Chapter 14: Magnetically Coupled Circuits – Electric Circuit Analysis

Chapter 14

Magnetically Coupled Circuits

CHAPTER OBJECTIVES
  • To introduce mutual inductance element
  • To explain dot polarity convention
  • To explain equivalent circuits for mutually coupled coil systems
  • To define a perfectly coupled linear transformer
  • To define an ideal transformer
  • To define and explain steady-state analysis of coupled coil systems
  • To explain transformers for impedance matching
  • To explain what is meant by single-tuned and Double-tuned Band-pass Filters
  • To explain transient response of coupled coil systems
  • To explain constant flux linkage theorem
INTRODUCTION

There are many different ways in which two or more circuits can get coupled without any conductive connections among them. Two separate circuits interacting with each other by means of optical signals is an example.

Another form of coupling is through thermal effect. The heat produced by dissipation in one circuit may affect the temperature of components in a nearby circuit and change their parameters. Change in parameter values affects the circuit operation in the nearby circuit.

Circuits can be designed to cause mechanical motion of some part of a component when input is applied to the circuit. This mechanical motion may result in variation of some circuit parameter in a nearby circuit, thereby affecting its operation.

There is no conductive connection (i.e., a wire connection) between the interacting circuits. They are galvanically isolated – yet they interact.

A practically important interaction between two or more galvanically isolated circuits takes place through the medium of shared flux linkage in circuits. Current in a coil in one circuit produces magnetic flux at locations where coils belonging to other circuits are placed. This results in interaction between various circuits through magnetic fields and gives rise to mutual flux linkages and mutually induced electromotive forces (emfs). We take up transient and steady-state analysis of such circuits in this chapter.

14.1 THE MUTUAL INDUCTANCE ELEMENT

Flux linkage in a coil can be produced in two ways. The first method is to send a current through the coil. The flux linkage produced in the coil will be proportional to current through it and the proportionality constant is its self-inductance, L. We now call this inductance as self-inductance because we need to distinguish it from mutual inductance. However, the qualifier ‘self’ is often dropped in practice and the word ‘inductance’ alone is taken to mean ‘self-inductance’ by default.

The flux linkage will be time-varying when the current in the coil is a time varying one. Time-varying flux linkage produces induced emf in the coil. This induced emf, if viewed as a voltage drop across element, is given by v = L di/dt where L is the self-inductance of the coil.

The second way flux linkage in a coil can be produced is to place it near some other coil that is carrying a current. The flux produced by the current in the second coil links the first coil too due to the coils sharing same magnetic path. Now, two possibilities arise. The flux in the first coil due to the second coil may be crossing the area of cross section of the coil in the same direction as the flux produced in the coil by a current flowing in it did. Or, it could be in the opposite direction. That depends on how the coils are wound. The magnitude of flux linkage produced in the first coil by the current in the second coil is proportional to that current and the proportionality constant is defined as mutual inductance M12.

Both self-induced as well as mutually induced emfs are neglected in modeling a physical electrical system by two-terminal elements except in the case of those devices that have been specifically designed to strengthen induced emf components. The self-induced emf in such a device was modeled by a two-terminal inductance element in Chapter 1.

Fig. 14.1-1 A two-coil coupled system

Consider a two-coil system shown in Fig. 14.1-1. Assume for a moment that i2(t) is zero. Then, the induced emf in the first coil is entirely due to induced electric field created by its own current. Or, equivalently, the flux linkage in the first coil is entirely due to its own current. The flux linkage per ampere in the first coil per unit current in it will give its self-inductance. Let it be L1. Then, the voltage that appears across first coil with the second coil kept open is v1(t) = and the flux linkage of the first coil with the second coil kept open is Ψ1(t) = L1i1(t)Wb-T. Similarly, the voltage that appears across the second coil with the first coil kept open is v2(t) and the flux linkage of the first coil with the second coil kept open is Ψ2(t) = L2i2(t) Wb-T, where L2 is the self-inductance of the second coil. Self-inductance of a coil in a multi-coil system is measured by measuring the inductance across the coil with all other coils kept open.

Time-varying current flow in the second coil will produce induced electric field at all points inside the first coil. Therefore, there will be an induced emf in the first coil due to i2(t) in the second coil even when the first coil is kept open. This emf will appear as a potential difference across its terminals. This emf is called the mutually induced emf in coil-1 due to coil-2. It will be proportional to and the proportionality constant is defined as M12, the mutual inductance between coil-1 and coil-2. The induced electric field created at a point in coil-1 by an increasing current in the second coil may add to the induced electric field created at the same point coil-1 by an increasing current in coil-1 itself or may subtract from it. That depends on relative winding directions in the two coils.

The process of mutual induction may also be understood from the point of view of flux linkage. There is a flux linkage in the first coil due to the current in the second coil even when the first coil is kept open. This flux linkage is called the mutual flux linkage in coil-1 due to coil-2. The value of this mutual flux linkage will be proportional to i2(t) and the proportionality constant is the mutual inductance M12. Thus, M12 can be understood as the mutual flux linkage in coil-1 per unit current in coil-2. Rate of change of the mutual flux linkage gives the emf induced in coil-1 by current in coil-2. The flux linkage created in coil-1 by coil-2 may add to or subtract from the flux linkage created in coil-1 by its own current.

Hence, the voltage that appears across the coil-1 when both coils are carrying current will be and total flux linkage in coil-1 is Ψ1(t) = L1il(t) ± M12i2(t) Wb-T. The sign connecting the two terms, + or –, will be decided by whether the flux produced in the first coil by the second coil current is aiding the flux produced by its own current or opposing it.

Similarly, the voltage that appears across coil-2 is v2(t) and total flux linkage in coil-2 is Ψ2(t) = L2i2(t) ± M21i1(t)Wb-T where M21 is the mutual inductance between coil-2 and coil-1.

The value of mutual inductance M12 can be measured by measuring the voltage that appears across the open-circuited coil-1 with a known time-varying voltage applied to coil-2. Similarly, the value of mutual inductance M21 can be measured by measuring the voltage that appears across the open-circuited coil-2 with a known time-varying voltage applied to coil-1.

Mutual induction arises out of magnetic coupling between two coils. Now, we make an important assumption. We assume that the two-coil system is constructed in such a way that there is magnetic coupling only between them and there is no magnetic coupling between these coils and any other element in the circuit or with the circuit loop itself. This is possible only if the two-coil system is designed to confine the magnetic field almost entirely within itself. A closed core structure employing a magnetic material with high magnetic permeability will be needed in practice to achieve this. Both the coils will be wound around the same core. With this assumption of confinement of magnetic flux linkage entirely within the device itself, we can model a two-coil system by an ideal four-terminal element model. The symbol of the model is shown in Fig. 14.1-2. L1 and L2 are the self-inductance of the coils. M is the mutual inductance between them. The two parallel lines between the coils indicate that they share a common core. The two ‘dot’ points marked by the side of coils help to decide the polarity of mutual emf in relation to the self-induced emf. Increasing current entering the dot point in one coil generates a mutual emf in other coil with positive polarity at its dot point.

