.5 inch) in diameter, a length of 0.076 meter (3 inches), and uses iron as a core material which has a permeability of 1,000. What is the inductance of the coil?
First of all, the cross sectional area of the coil is determined.
Variable Inductors
The fact that the permeability of the core material determines the value of a coil is put to use in manufacturing variable inductors. Iron is the most common core material used in the manufacture of inductors. Iron has a much higher permeability than air. Variable inductors are made by constructing an inductor with an iron core that can be inserted various lengths into the inductor.
Figure 8.19 shows a typical variable inductor. The portion of the coil with the iron core has a much higher value of inductance than does the section with the air core. The two inductances add together to determine the overall inductance of the coil. As the iron core, sometimes called a slug, is screwed into the coil, the inductance increases. As the iron core is extracted, the inductance of the coil decreases. Thus, by varying the amount of iron core within the coil, a variable inductor has been created which can be “tuned” to a specific inductance. This type of variable inductor is called a slug–tuned or permeability–tuned inductor.
INDUCTANCE IN A CIRCUIT
Effect of an Inductor on a Changing Current in a Circuit
When considering the electrical action of an inductor in a circuit it is helpful to compare its action to that of a resistor. Recall that in a circuit containing only resistors, as shown in Figure 8.20, that when a voltage is applied by closing the switch, the full value of current flows almost immediately, and it continues to flow through the resistors as long as the voltage is applied.
If, however, the resistor R1 is replaced with an inductor, as shown is Figure 8.21, when the switch is closed the full value of current does not flow immediately. Instead, when the current first begins to increase in value from zero, the inductive action sets up a counter EMF that opposes the current change. When the switch is first closed, in fact, the rapid change in current that starts to flow causes a rapid change in magnetic flux that produces a high counter EMF. Therefore, very little current is allowed to flow initially. At this point in time, the inductance appears as an open circuit as illustrated in Figure 8.22.
This can be compared to the physical property of inertia. Initially, it is difficult to get the current moving, but once moving, and if forces are not changed, it flows smoothly.
When the self–inductance has stopped the initial flow of current, the self–inductance of the coil ceases since it depends on a change of current. The current again attempts to increase. As it increases slightly, the self–inductance again hinders the flow of current in the circuit. Then, when the self–inductance of the inductor has stopped the increase in current, the self–inductance ceases and the current attempts to increase again. This back–and–forth action continues smoothly until the current in the circuit reaches its maximum value. The maximum current in this circuit is determined by the value of R1 and the internal resistance of the coil.
Internal (Winding) Resistance of a Coil
Every coil has some internal resistance bacause of the length of the wire used. The internal resistance of coils range from fractions of an ohm to several hundred ohms depending on the length and size (gauge) of wire used. You can measure this internal resistance easily with an ohmmeter.
Exponential Change of Current
Figure 8.23 shows graphically the action of a resistive–inductive (RL) circuit. At first, time t1, the voltage across the inductor is at maximum since the inductor appears as an open circuit initially. Current is zero. All of the applied voltage appears across the inductor as counter EMF. Gradually, the current begins to increase until it reaches its maximum value. As the current increases, the counter EMF across the inductor decreases.
The increase of current and decrease of voltage is said to be exponential. However, if one examined the circuit action in very small periods of time, it would appear as small steps where current increases, counter EMF opposes it, current increases again, counter EMF opposes it again, and this action continues until the current attains its final value. As shown in Figure 8.24, it is, as stated, a step function. But the steps are infinitesimally small, and to us the action appears as a smooth exponential rise in current.
When the current has reached maximum value and ceases to change any further, at time t2, the counter EMF across the inductor is zero volts since there is no more change in current. The fact that the current stopped changing here is significant because there must be a change in current value to cause the magnetic flux lines to expand or collapse so that they cut through the coil to induce the counter EMF.
Rate–of–Change Equation
An equation that mathematically describes the current–voltage relationship for an inductor is
ELrepresents self–induced EMF. Recall that a Δ symbol designates a change in a quantity.
Thus, the value 
is the change in current measured in amperes through the inductor divided by the change in time measured in seconds in which the change in current occurs. 
is the rate of change of current. If this equation is substituted as the voltage across the inductor in the dc circuit of Figure 8.21, and a dc voltage is applied when the switch is closed, then the solution for the circuit current will be the exponential curve shown in Figure 8.21.
But what will happen in the circuit of Figure 8.21 if the dc voltage is replaced with an ac voltage source? Figure 8.25 shows the circuit with an ac voltage source. In this circuit, the sinusoidal voltage is continually changing in magnitude and polarity, causing the magnetic field of the coil to continously expand and collapse.
If the phase relationship of the voltage across the inductor, which is the counter EMF, is compared with the current in the inductor, it will be found that they are 90 degrees out of phase as shown in Figure 8.26. More specifically, the voltage leads the current by 90 degrees.
