Ultimo 20 supch13.pdf

CHAPTER 13 - MAGNETICALLY COUPLED CIRCUITS

List of topics for this chapter :Mutual InductanceEnergy in a Coupled CircuitLinear TransformersIdeal TransformersThree-Phase TransformersPSpice Analysis of Magnetically Coupled CircuitsApplications

MUTUAL INDUCTANCE

Problem 13.1 Given the circuit in Figure 13.1 and 1k = , find 1I and 2I .

Figure 13.1

Carefully DEFINE the problem.Each component is labeled completely. The problem is clear.

PRESENT everything you know about the problem.We know all values of the independent sources. We also know the values of all the elements.

In order to find the equivalent circuit containing the induced voltages, we need to know themutual inductance, M.

We know that the coupling coefficient is

1LL

Mk

21

==

Then,

2121 LLLLkM ==

10 10

+

100 V 100 V+

I1 I2

jM

j10 j10

Establish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.The three solution techniques that can be used are nodal analysis, mesh analysis, and basiccircuit analysis. The goal of the problem is to find two currents. Hence, mesh analysis willbe used.

ATTEMPT a problem solution.From the circuit in Figure 13.1, we can see that 10jLjLj 21 == .

Thus,10LL 21 == and LLL 21 == .

Hence,LLLM 21 == and 10jLjMj == .

So, the equivalent circuit is

Now, using mesh analysis,Loop 1 : 0I10jI10jI1010- 211 =++

10I10jI)10j10( 21 =+1IjI)j1( 21 =+

Loop 2 : 010I10I10jI10j- 221 =+++10-I)10j10(I10j- 21 =++

1-I)j1(Ij- 21 =++

In matrix form,

=

+

+

1-1

II

j1j-j-j1

2

1

or

+

+

=

1-1j1j

jj1

II

2

1

where 2j1j)j2j1(-j)()j1( 2222 +=++=+= .

10 10

+

100 V 100 V+

I1 I2j10 j10

++

j10I1j10I2

Now,

+

+

+

++

+

=

1-1

2j1j1

2j1j

2j1j

2j1j1

II

2

1

Therefore,

=

=

+=

+

+=

43.63501

2j11

2j1jj1

I1 A43.63-4472.0

=

=

+=

+

+=

43.6351801

2j11-

2j1)j1(j

I2 A57.1164472.0

EVALUATE the solution and check for accuracy.Use KVL to check the solution.

The equation produced by KVL of the left loop is0I10jI10jI1010- 211 =++

The equation produced by KVL of the right loop is0I10I10jI10j10 221 =++

Inserting the values for 1I and 2I results in valid equations. Thus, our check for accuracywas successful.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to ALTERNATIVE solutions and continue through the process again.This problem has been solved satisfactorily.

=1I A43.63-4472.0 =2I A57.1164472.0

Problem 13.2 [13.1] For the three coupled coils in Figure 13.1, calculate the totalinductance.

Figure 13.1

8 H6 H 10 H

2 H

4 H 5 H

For coil 1, 4246MML 13121 =+=+For coil 2, 1-548MML 23212 ==For coil 3, 75210MML 32313 =+=+

=+= 714LT H10

or 122312321T M2M2M2LLLL +++=)2)(2()5)(2()4)(2(1086LT +++=

=+++= 41081086LT H10

Problem 13.3 For the frequency domain circuit shown in Figure 13.1, determine the valueof )t(vout for )t377cos(10)t(vin = and a coupling coefficient 8.0k = .

Figure 13.1

Before an equivalent circuit can be drawn, we must determine the value of M . Using 8.0k = ,

21LLMk = .

Because the circuit is in the frequency domain rather than the time domain, we know the value ofL rather than the value of L . So, transform the equation for k to include . Then,

)L)(L(M

k21

= .

Hence,8)20)(5()8.0()L)(L(kM 21 ===

We also need to transform the voltage source from the time domain to the frequency domain.Let's assume a reference of

)t377cos(A + .Then,

= 010Vin .

I2jM5

Vin +

I1 j5

j20 +

Vout

The circuit can be redrawn as

Using the dot convention, we can draw an equivalent circuit to incorporate the induced voltagesfrom the coupling effects.

With this circuit, we can use mesh analysis to find the answer in the frequency domain.Loop #1 : 0I8j)II(5jI510- 2211 =++Loop #2 : 0VI20j)II(8j)II(5jI8j out221122 =+++

This is a system of two equations and three unknowns. We need a constraint equation. Due tothe open circuit, it is obvious that 0I2 = .

Combining like terms and introducing the constraint, the equations become10I)5j5( 1 =+

and1out I13jV =

Clearly,

=

=

+= 45-2

4525010

5j510I1

and=== 45213)45-2)(9013(I13jV 1out

Using the reference assumed above, the voltage converts to the time domain as=)t(vout V)45t377cos(213 ++++

I2j85

100 V +

I1 j5

j20 +

Vout

+

Vout

I2

5

100 V +

I1 j5

+ j5I1

+

j20 j8 (I1 I2)

Problem 13.4 Given the circuit in Figure 13.1, find the coupling coefficient, k, and thevoltage across the 1- resistor.