Fig. 14.1-2 A four-terminal element model for a two-coil system

14.1.1 Why Should M12 Be Equal to M21?

Consider an experiment with the two-coil system shown in Fig. 14.1-2. We want to set up a DC current of I1 in coil-1 and a DC current of I2 in coil-2 starting from zero-current condition in both coils. We do this by applying an independent current source iS1 to coil-1 and the independent current source iS2 to coil-2. Refer to Fig. 14.1-3.

The current in the first coil ramps up linearly with a slope of I1/T A/s. This results in v1(t) = L1I1/T V and v2(t) = M21I1/T V during [0,T] interval. But the current in the second coil during that interval is zero and hence there is no power delivered to the coil-2 during this interval. There is power delivered to coil-1 during this interval. The total energy delivered to coil-1 during [0,T] is L1I12/2 J.

The current in coil-2 ramps up linearly during the interval [T ′, T ′ + T] with a slope of I2/T A/s. This results in v1(t) = M12I1/T V and v2(t) = L2I1/T V during [T ′, T ′ + T] interval. The total energy delivered to coil-2 is L2I22/2 J. But, now the first coil current is at a constant level of I1 and hence the mutually induced voltage across the first coil results in additional energy input into the first coil from the current source iS1. This additional energy input is

Fig. 14.1-3 Current source waveforms and voltage appearing across them during the current-build up in a two-coil system.

Therefore, the total energy delivered to the two-coil system in building up the currents is =

Now, assume that we built up the currents in the coils to the same levels using a current source iS1 which had a waveform that is delayed by T ′ second for coil-1 and a current source iS2 which had a waveform that started at t = 0 for coil-2. Then, the total energy required to build up the currents will be

The total energy delivered to the system must be the same in these two situations. Otherwise, we can build up the currents using the procedure resulting in lower energy input and subsequently reduce the currents to zero using the other process (with suitably constructed current sources) to receive some net energy out of the system. That is an attempt to violate conservation law for energy. Therefore,

Now, we know that the total energy storage in a two-coil system shown in Fig. 1.8-2 is given by Two-coil system is a passive system. Therefore, the total energy storage in it cannot be non-negative for any permissible voltage–current condition. Therefore,

This has to be true for any (i1(t), i2(t) ). Let us choose This implies that i1(t) and i2(t) have opposite polarities for this choice. Therefore, i1(t) i2(t) will be negative. Therefore, has to be ≥ M for the last inequality to be true. Therefore, M

The equality sign applies only if the magnetic coupling is so tight that there is no leakage of magnetic flux from the common core and windings. That is, the fluxes linking the two coils are equal. The coupling coefficient goes to unity under this situation. No physical coil system can have k equal to 1. However, coil systems in power transformers approach this value very closely.

14.1.2 Dot Polarity Convention

One does not want to physically examine every coupled coil-set to find out the relative winding directions in order to decide the sign connecting the self-induced voltage drop and mutually induced voltage drop in each coil. Therefore, the manufacturer marks this information in the form of two dots – one each on each coil in a coupled set of two coils. He/She has a choice of two terminals in the first coil to put a dot on. He/She chooses that arbitrarily. That choice will fix the terminal of the second coil at which the second coil dot has to be marked. The following rule helps us to write the voltage equation in a set of coupled coils.

 

Increasing current entering a dot point in one coil will produce a mutual emf with its positive polarity at the dot point in the other coil.

Increasing current leaving a dot point in one coil will produce a mutual emf with its negative polarity at the dot point in the other coil.

With this interpretation for dot polarity assignment, the mutual inductance between a pair of coils is always a positive quantity.

Self-induced voltage is always positive at the current entry point for both coils.

Various possibilities for mutual emf polarity in a two-coil coupled system are shown in Fig. 14.1-4. A double-headed arrow indicates magnetic coupling with M value marked by the side.

Fig. 14.1-4 Dot polarity convention and sign of mutually induced voltage in a two-coil system

But markings can fade with time. And people can forget to do things they are supposed to do. Therefore, the following experimental method can help us to put the dots if they are not known already.

Let A and B be the two terminals of the first coil and C and D be the two terminals of the second coil. Join B and D. Apply a sinusoidal source of suitable amplitude and frequency between A and B as shown in Fig. 14.1-5. Use three rms reading voltmeters to record the readings V1, V2 and V3. Then, V2 will be close to either V1 + V3 or | V1V3|. The dot point assignment for these two cases are shown in Fig. 14.1-5.

Fig. 14.1-5 Experimental determination of dot point assignment in a two-coil system

Dot polarity convention can be applied in the case of more than two coupled coils too. Consider three mutually coupled coils with self-inductance L1, L2 and L3. Let M1 be the mutual inductance between coil-1 and coil-2, M2 be the mutual inductance between coil-2 and coil-3 and M3 be the mutual inductance between coil-1 and coil-3. We need three pairs of dots to indicate the relative polarity of mutually induced emf in the three coils. Each coil has a self-induced emf and two mutual emf from the remaining two coils. A possible dot polarity assignment is shown in Fig. 14.1-6.

Fig. 14.1-6 Dot polarity assignment in a three coil system

The KVL equations (assuming that there is no resistance in the coils) can be obtained as follows:

14.1.3 Maximum Value of Mutual Inductance and Coupling Coefficient

The maximum value of mutual inductance that a two-coil coupled system can have is the geometric mean of self-inductances of the two coils – i.e., This maximum value is realised when the flux produced by a current in one coil links totally with the other coil. This is impossible to achieve physically and can only be approached in practice. Winding both the coils over a common core made of some high permeability material like laminated iron makes M very close to (but less than) in practice.

The ratio between the actual mutual inductance realised to the theoretical maximum value is defined as coupling coefficient and is symbolised by k. Coupling coefficient is a positive number between 0 and 1. It is difficult to achieve anything more than few tenths for k-value in air-cored coils; however, coils wound on common core made of iron can have k-value very close to unity.

14.2 THE TWO-WINDING TRANSFORMER

A system of two coils with constant values of L1, L2 and M with two pairs of terminals identified for application of excitation and/or measurement of response is called a two-winding linear transformer. Such a two-winding transformer will be referred to simply as a transformer in this section.

Fig. 14.2-1 shows a two-coil system with self-inductance values of L1, L2 and mutual inductance value of M. Also shown are two independent voltage sources driving the circuit at two pairs of terminals.

Fig. 14.2-1 A transformer with voltage source drive at both ports

The mesh equations of the circuit are written assuming that the coils have zero winding resistance values.

The current i2 leaves the dot in the second coil and therefore the mutually induced voltage in the first coil appears with negative polarity at the dot. Hence, the negative sign for mutually induced voltage in the first mesh equation. The current i1 enters the dot of the first coil and therefore the mutually induced voltage in the second coil appears with positive polarity at the dot. When mesh equation is written by traversing the loop in clockwise direction the self-induced voltage and mutually induced voltages will enter with opposite signs. Hence, the negative sign for mutually induced voltage in the second mesh equation.