The voltage across the inductor, EL, in the circuit of Figure 8.25 can be calculated using equation 8–3. But before this is done one thing should be noted. The rate–of–change of current 
, may be abbreviated ROC. Since any voltage that appears across the inductor is equal to the inductance times the rate of change of current through the inductor, equation 8–3 can be rewritten:
Discussion of an example will show how this equation describes the current–voltage relationship for an inductor. The sine wave of Figure 8.27 represents the current passing through the inductor. Since EL= L(rate of change of current), when the rate of change of current of the sinusoidal waveform is zero at its peak, the value of the voltage is zero. When the rate of change of current of the sinusoidal waveform is maximum as it crosses the zero axis either increasing positively or increasing negatively, the value of the voltage is a positive or negative maximum. Since the current changes sinusoidally, the rate–of–change of current changes sinusoidally and so does the voltage. By connecting these points with a smooth sinusoidal waveform as shown in Figure 8.27, the voltage waveform can be plotted. It can be seen that, as stated previously, the voltage leads the current by 90 degrees as shown in Figure 8.28. Therefore, it should give you a clear idea of the relationship of a current through and voltage across an inductance.
SERIES AND PARALLEL INDUCTORS
When inductors are connected in series or in parallel, it may be necessary to know their total equivalent inductance.
Series Inductors
To calculate the total inductance of series–connected inductors, simply add their individual inductance values as you would add up series resistance. Therefore,
For example, the circuit of Figure 8.29 contains two inductors, series–connected. Their total inductance value is simply their sum:
The effects of the inductors are additive because in a series circuit the current must flow through both inductors. Therefore, both inductors will respond to any changes in this current, and the effects are additive.
Parallel Inductors
When inductors are parallel to one another the total inductance is calculated using one of two methods.
If three or more inductances are in parallel then the total inductance is calculated by this reciprocal equation.
For example, suppose three inductors of 3 mH, 6 mH, and 12 mH are in parallel. The total inductance is
If two inductors are in parallel as shown in Figure 8.30, then total inductance is calculated by using the reciprocal addition process that was used to add parallel resistances. Like the parallel resistance equation for two resistors, the total inductance for two series inductors is calculated by using a product–over–sum equation.
If the two inductors are 3 mH and 6 mH, then, the total inductance of their combination is
Note that the combination of parallel inductances, LT, is always less than the smallest individual inductance value. This result is similar to that obtained for a combination of two parallel resistances.
It is interesting to note that the solution for total inductance could be accomplished by taking the inductances “two at a time” and use the product over sum equation. For example, a total inductance of 2 mH, as previously calculated, can be used for the combination of 3 mH and 6 mH in parallel. This resultant can then be used in a product over sum equation with the 12 mH inductor to determine the final total inductance. Here is the calculation:
MUTUAL INDUCTANCE
Although the total inductance of coils connected in series or parallel can be determined using the same technique used for determining the total resistance of resistors connected in series or parallel, inductors have a property not at all similar to resistive circuits. This property is called mutual inductance. When the current in an inductor increases or decreases, the magnetic flux field surrounding that inductor changes. The varying flux can cut across any other coil which may be nearby, causing induced voltage to exist in it.
For example, series–connected coils L1and L2in the circuit of Figure 8.31 are close to one another. An ac voltage source is connected to L1and L2. The current through L1produced by the ac source causes a magnetic field to exist around L1which also cuts through the windings of L2, inducing a voltage in its windings. Thus, a voltage exists across L2as a result of the magnetic field of L1cutting through its windings. At the same time L1is producing this voltage across L2, the current flowing through L2produces a field around L2that induces a voltage in L1. This effect is known as mutual inductance.
Mutual inductance is measured in henrys and is designated by the symbol LM. The amount of mutual inductance between two coils, L1and L2, can be calculated by this equation:
Coefficient of Coupling
The coefficient of coupling is the fraction of total magnetic flux lines produced by both coils that is common to both coils. Stated as an equation:
For example, in a circuit there are two coils, L1and L2, which produce 10,000 flux lines, all of which link both coils. Therefore, using equation 8–8 the coefficient of coupling can be calculated:
In this case, there is unity coefficient of coupling and k = 1. However, if L1and L2produce 10,000 lines of flux and only 5,000 are common, the coefficient of coupling is
Therefore, in this case, the coefficient of coupling is 0.5. There are no units for k since it is a ratio of two amounts of magnetic flux.
A low value of k is called “loose coupling”; a high value of k is called “tight coupling.” Coils wound on a common iron core can be considered to have a coefficient of coupling of one or “unity coupling.” Figure 8.32 summarizes the types of coupling and range of coefficients of coupling.
The value of k can be varied by factors relative to the coils. For example, as shown in Figure 8.33, k becomes larger (with a maximum of one) as the coils are placed closer together; are placed parallel to one another, as opposed to perpendicular; or are both wound on a common iron core. Conversely, k becomes smaller as the coils are placed farther apart; are placed perpendicular to one another; or are both wound on separate iron cores.
Calculating Mutual Inductance
Mutual inductance is calculated using equation 8–7Figure 8.34 shows a circuit with two inductors, L1and L2. L1equals 20 henrys and L2equals 5 henrys. The coefficient of coupling is 0.4. The coils’ mutual inductance is calculated as follows:
Series–Connected Inductors with Mutual Inductance
When two series–connected inductors are separated by a sufficient distance, are mounted at right angles to one another, or are well–shielded, their mutual inductance is negligible. Their equivalent total inductance is the sum of the inductance values as described previously. However, if series–connected inductors are situated so that their magnetic fields interact, the mutual inductance between the two coils must be considered. Their total inductance, then, is determined by this equation:
When the inductors are situated such that their magnetic fields aid each other the plus sign is used.
When the fields oppose, the minus sign is used.
The 2 is used because magnetic fields of L1and L2interact and affect each other.
For example, a 10–henry and 20–henry coil are connected in series as shown in Figure 8.35 such that their magnetic fields aid one another, and they have a mutual inductance of 5 henrys. Their total inductance is calculated:
Figure 8.35 Example Circuit for Calculating Total Inductance of Series–Aiding and Series–Opposing Inductors
If the same two coils are oriented such that their magnetic fields oppose one another, their total inductance is calculated:
Parallel–Connected Inductors with Mutual Inductance
When two coils are parallel–connected without any mutual inductance, their equivalent total inductance is found by the reciprocal addition or product–over–sum method discussed earlier. However, if their magnetic fields interact, their equivalent total inductance is found by one of the following equations:
Parallel–aiding connection — If the two coils are oriented such that their magnetic fields aid each other.
Parallel–opposing connection — If the two coils are oriented such that their magnetic fields oppose each other.
Using the same 10 henry and 20 henry coils with a mutual inductance of 5 henrys due to an aiding field, but connected in parallel, the total inductance using equation 8–12 is
TRANSFORMERS
A device in which the property of mutual inductance is put to practical use is the transformer. A typical transformer is shown in Figure 8.36. A typical standard transformer consists of two separate coils, wound on a common iron core as shown in the schematic of Figure 8.37 and considered to have a coefficient of coupling of one. One coil is called the primary; the other is called the secondary. As a result of mutual inductance, a changing voltage across the primary will induce a changing voltage in the secondary. Thus, if the primary winding is connected to an ac source and the secondary to a load resistor, the transformer is able to transfer power from the primary to the secondary to the load resistance as illustrated in Figure 8.38. By having more or fewer turns in the secondary as compared to the primary, the primary voltage may be either stepped–up or stepped–down to provide the necessary operating voltage for the load.
Turns Ratio versus Voltage
Recall that if a coil has a larger number of turns, a larger voltage is induced across the coil. With a smaller number of turns the voltage is less. Therefore it is easy to see that by having more or fewer turns in the secondary as compared to the primary, as shown in Figures 8.39 and 8.40, the voltage may either be stepped up or stepped down to provide the necessary operating voltage for the load.
The ratio of the number of turns in a transformer secondary winding to the number of turns in its primary winding is called the turns ratio of a transformer. The equation for turns ratio is:
In the transformer schematic shown in Figure 8.41, the number of turns in its primary is 10 and the number of secondary turns is 5. Using equation 8–14, the turns ratio of the transformer can be calculated.
Transformers have a unity coefficient of coupling. Therefore, the voltage induced in each turn of the secondary winding (EiS) is the same as the voltage self–induced (EiP) in each turn of the primary, as shown in Figure 8.42. The voltage self–induced in each turn of the primary equals the voltage applied to the primary divided by the number of turns in the primary. This can be written:
Figure 8.43 shows a schematic of a transformer in which there are 8 turns in the primary and 8 volts ac is applied to it. Using equation 8–15, the voltage self–induced in each primary turn can be calculated.
In this example, one volt is induced in each turn of the primary.
If each turn of the secondary has the same voltage induced in it, then the secondary voltage is equal to the number of secondary turns times the induced voltage. This can be written
The transformer shown in Figure 8.43 has 4 turns in its secondary. Using equation 8–16, the secondary voltage can be calculated.
The transformer’s secondary voltage is 4 volts — 4 turns times 1 volt per turn.
In another example, shown in Figure 8–44, there are 1,000 turns in the primary winding of the transformer and there are 10,000 turns in its secondary winding. Thus, the turns ratio is
Therefore, the secondary voltage would always be 10 times greater than the primary voltage. If the primary voltage is 10 volts ac, then the secondary voltage will be
Transformer secondary current is a function of secondary voltage and load resistance. If a 1 kilohm load is placed across the secondary as shown in Figure 8–45, then the secondary current, by Ohm’s law, will be
The secondary current is 100 mA. The transformer secondary acts as an ac voltage source to the load.