Fig. 14.2-2 Two circuit models for a transformer [a] model using linear dependent sources [b] Conductive equivalent model

Refer to Fig.14.2-2. The reader may easily verify that the circuit in Fig. 14.2-2 (a) and the circuit (b) have the same mesh equations as those of the circuit in Fig. 14.2-1. Hence, the coupled set of coils may be replaced by two decoupled coils and two linear dependent sources as in circuit model Fig. 14.2-2 (a) as far as the v–i behaviour at the terminals is concerned. This circuit model preserves the conductive decoupling – i.e., the galvanic isolation – that exists between the two sides of the circuit.

A coupled set of two coils can also be replaced by a T-shaped equivalent circuit comprising three pure decoupled inductors as in circuit model in Fig. 14.2-2 (a) as far as the v–i behaviour at the terminals is concerned. This circuit model hides the galvanic isolation that is present in the transformer. Therefore, it is called the conductive equivalent circuit of coupled coils. Note that one inductance (either L1 M or L2 M) can be negative-valued for sufficiently large value of coupling coefficient. There is no negative inductance in the physical world. But then, the inductors that appear in the conductive equivalent circuit of coupled coils are not physical inductors – they are mathematical inductors that are arranged to result in same set of circuit equations as those of the coupled coils. Therefore, they can assume negative values – we do not have to construct them!

Note that a certain dot polarity was assumed for the equivalent models established above. The equivalent models for the second relative dot polarity are shown in Fig. 14.2-3.

Fig. 14.2-3 Circuit models for coupled coils

The circuit in Fig. 14.2-3 (b) shows the dependent source based circuit model for the two-coil system in (a) and the circuit in Fig. 14.2-3 (c) shows the conductive equivalent circuit of the two-coil system in Fig. 14.2-3 (a). Note the reversal of polarity in the case of dependent sources and change in inductance values in the case conductive equivalent circuit.

Differentiation in time is replaced by multiplication by jω in the phasor equivalent circuit for sinusoidal steady-state analysis in the equivalent circuit employing dependent sources. Winding resistance can be included as series resistors on both sides in both equivalent circuits.

Example: 14.2-1

The applied voltage is vS(t) = 100√2 cos103t V in the circuit shown in Fig. 14.2-4. (a) Find the primary (the side with source connection) and secondary (the side with no source connection) currents and the average power delivered to the resistors for (i) k = 0, (ii) k = 0.1, (iii) k = 0.5 and (iv) k = 1. (b) Repeat calculations with k = 1 for L1 = 25 H and L2 = 100 H.

Fig. 14.2-4 Circuit for Example: 14.2-1

Solution

  1. The conductive equivalent circuit for coupled coils is used to obtain the phasor equivalent circuits for the four values of coupling coefficients. The phasor equivalent circuits are shown in Fig. 14.2-5.

    The solution process is illustrated for k = 0.1.

    Vs = 100∠0° V rms. The two mesh equations are expressed in matrix form as

    Solving these equations, we get I1 = 3.737∠–67.86° A rms and I2 = 0.174∠–46° A rms.

    Average power dissipated in 10 Ω resistor is 3.7372×10 = 139.65 W

    Average power dissipated in 40 Ω resistor is 0.1742 × 40 = 1.21 W

    Average power delivered by source = 100 × 3.737 × cos67.9° = 140.83 W.

    Impedance seen by source = Zin = 100∠0° ÷ 3.737∠67.86° = 10.09 + j24.78 Ω

    Fig. 14.2-5 Phasor equivalent circuits of the circuit in for different values of coupling coefficient

    The table that follows summarises the results of calculations for all four values of k.

  2. We observe from the table of results that as the coupling coefficient increases (i) the primary and secondary currents increase in magnitude and become more active in content than reactive, (ii) power delivered to the resistor in the secondary side increases, (iii) the power factor of circuit increases and (iv) the source experiences a decreasing input impedance that becomes more and more resistive.

    Now, we raise the self-inductance level of the coils by 1000 fold. But we keep the ratio between them at 4 as before. The mesh equations for k = 1 are

    We observe that the reactive terms in the matrix entries overpower the resistive terms. The solution is obtained as I1 = 5∠–0.01° A rms and I2 = 2.5∠0.01° A rms.

    Average power dissipated in 10 Ω resistor is 52 × 10 = 250 W

    Average power dissipated in 40 Ω resistor is 2.52 × 40 = 250 W

    Average power delivered by source = 100 × 5 × cos0.01° = 500 W.

    Impedance seen by source = Zin = 100∠0° ÷ 5∠–0.01° = 20 + j0.004 Ω

    The input impedance as seen by the source is almost purely resistive at 20 Ω. The resistor that is in series with the source in the primary side must be contributing 10 Ω to the input impedance. Therefore, the transformer and its secondary load of 40 Ω resistor contributes 10 Ω to input impedance. Thus, the equivalent impedance of the transformer plus 40 Ω load seems to be an almost pure resistance of 10 Ω. The ratio between 10 Ω and 40 Ω happens to coincide with the ratio L1: L2 – that is a coincidence that calls for further analysis.

    Let us calculate the voltage appearing across primary and secondary windings of the transformer.

    Voltage across primary winding = 100∠0° –10 × 5∠–0.01° ≈ 50∠0° V rms

    Voltage across primary winding = 40 × 2.5∠0.01° ≈ 100∠0° V rms

    Thus, primary voltage: secondary voltage ≈ 1:2 = : and using the solution for currents, primary current: secondary current ≈ 2:1 = : That makes it too many coincidences to pass over lightly. Therefore, we analyse the transformer with k = 1 in greater detail now.

14.3 THE PERFECTLY COUPLED TRANSFORMER AND THE IDEAL TRANSFORMER

Figure 14.3-1 shows a two-winding transformer with primary self-inductance of L1 and secondary self-inductance of L2 with coupling coefficient of unity. We neglect the winding resistance for the time being. The two circuits shown in this figure differ only in the dot polarity in the transformer.

Fig. 14.3-1 A passively terminated two-winding transformer with k = 1

The primary is driven by a source =cos ωt V and secondary is passively terminated in an impedance ZL = RL + j XL Ω. Passive termination implies that there are no sources of any kind in the load circuit.

The mesh currents are Ip and Is themselves and the mesh equations for circuit in Fig. 14.3-1 (a) in matrix form are

Since k = 1, the value of M = . Solving for primary and secondary current phasors using this value for M, we get

We have used the fact that the source voltage phasor V∠0° itself is the primary voltage phasor Vp to make substitutions in the equation above.

We note that the secondary voltage and primary voltage are in phase and related by a real scaling factor of . However, this number is nothing but where n1 and n2 are the number of turns employed in primary and secondary windings. This is so since self-inductance of a coil is proportional to the square of the number of turns in the coil. We represent this ratio as turns ratio ‘n’.

 

Therefore, secondary voltage magnitude in a transformer with unity coupling coefficient is turns ratio times the primary voltage magnitude.