Primary–to–Secondary Current Relationship
Ideal transformers, with coefficient of coupling considered to be one, and with no real power consumed in the windings or the core can be considered to have no loss, as shown in Figure 8.46. Therefore, the power in the primary is considered to be the same as the power in the secondary, PP= PS. Since P = EI,
Note that the current relationship is the inverse of the voltage relationship. Thus, if the voltage is stepped up in a transformer by a factor of 10, the current must have been stepped down the same factor. This may be stated another way using equations 8–17 and 8–18.
Thus, in the example shown in Figure 8.45, if EPis 10 volts, ESis 100 volts, and if IS, the secondary current, is 100 milliamperes, the primary current, IP, may be calculated by:
The primary current in the transformer is one ampere.
Performing the following calculations it can be determined that both the primary and secondary power are equal; both are 10 watts.
Power Ratings of Transformers
It is useful to note that many manufacturers rate transformers in volt-amperes (VA) rather than in units of watts. For example, a transformer that provides a secondary ac output voltage of 24 V with a maximum secondary current rating of 1 ampere would be considered to have a 24–VA capacity rather than a 24–watt transformer.
In the previous example, the primary and secondary power values might be expressed more appropriately as 10 VA rather than as 10 watts:
As an industrial example, if a transformer with a nominal secondary output of 240 VAC is rated as a 50-kVA transformer by a utility company, it means the transformer maximum secondary current can be calculated using the power equation as follows.
For the remainder of this chapter, however, we will use watts (W) for units of power of the transformer instead of volt-amperes (VA) to avoid confusion. It is important to realize that the two are interchangeable when referring to transformer power.
The transformer, then, either steps up or steps down the voltage and current, but conserves power from the primary to the secondary.
However, transformers do not affect the frequency of the ac voltage they act upon. If the frequency of the primary voltage and current is 60 hertz, then the secondary voltage and current will have a 60 hertz frequency.
Recall that a transformer will not operate with a dc voltage. That is because dc voltage is non-changing and cannot produce an expanding or collapsing magnetic field to cut the secondary windings to induce a secondary voltage.
Efficiency
Previously it was stated that the power in the transformer secondary is considered to be the same as that in the primary. This, however, is not precisely true with an actual operating transformer. When operating, the transformer’s secondary power is usually slightly lower than that in the primary. As shown in Figure 8.47, this is caused by core and winding losses such as hysteresis and eddy currents in the transformer. In fact, these power losses are evidenced by the heatgiven off by a transformer during operation.
How well a transformer delivers power from the primary to the secondary circuit is called its efficiency.Efficiency is defined as the ratio of the power in the secondary circuit divided by the power in the primary circuit. Because efficiency is usually stated as a percent, it can be written
To determine the actual percent efficiency of a transformer, primary voltage and current and secondary voltage and current are measured. Then the primary and secondary power are calculated, and using equation 8–20, the percent efficiency is calculated. Typical transformer efficencies are about 80 percent or more (but less than 100 percent).
Isolation
Another feature of the transformer is that it provides electrical isolation between primary
and secondary windings, as shown in Figure 8.48. That is, there is no electrical connection between the primary and secondary windings of the transformer. (Only the electromagnetic field links the secondary with the primary.) When a circuit is connected directly to the power source as shown in Figure 8.49a, no electrical isolation exists between the circuit and power source, which can present a shock hazard. But any circuit connected to the secondary of a transformer, as shown in Figure 8.49b, is electrically isolated from the primary, and thus, the power source. This prevents a shock hazard from existing regardless of the connection of the primary to the electrical ac power line. In fact, transformers designed to simply provide isolation with no voltagechange from primary to secondary (i.e., a 1:1 turns ratio) are referred to as “isolation transformers.” A schematic of an isolation transformer is shown in Figure 8.50.
Autotransformers
Transformers do exist, however, which do not provide isolation. These use a single winding and are referred to as autotransformers. As shown in Figure 8.51, an autotransformer utilizes a single tapped winding with one lead common to both input and output sides of the transformer. Therefore, there is no isolation. The voltages present across the primary and secondary portions of the transformer are proportional to the number of turns across which they appear. These are calculated the same as in a transformer with isolated winding by using equation 8–17.
The obvious advantage of this type of transformer is that only one winding is needed. But a less obvious advantage is that an autotransformer provides a more constant output voltage under varying load conditions than does the two-winding type of transformer. This is because the primary and
secondary currents are 180 degrees out of phase and tend to cancel in the portion of the winding that is common to both primary and secondary.
Variable-Output Transformers
Some manufacturers produce a type of autotransformer that has a variable output voltage. As shown in Figure 8.52, this is accomplished by making the secondary tap a wiper-type of contact (much like a wire-wound variable resistor). By varying the position of the wiper contact, various output voltages are obtainable. Of course, the same effect could also be produced using a variable tap on the secondary of a two-winding transformer as shown in Figure 8.53.
Multiple-Secondary Transformers
Transformers are also produced which have multiple-secondary and center-tapped secondary windings in order to provide for circuits requiring several different voltage levels. A schematic for a typical multiple-secondary transformer is shown in Figure 8.54.
Transformer Lead Color Code
Transformer leads are usually color coded using a standarized EIA wire color coding technique. A chart showing the standard EIA color code is provided in the appendix. Not all manufacturers use this particular color code so there will be some variation.
Transformer Specifications
Manufacturers provide specifications for transformers. The specifications enable a user to select a transformer that best meets the requirement of the application. Transformer specifications usually include primary voltage and frequency, secondary voltage(s), impedance, dc winding resistances, and current capabilities. For example, the power transformer of Figure 8.54 has the following specifications:
High-voltage secondary: 240 V-0-240 V (center-tapped) 150 mA
Low-voltage secondary:6.3 V, 2 A
Low-voltage secondary:5 V, 3 A
Power transformers are multiple secondary winding transformers with both high and low voltage secondaries. Typical power, audio, and filament transformers are shown in Figure 8.55 Power transformers originally were developed for use with vacuum tube circuits in which high voltage for power supply levels and low voltage for vacuum tube filaments (heaters) were needed. The primary ratings specify the voltage and frequency at which the transformer is designed to be operated. The secondary ratings specify the voltages available from the various secondary windings as well as the maximum current which the secondaries can supply.
Audio transformers are designed for input/output audio applications and are rated according to their primary and secondary impedances, power capabilities (wattage), and turns ratio. They have only a single secondary winding.
Filament transformers are single secondary, low voltage, high current (several amperes, typically) transformers rated according to their primary voltage and frequency, secondary output voltage, and maximum output current capabilities.
INDUCTIVE REACTANCE
Now that inductance, self-inductance, and transformer action have been discussed, the next step is a discussion of the effect of an inductor in an ac circuit.
Inductance is measured and specified in units of henrys. An inductor’s effect in a circuit depends on the inductance and is expressed in a quantity called inductive reactance. Inductive reactance is a quantity that represents the opposition that a given inductance presents to an ac current in a circuit, such as is shown in Figure 8.56. Like capacitive reactance, it is measured in ohms and depends upon the frequency of the applied ac voltage and the value of the inductor. Inductive reactance can be expressed as follows:
The constant of 2π comes from the number of radians in one cycle of a sinusoidal ac waveform. Because of this, this equation is valid only for calculating the inductive reactance of an inductor with sinusoidal alternating current applied.
Derivation of the Inductive Reactance Equation
This derivation is similar to the derivation of the capacitive reactance equation (6–5) previously illustrated in Chapter 6.
A sine wave may be evaluated for its average rate of change of current as shown in Figure 8–57. If the peak value of a current sine wave is Ipk, then the average rate of change of the entire sine wave may be determined by evaluating its change in voltage divided by its change in time (¼ of a second) during only the first 90 degrees. In equation 8–21b, the average rate of change of a sine wave is given as the change in current divided by the change in time. The change in current from 0 to 90 degrees is ¼ of the period.
Because the reciprocal of the frequency is equal to the period, the equation may be rewritten as
The right side of equation 8–21c can be rearranged to obtain
As shown in the first few pages of Chapter 2, the average value of a sinusoidal voltage or current during its first half or quarter cycle may be written as 0.637 of its peak value.
Recall that equation 8–3 describes the behavior of an inductor.
Equation 8–3 could also be written in terms of average voltage and average rate of change of current as
Substituting the average current (equation 8–21e) and the average rate of change of voltage (equation 8–21d) expressions into equation 8–21f yields
Rearranging equation 8–21g, to solve for Epkresults in:
Because inductive reactance is the opposition of an inductor to the flow of alternating current, it may be stated in an Ohm’s law format as
The values of inductive voltage and current may be specified in peak terms so that
Substituting the expression for Epk from equation 8–21h into equation 8–21i produces the following result:
When Ipk is factored out, the result is equation 8–21a.
Figure 8.58 shows a simple inductive circuit. The inductor’s value is 10 millihenrys. Applied frequency is 5 kilohertz. Using equation 8–21a, inductive reactance, XL, is calculated:
The value of current in the circuit can then be predicted by dividing the applied voltage by the inductive reactance:
Effect of Changing Frequency or Inductance on Inductive Reactance
Note from equation 8–21a that if either the frequency or the inductance is increased the inductive reactance increases. Figure 8.59 shows graphically how a change in either the frequency or inductance changes the inductive reactance, XL. Note that the inductive reactance increases linearly with frequency and inductance. As the frequency or inductance increases, the inductor’s opposition to the flow of current increases.
These plots of inductive reactance versus frequency and inductive reactance versus inductance shown in Figures 8.60 and 8.61 will be examined more closely to help you understand these relationships more clearly.
Figure 8.60 shows the inductive reactance versus frequency for an inductance of 10 millihenrys. It can be seen that as frequency increases so does the inductive reactance. For example, at a frequency of 159 hertz, the inductive reactance is 10 ohms. However, at a frequency of 1,590 hertz the inductive reactance is now 100 ohms. Inductive reactance is directly proportional to frequency.