The equation for Ip is the sum of two terms. Each term involves a division of primary voltage phasor by impedance. Therefore, the equation for Ip suggests that the input impedance of the circuit is a parallel combination of two impedances – an inductor of L1 and another impedance of value

 

Therefore, a transformer with unity coupling coefficient reflects the secondary load impedance ZL onto the primary side as a turns-ratio transformed impedance of value ZL/n2 where n = secondary turns/primary turns.

Thus, the entire transformer and load circuit can be replaced by the equivalent circuit shown in Fig. 14.3-2 as far as the input side v–i behaviour under sinusoidal steady-state condition is concerned. We will see later that this is true not only for sinusoidal steady-state analysis but also for time-domain analysis too.

Fig. 14.3-2 Input equivalent circuit of transformer with k = 1

The primary winding resistance is included as a series resistance in the primary side, whereas the secondary winding resistance is included as a part of RL in the load for analysis purposes.

Now, we consider the circuit in Fig. 14.3-1 (b). Reader may easily verify that the mesh equations for this circuit will be,

Both the off-diagonal terms in the mesh impedance matrix have changed sign. The solution for primary and secondary current phasors with k = 1 will be,

There is no difference in solution except for 180° phase shift in secondary voltage. That is anyway expected due to the change in relative polarity of windings. The voltage magnitude ratio remains at turns ratio n.

But the most important aspect to be noted is that the same equivalent circuit shown in Fig. 14.3-2 will describe the transformer in this case too. See the equation for primary current phasor. In fact, this is true even if k ≠ 1.

 

The input impedance of a passively terminated transformer is independent of its dot point assignment.

There are many important applications of transformers with k ≈ 1 in which we make use of the inductor L1 that appears in the input impedance to advantage. A tuned amplifier widely employed in communication circuits is such an application example.

There are many applications of transformers with k ≈ 1 in electrical power engineering and electronics engineering where that inductance in the input impedance of transformer is not desirable. Therefore, we minimise the effect of that inductance by making L1 (and consequently L2 and M too) very large compared to the maximum value of ZL that we may use. This is done by using a large cross-section iron core and a large number of turns in both windings keeping the turns ratio at the desired value. An ideal transformer is an idealised model for such a transformer.

 

An ideal transformer is a two-winding transformer with perfect coupling (k = 1) and infinite self and mutual inductance values (L1 → ∞, L2 → ∞, M → ∞). Complex power is conserved in an ideal transformer.

Thus, the equations for an ideal transformer are:

Complex power is conserved in an ideal transformer. That is, an ideal transformer absorbs zero active power, zero reactive power and zero complex power. Whatever active power and reactive power go into the primary winding go out of secondary winding into the load. An ideal transformer simply changes the voltage and current levels at which complex power transfer takes place. Moreover, an ideal transformer is an impedance transformer. It scales the secondary load impedance by a scaling factor of and presents it at the primary input. In fact, those two are the major applications of transformers – efficient voltage/current level translation in power engineering and impedance level translation (called impedance matching) in electronics engineering.

14.4 IDEAL TRANSFORMER AND IMPEDANCE MATCHING

Consider a power (or energy) delivery application in which a power delivery circuit delivers power to a load circuit. A typical case in point is an audio power amplifier. The power delivery block here is a cascade of low-power and high-power transistor amplifiers that amplify the audio input. The power delivery circuit maybe equivalenced to its Thevenin’s equivalent in the form of a signal source with a series impedance. This impedance is called the output impedance of the power delivery circuit. As per Maximum Power Theorem, the load impedance has to be the conjugate of output impedance of power delivery circuit for maximum power transfer into the load. But we may not be able to make the load impedance satisfy this requirement since the load is designed with other factors in mind. For instance, loud speakers that are designed for good acoustic performance generally have low impedance and amplifiers that are designed for efficient and distortion-free amplification of audio signals have high output impedance. Thus, we have to interpose a circuit element that will translate the load impedance to the value demanded by maximum power transfer condition. A tightly coupled two-winding transformer with small winding resistance can do this. Transformers used for this purpose are called Impedance Matching Transformers.

Example: 14.4-1

The sinusoidal voltage source that is driving a load of impedance 25 + j10 Ω at 1 MHz has a source impedance of 1–j1 Ω at 1MHz. An ideal transformer is introduced between the source and load such that the power transferred to load is a maximum. Find the turns ratio of the transformer and value and nature of additional components needed, if any.

Solution

The circuit diagram detailing the application context is shown in Fig. 14.4-1.

Fig. 14.4-1 Circuit for impedance matching example in Example: 14.4-1

The transformer should reflect the load impedance in such a way that the conjugate impedance criterion for maximum power transfer is met with. Therefore, the 25 Ω resistance in the load should appear as 1 Ω resistance when reflected to primary. This fixes the turns ratio n as

But with n = 5, the j10 Ω inductive reactance in the load will appear as j0.4 Ω to the primary side. As per conjugate impedance matching, we want it to appear as –(–j1) = j1 Ω to the primary side. Turns ratio cannot be adjusted any more. Therefore, we need to connect an extra reactance in series with the secondary winding such that the reflected total reactance becomes j1 Ω. Therefore, we need an extra inductive reactance of (j1 – j0.4) × 52 = j15 Ω in series with the secondary winding. The final matched network is shown in Fig. 14.4-2.

Fig. 14.4-2 Solution for impedance matching problem in Example: 14.4-1

14.5 TRANSFORMERS IN SINGLE-TUNED AND DOUBLE-TUNED FILTERS

Tuned amplifiers are transistor amplifiers that combine amplification with band-pass filtering. They are used extensively in electronic communication circuits.

In carrier modulated communication systems, some feature (amplitude, frequency or phase) of a high-frequency sinusoidal carrier waveform is made to vary according to the information-bearing signal (for example, an audio signal). The resulting signal will usually have its frequency content concentrated in a small band around the carrier frequency. For instance, a medium wave radio station transmitting at 1MHz will transmit a signal that contains frequency components in the frequency band between 995 kHz and 1005 kHz. The band-pass filters in the RF stage in the radio receiver have to be tuned to 1 MHz centre frequency with a bandwidth of 10 kHz to receive this station. Tuned amplifiers make use of inductances in the signal coupling transformer and variable capacitors to realise the required band-pass filter.

14.5.1 Single-Tuned Amplifier

Single-tuned amplifier uses a tightly coupled transformer and a capacitor. The capacitor is connected in parallel to the primary of the transformer. Next stage amplifier is connected to the secondary and it may be modeled by the input resistance of the next stage. The transformer primary in parallel with the capacitor forms a parallel resonant circuit and the transistor drives the signal current into this parallel combination.

Example: 14.5-1

An RF transformer with k ≈ 1 and n = 0.5 is passively terminated by a resistor of 800 Ω. It is driven by a current source iS(t) = Im cos ωt A in parallel with a capacitor of 220 pF at the primary side. The primary self-inductance of the transformer is 115 μH. Derive an expression for the ratio of output voltage phasor to input current phasor.

Fig. 14.5-1 Circuit for Example: 14.5-1

Solution

The phasor equivalent circuit of the circuit is shown in Fig. 14.5-2. The transformer is replaced by its equivalent circuit in this diagram by reflecting the 800 Ω in the secondary as 800/(0.5)2 = 3.2 kΩ in the primary.

Fig. 14.5-2 Phasor equivalent circuit for the circuit in Fig. 14.5-1

The ratio of output voltage phasor to input current phasor is just the impedance of parallel combination of inductor, capacitor and resistor. Using the parameter symbols R, L and C,

Substituting the numerical values for circuit parameters, where f is the cyclic frequency of input in MHz. The magnitude and phase of this ratio function is given by

is the angular frequency at which the susceptance of inductor and capacitor will be equal in magnitude but opposite in sign. Therefore, these two susceptance values will cancel each other at that frequency and leave R as the impedance value. At all other frequencies, some susceptance – inductive or capacitive – will shunt R thereby bringing down the magnitude of impedance below R.

Thus, impedance is a maximum of R at Normalising the impedance function by dividing it by value of R,

Fig. 14.5-3 shows the plots for these functions.

Fig. 14.5-3 Magnitude and phase plots of normalised impedance function in Example: 14.5-1

The plots reveal that it is a highly frequency-selective circuit. The circuit produces large magnitude output if the frequency of sinusoidal current source is at 1 MHz or nearby. The response is small if the frequency is far away from 1 MHz on either side. Such a response characteristic is called a band-pass characteristic or tuned characteristic. A circuit with band-pass nature can extract some frequency components selectively from a mixture of sinusoidal waveforms at different frequencies presented to it at the input. Such circuits find wide application in communication and signal processing circuits. The circuit we discussed in this example illustrates the principle of tuned amplifiers that are used in all sorts of electronic communication circuits starting from the common transistor radio receiver. The current source is realised by a transistor circuit in a tuned amplifier.

Single-tuned amplifier is an example where the inductance of a transformer is intentionally designed to resonate with a chosen capacitor at a desired frequency.

The cut-off frequencies of a band-pass filter are the frequencies at which the magnitude of transfer gain of the filter is times the value at centre frequency. They are 0.9 MHz and 1.1 MHz in this example. The difference between the two cut-off frequencies is defined as the bandwidth of the band-pass filter. Hence, the band-pass filter in this example has centre frequency of 1MHz and bandwidth of 0.2 MHz.

14.5.2 Double-Tuned Amplifier

Consider a modulated waveform with a carrier frequency of 1MHz with signal frequency components in the range between 0.9MHz and 1.1MHz. The single-tuned amplifier in the previous example can amplify this modulated signal since the centre frequency and bandwidth of that amplifier are 1MHz and 0.2 MHz, respectively. However, the signal frequency components in 0.9MHz to 1.1MHz range get a variable gain – varying from 100% to 70.7%. This will lead to waveform distortion in the demodulated waveform. The signal after demodulation will be a distorted version of signal that was transmitted. The gain offered to components in the frequency range 0.9MHz to 1.1MHz must be a constant if this kind of signal distortion is to be avoided. We can achieve this by placing two resonant humps at two frequencies that are nearby in frequency response of the tuned amplifier. This is achieved in a double-tuned amplifier.

A double-tuned amplifier is a voltage-driven amplifier and uses a weakly coupled transformer and two capacitors – one in series with the primary and one in series with the secondary and load. We consider a 1:1 transformer with equal capacitors in the primary and secondary for illustration. The secondary is loaded resistively with a resistor equal in value to the source resistance in the primary side. We ignore the winding resistances for simplifying the analysis. The double-tuned stage is shown in Fig. 14.5-4. Transistor circuitry provides the input voltage signal to this filter.

Fig. 14.5-4 Doubled-tuned band-pass filter

We first study the AC steady-state input impedance of the transformer to understand how two tuning frequencies appear in the circuit. We assume that the resistor R is small compared to and ignore the resistor in this study. See Fig. 14.5-5.

Fig. 14.5-5 Simplified circuit for study of input impedance

The circuit in Fig. 14.5-5 (a) is translated to the phasor equivalent circuit in Fig. 14.5-5 (b) using conductive equivalent circuit for the transformer. We require the input impedance of the circuit in Fig. 14.5-5 (b) for sinusoidal steady-state. This may be obtained by applying impedance series parallel combination rules. The result is given in the following:

We have used the relation kL in arriving at this impedance function. The impedance is always reactive. It is an inductive reactance for It is a capacitive reactance for It is an inductive reactance again for It becomes zero at It changes from an infinite valued inductive reactance to infinite valued capacitive reactance as ω crosses from left to right.

The plot of this reactance function is shown in Fig. 14.5-6. Also shown in the same figure in dotted curve is the magnitude of capacitive reactance of the capacitor connected in series with the primary. L = 115μH, C = 200pF and k = 0.5 for the circuit for which these reactance curves were prepared. The resonant frequency of L and C is 1 MHz.

Fig. 14.5-6 Input reactance in a double-tuned circuit with no load resistance

The input impedance of the circuit in Fig. 14.5-5 (a) is the sum of solid curve and negative of the dotted curve in Fig. 14.5-6. This sum will go zero at frequencies corresponding to A and B in Fig. 14.5-6. The corresponding frequencies are 0.835 MHz and 1.45 MHz. The circuit is resonant at these two frequencies.

When the load resistance R is connected in the secondary, the resonant frequencies will shift. However, the shift will be small if

The circuit in Fig. 14.5-4 can be analysed to obtain the steady-state frequency response function H(jω) relating the output voltage to input voltage. It is possible to show that the frequency response function between load voltage and source voltage is

The term within the square brackets in the denominator goes to zero at and if The gain magnitude of H(jω) will exhibit a peak value of 0.5 at these two frequencies.

The plot of magnitude of frequency response function between source voltage and output voltage for single-tuned filter and double-tuned filter is shown in Fig. 14.5-7. The source resistance was 700 Ω and load resistance was 700 Ω in both cases. Desired centre frequency is 1.05 MHz and desired bandwidth is 165 kHz. The single-tuned design makes use of a 1:1 tranSsformer with k ≈ 1, primary and secondary inductance of 8 μH each and a 2.8nF tuning capacitor connected across primary. Secondary is loaded with a 700 Ω resistor. Gain magnitude for this filter is shown in dotted curve in Fig. 14.5-7. The double-tuned filter uses a 1:1 transformer with k = 0.11, primary and secondary inductance of 1.045mH and two capacitors of value 22pF each in series with the windings. A 700 Ω in series with the capacitor and secondary winding load the filter. Gain magnitude for this filter is shown in solid curve in Fig. 14.5-7.

Two resonant peaks due to double tuning can be seen clearly in the frequency response plot. Note that while keeping the centre frequency and bandwidth the same as in single-tuned filter, pass-band gain variation in double-tuned filter is lesser than in the case of single-tuned filter. The skirts on either side of pass-band are steeper. The stop-band performance of double-tuned filter too is superior to that of single-tuned filter.