In Figure 8.61, which plots inductive reactance versus inductance at a frequency of 159 hertz, it can be seen that as inductance increases so does the inductive reactance. For example, with an inductance of 0.01 henrys (10 millihenrys), inductive reactance is 10 ohms. However, if the inductance is increased to 1 henry, the inductive reactance is now 1 kilohm. Inductive reactance also is directly proportional to inductance.
The basic equation for inductive reactance may be rewritten in two other forms:
Equation 8–22 can be used to determine the frequency at which a given inductance will produce a certain reactance. Equation 8–23 can be used to determine the inductance that will have a given reactance at a specified frequency. For example, equation 8–22 can be used to determine the frequency at which an 8.5 henry inductor will have an inductive reactance of 5 kilohms.
Equation 8–23 can be used to determine the value of inductance needed to produce an inductive reactance of 10 kilohms at a frequency of 300 kilohertz.
Compatibility of Reactance Equation with Rate-of-Change Equation
It can be shown that the inductive reactance equation is compatible with the previous discussion of inductance where the voltage across an inductance is equal to L times a rate-of-change-of-current. Figure 8.62 shows a simple series ac circuit containing a resistance and inductance. The inductive reactance is determined using equation 8–21a. If in this series circuit the frequency of the applied voltage increases, then inductive reactance increases and the voltage drop across the inductor increases because the inductive reactance is a larger portion of the total impedance in the circuit.
When the voltage across the inductor is expressed using the rate-of-change-of-current equation as shown in Figure 8.63, the voltage increases as the rate-of-change of current increases. The rate-of-change of current increases as frequency increases. Therefore, the voltage across the inductance increases as frequency increases.
Thus, both equations prove that in an inductive circuit, as frequency increases, the voltage across the inductor increases. This analysis also proves that the inductive reactance and the rate-of-change equations state the same basic principle.
Like a capacitor, an inductor can be thought of as a variable resistor whose opposition to current flow (its inductive reactance) is controlled by the applied frequency. This is shown in Figure 8.64.
SERIES AND PARALLEL INDUCTORS IN AC CIRCUITS
Thus far, only circuits with a single inductor have been discussed. More than one inductor can be used in a circuit. Therefore, an understanding of series and parallel combinations of inductors with applied voltage and frequency is important.
Series Inductors
Figure 8.65 shows a circuit in which two inductors, one 40 millihenrys, and the other 20 millihenrys, are connected in series. The applied voltage is 40 volts at 10 kilohertz.
First the inductive reactances of the two inductors will be calculated using the basic inductive reactance equation 8–21a.
In this circuit the inductive reactance of L1is twice the inductive reactance of L2. Thus, if the inductance increases by a factor of two, the inductive reactance increases by a factor of two.
Now that the individual reactances have been calculated, the reactance ohms are treated like resistive ohms, and series reactances are added.
Therefore, the total reactance for the circuit is 3,768 ohms. This total reactance is the total opposition that the circuit presents to alternating current flow at the applied frequency.
An alternate method of finding total reactance is to first calculate total inductance.
Assuming no mutual inductance, the total inductance is 60 millihenrys. Now the total inductive reactance can be calculated using the basic inductive reactance equation:
This is the same value of total inductive reactance calculated using the other method.
Because the total inductive reactance has been determined, it can be used to calculate the total current in the circuit. The total current, by Ohm’s law, is equal to the applied voltage divided by the total inductive reactance.
Now the rules of a series circuit can be used to solve the voltage drop across L1 and L2Since it is a series circuit, the total current is the same throughout the circuit.
In a purely inductive circuit, the voltage drops add to the total applied voltage as in a series resistive circuit.
Parallel Inductors
Figure 8.66 shows a circuit in which there is a parallel combination of two inductors. The voltage, frequency, and inductance are the same as those used in the previous discussion of series inductors. Keeping these the same should help you better understand the differences between series inductor and parallel inductor circuits. Since inductance of the coils is the same and applied frequency is the same, the inductive reactance of each of the inductors will be the same:
The branch current for each branch can be determined by dividing the voltage across the branch by the opposition in the branch.
The total current is simply the sum of the branch currents.
To find the total inductive reactance, simply divide the applied voltage by the total current.
There are two alternate methods that can be used to find the total inductive reactance. The first method is to treat the inductive reactances as if they were parallel resistances and use the product-over-sum method of solution.
The other method is to first determine the total inductance of the circuit, then calculate the total inductive reactance using the total inductance. Using this method,
Power in Inductive Circuits
In inductive circuits, power can be determined using calculations similar to the calculations used to determine power in resistive or capacitive circuits. Remember that in resistive circuits, electrical energy is converted into heat energy. However, in inductive circuits, the electrical energy is stored temporarily in the magnetic field around the inductor. This is similar to the storage of energy in an electrostatic field in a capacitor. Thus, inductors store electromagnetic energy temporarily; capacitors store electrostatic energy temporarily.