Fig. 14.5-7 Magnitude of frequency response function for single-tuned and double-tuned band-pass filters

14.6 ANALYSIS OF COUPLED COILS USING LAPLACE TRANSFORMS

In the first part of this section, we show that (i) the input impedance function of a passively terminated two-winding transformer is independent of relative polarity of windings, (ii) the secondary voltage of a transformer with unity coupling coefficient (k = 1) and zero winding resistance is turns ratio times primary voltage, quite independent of the waveshape of input, (iii) a transformer with k = 1 reflects an impedance Z(s) connected in the secondary side to the primary side as Z(s)/n2 where n is the ratio of secondary turns to primary turns and (iv) a practical two-winding transformer with finite inductance values, imperfect coupling (k < 1) and non-zero winding resistance will be a band-pass system with a lower cut-off frequency decided by winding resistance and an upper cut-off frequency decided by coupling coefficient.

In the second part of this section, we take up the issue of instantaneous changes in currents in a coupled-coil system and show that such changes are possible in inductors involved in a coupled-coil system under certain specific circuit conditions. We show that a coil with zero winding resistance kept shorted will expel magnetic flux in the common magnetic path. This is the principle that works behind electromagnetic shielding of electrical and electronic equipment.

In the third part of this section, we look briefly at a practical problem involving coupled-coils – that of breaking the primary side circuit while it is carrying current. This turns out to be a stressful operation for the load connected at the secondary side and specific design measures are incorporated in the load circuit to protect it against failure under this condition.

14.6.1 Input Impedance Function of a Two-Winding Transformer

Consider a two-winding transformer with zero winding resistance. Let the self-inductance of primary and secondary windings be Lp and Ls. Let n be the ratio of secondary turns to primary turns. Let the secondary be terminated in a load circuit that has an input impedance of Z(s). See Fig 14.6-1.

Fig. 14.6-1 A passively terminated two-winding transformer [winding resistance negligible]

We do not specify the relative polarity. Instead, we cover both options by using proper signs for the inductance values in the equivalent circuit shown.

Now, the primary-side input impedance function Zp(s) can be derived as shown below.

We observe that this impedance function is independent of relative polarity of windings.

Now, assume that the coupling coefficient is unity. Thus, for a perfectly coupled two-winding transformer with zero winding resistance, the input impedance function is given by,

This impedance is shown in Fig. 14.6-2.

Fig. 14.6-2 Reflection of secondary impedance function to primary with k = 1

Now, we can incorporate the primary winding resistance rp as extra impedance in series at the input and the secondary winding resistance rs as a part of the load.

Fig. 14.6-3 Primary-side input impedance in a perfectly coupled transformer

A perfectly coupled transformer is called an ideal transformer if Lp, Ls and M have infinitely high values. Practical iron-cored transformers can be modeled approximately by an ideal transformer. Such an ideal transformer will have input impedance that is a turns-ration ratio transformed version of the secondary load impedance. We have discussed the application of such transformers in impedance matching applications earlier in this chapter.

14.6.2 Transfer Function of a Two-Winding Transformer

Now, we look at the voltage transfer function of a two-winding transformer. Let Ip(s) and Is(s) be the Laplace transforms of the primary and secondary mesh currents in the s-domain equivalent circuit shown in Fig. 14.6-1. Then, the mesh equations in matrix form is

LpLs = M2 for a perfectly coupled transformer ( i.e., k = 1). Therefore, for a perfectly coupled transformer with rp = rs = 0, the voltage transfer function is given by

Eqn. 14.6-2 indicates that there is a polarity inversion in secondary voltage with respect to primary voltage for one relative polarity in windings.

Various factors prevent a practical transformer from meeting this ideal. Winding resistance and imperfect coupling are two such factors. If the mesh equations are written again with rp and rs included, the transfer function can be shown as

for a special load – resistive load. Let Z(s) = R. Then,

This is a second order band-pass function (note the s term in the numerator polynomial).

We express this transfer function as the product of the ideally expected value ( = ±n) and a factor involving s as follows.

We want to determine the low-frequency cut-off and high-frequency cut-off for the sinusoidal steady-state frequency response function of this transfer function. In general, these two frequencies will be some complex functions of k, n, rs, rp, R and Lp. However, we can use some approximations in the case of a practically important iron-cored transformer.

In an iron-cored transformer, the coupling coefficient is nearly unity. The load resistance is usually much larger than the transformer’s winding resistance. Hence, the effect of leakage flux and the effect of winding resistance are only second-order effects. The aggregate result of two second-order effects can be taken as the product of effects when each is acting alone. Therefore, when we consider the effect of winding resistance we assume that k = 1 and when we consider the effect of imperfect coupling we assume that rp = rs = 0.

Further algebraic manipulation of the above expression leads to

β is a voltage reduction factor due to part of the applied voltage getting dropped across the winding resistance. α is the pole of the first order transfer function. The transfer function is seen to be a first order high-pass function. The DC gain (evaluated by substituting s = j0) – i.e., the ratio between the steady-state response under DC conditions to the DC input value – is zero. The upper cut-off frequency is α rad/s where

for practical values of R

We observe that both the voltage reduction factor β and the cut-off frequency α can be obtained from the equivalent circuit shown in Fig. 14.6-3 with Z(s) replaced with R.

 

The winding resistances prevent a two-winding transformer from passing DC voltage and low frequency AC voltage from primary to secondary. Hence, a signal containing low-frequency content will suffer waveform distortion when it goes through a transformer.

The high-frequency behaviour of a tightly coupled transformer is studied after neglecting the winding resistances. Then, Eqn. 14.6-6 simplifies to

This is a low-pass transfer function with a cut-off frequency of rad/s.

 

Imperfect magnetic coupling prevents a two-winding transformer from passing high frequency AC voltage from primary to secondary. The two-winding transformer exhibits low-pass behaviour in the high-frequency end due to imperfect magnetic coupling. Therefore, a signal containing high-frequency content will suffer waveform distortion when it goes through a transformer.

The product of right sides of Eqn. 14.6-7 and Eqn. 14.6-8 provides an approximation for Eqn. 14.6-6 in the case of a tightly coupled two-winding transformer with winding resistances very small compared to load resistance. Thus,

This transfer function has a band-pass frequency response function with lower cut-off frequency of α and upper cut-off frequency of λ. Primary winding resistance and primary winding inductance decide the lower cut-off frequency. The upper cut-off frequency is decided by coupling coefficient, primary winding inductance and load resistance.

The bandwidth of a transformer can be increased by reducing the winding resistances and increasing the coupling coefficient. Bandwidth of a transformer is load dependent.

14.7 FLUX EXPULSION BY A SHORTED COIL

One does not make a good transformer and keep its secondary shorted! However, there are situations and applications in which an electrical system can be modeled as a transformer with a low resistance secondary winding kept shorted. Therefore, we look at the input impedance of a transformer with the secondary winding shorted as in Fig.14.7-1.