Recall that the basic power equation is
This power equation can be used to calculate the power in inductive circuits. The power in inductive circuits is measured in units called VAR as it is in capacitive circuits. VAR stands for Volts Amperes Reactive. The power in an inductive circuit is called reactive power since the opposition to current in the circuit is strictly reactive.
Power Calculations for Series-Inductive Circuits
In the series-inductive circuit discussed previously, and repeated in Figure 8.67, the voltage drop across L1is 26.67 volts. The current through L1and L2is 10.6 milliamperes. Reactive power for L1is calculated:
Reactive power for L2is calculated:
By adding these two values of reactive power, the total reactive power of the circuit can be determined. That is,
Total reactive power of the circuit could also have been calculated by multiplying the total applied voltage times the total current. In this case,
Power Calculations for Parallel-Inductive Circuit
Similar calculations can be performed to obtain the reactive power for the parallel inductive circuit shown in Figure 8.66. Recall in that circuit IL1= 15.9 milliamperes and IL2= 31.8 milliamperes. Remember the voltage across each branch is the applied voltage. The reactive power of L1is:
Also, the total reactive power in a parallel circuit equals the total applied voltage times the total current.
SUMMARY
This chapter has been an introduction to the inductor, how it is structured, its schematic symbol, its typical units of inductance, and how it functions in typical circuits. The phase relationships of the voltage and current in an inductive circuit were discussed. Mutual inductance and how it is put to use in transformers, and how to make voltage and current calculations for transformer circuits were also discussed. Series and parallel inductive problems were solved, and reactive power calculations were described.
1. Describe the action of an inductor in a circuit.
Solution:A magnetic field surrounds any wire carrying current. As current increases through a wire, the magnetic field expands through the wire inducing a counter current which opposes the increase in the initial current. As current decreases in a wire, the magnetic field collapses through the wire inducing current in the same direction and aiding the current which is trying to decrease, thus opposing the decrease of current. When the wire is wound into a coil, the magnetic field produced by each turn of wire in the coil interacts with adjacent turns increasing this inductive effect. This coil of wire is called an inductor. If it is placed in a circuit such that a changing current passes through it, it will oppose the change (increase or decrease) of current.
3. If the current through an 8 millihenry-coil is changing at the rate of 10 milliamperes every 5 seconds, determine the rate of change of the current in amperes per second, and the voltage (CEMF) induced across the coil.
4. If two coils are connected in series as shown, determine their total inductance with no mutual inductance, and their mutual inductance and total inductance considering mutual inductance (aiding and opposing) if k = 0.4.
a. Given the circuit shown solve for the total inductance of the parallel-connected inductors if there is no mutual inductance.
b. Determine their mutual inductance and total inductance (aiding and opposing) if mutual inductance exists with a coefficient of 0.2.
6. If the primary voltage applied to a transformer is 120 VAC and the secondary voltage output is 480 VAC, determine the turns ratio for the transformer and state whether it is a step-up or step-down transformer.
7. Given the transformer with turns-ratio and load-resistance specified, determine the following values: Esec, Isec, Ipri, Ppri, and Psec. (Assume 100 percent efficiency.)
8. If the primary voltage is 120 VAC with a primary current of 10 mA and the secondary voltage is 12.6 VAC with a secondary current of 85 millamperes, determine the percent efficiency of this transformer. Explain the loss of power between primary and secondary.
b. The power loss (10.7 percent of the primary power) between primary and secondary is due to eddy currents, hysteresis and winding resistance heat loss.
9. Calculate the inductive reactance of the inductors at these specified frequencies:
a. 10 millihenry coil operated at a frequency of 5 kilohertz:
b. An 8.5 henry coil operated at a frequency of 60 hertz:
c. A 45 microhenry coil operated at a frequency of 1,250 kilohertz:
10. Calculate the value of the inductor needed to produce the reactance specified at the given frequency:
a. A reactance of 1 megohm at a frequency of 40 kilohertz:
b. A reactance of 47 kilohms at a frequency of 108 megahertz:
11. Calculate the frequency at which the given inductors will have the specified reactance.
a. A reactance of 50 kilohms with a 4 millihenry inductor:
12. Solve for the values indicated using the circuit shown. (Assume LM= 0.)
LT = L1+ L2= 15 mH + 85 mH = 100 mH
b. XL1= πfL1= (6.28) (25 kHz) (15 mH) = 2,355 
= 2.36 k
c. XL2= πfL2= (6.28) (25 kHz) (85 mH) = 13,345 = 13.35 k
d. XLT= XL1+ XL2= 2.36 k +13.35 k = 15.7 k or
XLT= πfLT= (6.28) (25 kHz) (100 mH) = 15.7 k
f. EL1= IL1XL1= ITXL1= (1.02 mA) (2.36 k) = 2.4 V
g. EL2= IL2XL2= ITXL2= (1.02 mA) (13.35 k) = 13.6 V
h. PL1EL1IL1= EL1IT= (2.4 V)(1.02 mA) = 2.45 mVAR
i. PL2= EL2IL2= EL2IT= (13.6 V)(1.02 mA) = 13.87 mVAR
1. State a short definition of inductance.
2. The concepts of two men are used to explain CEMF for inductors. Who are they?
3. If an iron core is extracted from a coil, will the coil’s inductance increase or decrease? Why?
4. As the number of turns of wire used in a coil increases, does the value of its inductance increase or decrease?
5. If two coils are placed in proximity of one another and one coil produces 4,000 lines of flux, 3,500 of which cut the second coil, what is the coefficient of coupling of these two coils? k = ___
6. What is the range of values for the coefficient of coupling? __ to _
(upper and lower limits for k).
7. In the circuit shown, two coils are connected in series. Determine their total inductance with no mutual inductance. Then determine their mutual inductance and their combined inductance considering mutual inductance (aiding and opposing). k = 0.4, L1 = 4 henrys, and L2 = 9 henrys.
8. In the circuit shown, determine the total inductance of the two parallel-connected inductors if there is no mutual inductance. Then determine their mutual inductance and total inductance (aiding and opposing) if they have a coefficient of coupling of 0.2.
a. Sketch the magnetic field about the coil in the drawing. Indicate north and south poles.
b. Sketch the magnetic field about the conductor. Show its direction.
10. If the current through a coil is changing at the constant rate of 40 milliamperes every 10 seconds, determine the rate of change of the current in amperes per second. If the coil is rated at 5 millihenrys, determine the voltage across the coil.
11. What coefficient of coupling is desired for transformers?
12. If the primary voltage is greater than the secondary voltage of a transformer, is it known as a step-up or step-down transformer?
13. What two types of core losses in a transformer are associated directly with the core?
14. 14. If EP= 120 VAC and ES= 25.2 VAC, determine the turns ratio (NS:NP) of the transformer. Turns ratio = __:__
15. What type of transformer does not provide electrical isolation of primary to secondary?
16. If the primary voltage is 240 VAC with a primary current of 8 milliamperes and the secondary voltage is 50 VAC with a secondary current of 33 milliamperes, determine the percent efficiency of this transformer:
17. A transformer has a turns ratio (NS:NP) of 1:4.5, has 120 VAC applied to its primary, and has a 6.8 kilohm resistor as a load on its secondary. Determine the secondary voltage, the secondary current, and primary current. (Assume 100 percent efficiency.)
18. When 40 VAC is applied to the primary of a transformer, a secondary current of 8 milliamperes flows through a one kilohm resistor connected across the secondary. 2 milliamperes of primary current is present. Determine the percent efficiency of the transformer and its turns ratio.
19. Calculate XLfor a 2 millihenry coil operated at frequencies of 100 hertz, 5 kilohertz, and 1.2 megahertz.
20. From Problem 19, you see that as the frequency applied to an inductor increases, the inductive reactance of it __ (increases, decreases).
21. What is the value of an inductor needed to produce a reactance of 482 kilohms at a frequency of 5 kilohertz?
22. What is the frequency at which an inductor of 8.5 henrys will have an inductive reactance of 1 kilohms?
23. Solve for the indicated values using the circuit shown. (Assume LM= 0.)
24. Solve for the values using the circuit shown. (Assume LM= 0.)
1. Inductance is the property of a circuit that
2. Which of the factors listed below does not govern the value of a coil?
3. The rise or fall of current through an inductor in a circuit is said to be:
4. The voltage that appears across an inductor in a circuit is called _ and appears only when _ the inductor.
a. counter EMF; the current is constant through
b. voltage drop; the voltage changes across
5. The phase relationship of the voltage across an inductor and the current passing through it in an ac (sinusoidal) circuit is such that
a. the voltage lags the current by 90 degrees.
b. the current leads the voltage by 90 degrees.
6. Determine the mutual inductance of two inductors having a coefficient of coupling of 0.8 if their values are 16 millihenrys and 5 millihenrys.
7. If two inductors are series-connected and their values are 16 henrys and 25 henrys, determine their total inductance if they have no mutual inductance.
8. If the two inductors of Question 7 have a coefficient of coupling of 0.2, determine their total inductance aiding and opposing.
9. If a transformer has a turns ratio of 1:19 (NS:NP), an applied primary voltage of 120 VAC, 60 hertz, and a secondary load resistance of 3.3 kilohms, determine the quantities specified below. (Assume 100 percent efficiency.)
10. A transformer has a greater primary current than secondary current under load conditions. Is it a step-up or step-down transformer?
11. Using the inductive reactance equation and given the data specified below, solve for the unknown quantity.
12. Determine the requested voltages, currents, and power for these two circuits. (Assume LM= 0.)