Fig. 14.7-1 A transformer with secondary shorted

Zp(s) may be determined by mesh analysis. The result will be,

The impedance at DC frequency is rp Ω.

The impedance at high frequency is ≈ s(1 – k2)Lp, indicating that the transformer behaves as an inductance of (1 – k2)Lp H at high frequencies when its secondary is kept shorted.

Assume that the shorted coil has negligibly small resistance. Taking rs = 0, we get,

 

Zp(s) = rp + s(1–k2)Lp.

 

Now, if the transformer is perfectly coupled (i.e., k = 1), then the input impedance becomes equal to primary winding resistance at all frequencies. Hence, the short-circuit impedance of a good transformer with k = 1, rp = rs = 0 is zero at all frequencies.

The expressions for Ip(s) and Is(s) with r2 = 0 may be derived as

Currents entering the dots generate positive flux linkages in coils. Let Ψp be the flux linkage in primary coil and Ψs be the flux linkage in the secondary coil. Then,

Therefore, the secondary coil that started with zero flux linkage continues to hold zero flux linkage at all time after that. Thus,

 

A shorted coil with zero resistance expels flux completely at all frequencies in the magnetic path where it is located.

It does this by allowing suitable current to flow in it in order to cancel the mutual flux generated by current in the driven coil. Both primary and secondary currents rise exponentially with a time constant of in step response.

But what if the windings are perfectly coupled in addition? Then,

If vp(t) is a unit step function, the current rises instantaneously from 0 to 1/rp in the primary winding and from 0 to k/nrp in the secondary winding. Further, the flux linkages in both windings continue to be at zero.

There was no impulse voltage in the circuit. How was it possible for the inductor currents to change instantaneously without impulse voltages in the circuit?

Currents in individual coils can change instantaneously in a coupled-coil system. Faraday’s law indicates that the induced emf across a coil is proportional to the rate of change of flux linkage in it. Therefore, flux linkage of a coil cannot change instantaneously unless impulse voltage is applied to the coil. This reduces to the statement that current in an inductor cannot change instantaneously unless impulse voltage is applied in the case of inductors that are not magnetically coupled. However, there is a distinction between instantaneous change of flux linkage and current if the flux linkage in a coil can be produced by more than one current as in the case of coupled inductors. Therefore, the more general principle of ‘no instantaneous change in flux linkage in a coil unless impulse voltage is applied’ should be applied in the case of coupled-coils.

 

The flux linkage in a shorted coil with zero resistance remains constant in time. This statement is sometimes referred to as ‘Constant Flux Linkage Theorem’.

Note that in the step response of a perfectly coupled transformer with shorted secondary (rs = 0), the flux linkage of secondary winding remained at zero though the current increased instantaneously. The magnetic coupling was perfect, and the primary cannot have flux linkage in it in that case unless secondary has it. Therefore, the primary winding flux linkage also remained at zero despite step change in primary current.

We may derive a relationship between the instantaneous changes in primary and secondary windings as follows:

But, there cannot be instantaneous changes in flux linkages in the absence of impulse voltages. Hence, Δψp (t) and Δψs (t) are zero. Therefore,

The determinant of the matrix on the left side is nonzero for all k < 1. Therefore, Δip and Δis can only be zero with imperfect coupling. That is, there can be no instantaneous change in coil currents in a two-coil system if the coils are imperfectly coupled and there is no impulse voltage applied or supported somewhere in the system.

However, if k = 1, the determinant of the square matrix in the left side of the equation is zero and there can be a non-zero solution for Δip and Δis. In fact, any pair of values that satisfy the constraint that Δis = Δip/n will be permitted. The exact value by which the primary current jumps will be decided by the jump in primary voltage and the primary resistance. Since the flux linkage in primary winding does not change, all the primary voltage will have to be absorbed by the primary resistance at all t.

Its resistance compromises the effectiveness of flux expulsion from shorted coil. The expression for flux linkage in shorted coil when k ≠ 1 and rs ≠ 0 is

DC value of this ratio is M/rp. Therefore, DC flux will not be expelled under steady state. Similarly, low frequency AC flux will also manage to get into the shorted coil under steady state conditions. However, the ratio goes to low values at high frequency. Therefore non-zero resistance in shorted coil results in DC and low-frequency fluxes penetrating into the coil. High-frequency flux is expelled more or less effectively.

The principle of flux expulsion detailed in this section is employed in shielding sensitive electronic equipment from electromagnetic interference.

14.8 BREAKING THE PRIMARY CURRENT IN A TRANSFORMER

Let Ip and Is be the currents in primary winding and secondary winding of the transformer at the instant at which the switch S is opened in the circuit shown in Fig. 14.8-1.

Let us denote the instant of switching as t = 0. Then, the flux linkages in the windings just before switching are

and the flux linkages at t = 0 + are

Fig. 14.8-1 Breaking the primary current in a transformer

Since there is no impulse is the closed secondary loop, the flux linkage in the secondary winding will have to be continuous. Therefore,

Therefore,

This current is usually in a direction opposite to that of is (0).

The corresponding flux linkage in primary winding at The initial flux linkage in this coil was Ψp (0 ) = LpIp – MIs . Therefore, the instantaneous decrease in flux linkage that took place in this coil is

Hence, an impulse voltage of area content equal to (1 – k2)LpIp V-s will appear across the primary winding and will have to be supported by the switch. In fact, there will be arcing across the switch due to this very large voltage trying to establish across it.

However, arcing involves energy. Where does the energy come from? It is possible to show that the energy that gets dissipated across the switch is 0.5(1 – k2)LpI2p J. This comes from the magnetic energy storage in the coupled-coil system.

The sudden change of secondary current from IS to results in a sudden change across the load voltage from R Is to There are practical applications in which the second term dominates and makes the load voltage a negative voltage with enough magnitude to destroy the load if the load happens to be a sensitive electronic equipment.

This large negative voltage decays exponentially with a time constant of Ls/R s. By the time the transient is over, the remaining initial magnetic energy stored in the coil system would have got dissipated in R.

14.9 SUMMARY
  • Self-inductance of coil is the flux linkage in it when 1A flows in it with all other coils in the vicinity kept open. Mutual inductance between two coils is the flux linkage in one coil when it is kept open with 1A flowing in the other coil.
  • For two coupled coils with self-inductance L1 and L2, the mutual inductances M12 and M21 are equal. The maximum value mutual inductance can have is H. The ratio between actual value of mutual inductance and its maximum possible value is defined as the magnetic coupling coefficient.
  • A system of two coils with constant values of L1, L2 and M, with two pairs of terminals identified for application of excitation and/or measurement of response, is called a two-winding linear transformer.
  • Secondary voltage magnitude in a transformer with unity coupling coefficient is turns ratio times the primary voltage magnitude. A transformer with unity coupling coefficient reflects the secondary load impedance ZL onto the primary side as a turns-ratio transformed impedance of value ZL/n2 where n = secondary turns/primary turns.
  • An ideal transformer is a two-winding transformer with perfect coupling (k = 1) and infinite self- and mutual inductance values (L1 → ∞, L2 → ∞ M → ∞). The input impedance of such a transformer is ZL/n2. Complex power is conserved in an ideal transformer. Practical transformers that approach ideal transformer are employed in voltage/current level conversion and impedance matching applications.
  • The input impedance function of a passively terminated two-winding transformer is independent of relative polarity of windings.
  • The secondary voltage of a transformer with unity coupling coefficient (k = 1) and zero winding resistance is turns ratio times primary voltage, quite independent of the waveshape of input.
  • A transformer with k = 1 reflects an impedance Z(s) connected in the secondary side to the primary side as Z(s)/n2 where n is the ratio of secondary turns to primary turns, quite independent of the waveshape of input.
  • A practical two-winding transformer with finite inductance values, imperfect coupling (k < 1) and non-zero winding resistance will be a band-pass system with a lower cut-off frequency decided by winding resistance and upper cut-off frequency decided by coupling coefficient.
  • The flux linkage in a shorted coil with zero resistance remains constant in time. If the shorted coil has non-zero resistance, DC and low-frequency flux will penetrate into the coil. However, high-frequency flux will be expelled from the coil.
  • There can be instantaneous change in coil currents in a coupled-coil system even in a circuit that does not apply or support impulse voltage if the coils are perfectly coupled. However, there cannot be an instantaneous change in flux linkage of any coil, whether coupled or uncoupled, unless the circuit applies or supports impulse voltage.
14.10 PROBLEMS
  1. A current source with iS(t) = 10t A for 0 ≤ t ≤ 1 s and zero for all other t is connected to one of the coils of a two-coil system with the second coil kept open. The voltage across current source is seen to be a rectangular pulse of amplitude 10 V and duration 1 s. The voltage across the other coil is seen to be a rectangular pulse of amplitude 7 V and duration 1 s. Find the self-inductance of the first coil and mutual inductance between the coils.
  2. A two-coil system with self-inductance values of 1H and 4H carries 2A in 1H coil and 1A in 4H coil. The total stored energy in the system is seen to be zero under this condition. Find the mutual inductance between the coils and coupling coefficient between them.
  3. A two-coil system with L1 = 100 mH, L2 = 300 mH and k = 0.8 has i1(t) = 1.2 sin (100 π t) A and i2(t) = 0.5 cos (100 π t) A. Assuming the self-flux linkage and mutual flux linkage aid each other, (a) find time-domain expressions for flux linkage in coil-1 and coil-2 and (b) find time-domain expressions for voltage across both coils. (c) Repeat (a) and (b) assuming that the self-induced emf and mutually induced emf oppose each other in the coils.
  4. A sinusoidal source of sin100t V is applied at terminals A and B in the circuit in Fig. 14-10.1 with terminals C and D kept open. The coils are lossless. The rms reading meters read A1 = 1A, V1 = 200 V, V2 = 100 V, and V3 = 100 V. When the same source is connected across C and D with terminals A and B kept open, the ammeter A2 is seen to read 0.5A rms. Find the position of the missing dot, L1, L2, M and k.

    Fig. 14-10.1

  5. A sinusoidal source of sin1000t V is applied at terminals A and B in the circuit in Fig. 14-10.2 with terminals C and D kept open. The coils are lossless. The rms reading meters read V1 = 200 V, V2 = 100 V, V3 = 100 V, V4 = 200 V and V5 = 100 V. When the same source is connected across C and D with terminals A and B kept open, the voltmeters V2 and V4 are seen to read 100 V each. Obtain a dot polarity assignment for this coil system using three distinct symbols for coil 1-coil 2, coil 2-coil 3 and coil 1-coil 3 mutual coupling.

    Fig. 14-10.2

  6. Two coupled coils can be connected in series in two ways. The connection (a) in Fig. 14-10.3 is differential series connection and connection (b) is cumulative series connection. Find the equivalent inductance between A and B in both cases.

    Fig. 14-10.3

  7. Find the equivalent inductance between A and B in the two possible parallel connections of two coupled inductors in Fig. 14-10.4.

    Fig. 14-10.4

  8. The complex power input into the primary side of a lossless transformer with perfect coupling is 100 + j100 kVA at 230V rms and 50Hz. The load in the secondary side is seen to consume 90kVA of reactive power along with active power. (i) Find the complex power consumed by the load and its power factor. (ii) Find the primary winding self-inductance.
    1. What is the turns ratio and coupling coefficient of the transformer in the circuit in Fig. 14-10.5?
    2. Find the primary and secondary currents and voltage across 10 Ω resistor as functions of time if the source has 110 V rms value and 60 Hz frequency.

      Fig. 14-10.5

  9. Find all angular frequency values for which the input impedance of the circuit shown in Fig. 14-10.6 is purely resistive.

    Fig. 14-10.6

  10. Find the power dissipated in the resistors and complex power delivered by the source in the circuit in Fig. 14-10.7 if the voltage source has 110 V rms value at 60 Hz,

    Fig. 14-10.7

  11. What is the turns ratio n of the ideal transformer in the circuit in Fig. 14-10.8 if the reactive power delivered by the source (230 V rms at 50 Hz) is zero?

    Fig. 14-10.8

  12. What should the value of turns ratio n if maximum power is to be transferred to the 8 Ω load in the circuit in Fig. 14-10.9? If the source has 110 V rms value, what is the amount of power transferred to 8 Ω with this turns ratio?

    Fig. 14-10.9

  13. The sinusoidal voltage source that is driving a load of impedance 16–j10 Ω at 1 MHz has source impedance of 1 + j1 Ω at 1MHz. An ideal transformer is introduced between the source and load such that the power transferred to load is a maximum. Find the turns ratio of the transformer and value and nature of additional components needed, if any.
  14. A transformer used in a DC power supply has 100 turns in the primary and 25 turns in the secondary. The primary winding resistance is 0.1 Ω and secondary winding resistance is 0.15 Ω. The step response of input current with secondary open is found to have a rise time of 0.22 s. The coupling coefficient is 0.98. (a) Find the self-inductance of windings and the mutual inductance between them. (b) Find the input admittance function and voltage transfer function when a resistive load of 2 Ω is connected across the secondary. (c) Prepare the pole-zero plot for the transfer function. (d) Sketch the frequency response plots for output voltage across the load resistance and estimate the cut-off frequencies and bandwidth.
  15. Obtain the step response of output voltage for the transformer in the Problem 15 with a 2 Ω load in the secondary. What are the flux linkages in coils and total energy stored in the transformer under step response steady-state condition?
  16. The transformer used in the circuit in Fig. 14-10.10 is the same as the one in Problem 15. Switch was in closed position for a long time prior to t = 0 and it is opened at t = 0. (i) Find and plot i2(t), v0(t) and vS(t) for t ≥ 0 + and (ii) Calculate the energy dissipated in the switch and the 2 Ω resistance after at t = 0.

    Fig. 14-10.10