Figure 13.1

=k 1 =1V V01.0

ENERGY IN A COUPLED CIRCUIT

Problem 13.5 Given the circuit in Figure 13.1, 10VV 21 == volts, 10RR 21 == ohms,10LL 21 == , and 5M = , find the coupling coefficient, k, the currents in the primary and

secondary circuits, 1I and 2I , and the power absorbed.

Figure 13.1

The coupling coefficient is 21LL

Mk = .

Given values for M , 1L , and 2L , we need to modify the equation for k to be

)L)(L(M

k21

= .

=== 105

)10)(10(5

k 5.0

100

100 V +

j100

j100 j100 1 +

V1

R2R1

+

V1 V2+

I1 I2

M

L2L1

To find the currents, begin by finding an equivalent circuit that takes into account the couplingeffects, i.e., the induced voltages.

Use mesh analysis to find 1I and 2I .Loop 1 : 21 I5jI)10j10(10 +=Loop 2 : 21 I)10j10(I5j-10- ++=

In matrix form,

=

+

+

10-10

II

10j105j-5j-10j10

2

1

where )8j1)(25(25200j)5j-)(5j-()10j10)(10j10( +=+=++= .

+

+

+

=

10-10

)8j1)(25(10j105j

5j10j10

II

2

1

+

+

+

=

+

+

+

++

+

=

)8j1)(25(100j-100-50j)8j1)(25(50j-100j100

10-10

)8j1)(25(10j10

)8j1)(25(5j

)8j1)(25(5j

)8j1)(25(10j10

II

2

1

+

+

+

=

+

+

+

=

8j12j-4-8j12j4

)8j1)(25()j-2-)(50()8j1)(25()j2)(50(

II

2

1

Thus,

=

=

+

+= 88.820623.8

57.264721.48j12j4

I1 A31.56-5547.0

=

=

+= 88.820623.8

43.153-4721.48j12j-4-

I2 A69.1235547.0

10 10

+

10 V 10 V+

I1 I2j10 j10

++

j5I1j5I2

Now, find the power absorbed in the circuit. Look at the power absorbed by each element.

Starting with the primary circuit,W0769.3-))31.56-(0cos()5547.0((10)-cosIV-p 1V ===

W0769.3)10()5547.0(RIp 21211R ===W0)270cos(5385.1))31.56-(69.213cos()5547.0)(7735.2(-p1 ===

where 1p is the power absorbed by the induced voltage of 1L .

Ending with the secondary circuit,W0769.3-)69.1230cos()5547.0)(10(cosIVp 2V ===

W0769.3)10()5547.0(RIp 22222R ===W0)90cos(5385.1)69.12369.33cos()5547.0)(7735.2(-p2 ===


The voltage sources absorb 3.0769 watts, or deliver +3.0769 watts, the resistances absorb3.0769 watts, and the induced voltages absorb 0 watts. The inductors do not absorb power.

Problem 13.6 [13.13] Determine the currents 1I , 2I , and 3I in the circuit of Figure13.1. Find the energy stored in the coupled coils at 2t = ms. Take 1000= rad/s.

Figure 13.1

Transform the current source to a voltage source as shown below.

4 j5

8

390 A 200 V+

I2

I1

k = 0.5

j10 j10 I3

j5

8

1290 A 200 V+

k = 0.5

j10 j10 I3+

4

I2I1

21 LLM

k = or21 LL

kM =

5)10)(5.0()L)(L(kM 21 ===

Using mesh analysis,Mesh 1, 21221 I10jI)5j4(I5jI5jI)5j10j4(12j ++=+++= (1)

Mesh 2, 112 I5jI5jI)5j10j8(200 ++++=21 I)5j8(Ij1020- ++= (2)

From (1) and (2),

+

+=

2

1

II

5j810j10j5j4

20-12j

60j107 += , 296j60-1 = , 100j402 =

=

=1

1I A18.72462.2

=

=2

2I A97.48-878.0

== 213 III A89.74329.3

A)18.72t1000cos(462.2)t(i1 +=A)48.97t1000cos(878.0)t(i2 =

At ms2t = , == 6.114rad2t1000

-2.445)18.726.114cos(462.2)ms2(i1 =+=0.8391)48.976.114cos(878.0)ms2(i2 ==

The total energy stored in the coupled coils is21

222

211 iiMiL5.0iL5.0w ++=

Since 10L1 = and 1000= ,mH10LL 21 == , mH5L5.0M 1 ==

8391)-2.445)(0.)(5(0.8391))(10)(5.0(-2.445))(10)(5.0(w 22 ++==w mJ15.23



Figure 13.1

As seen in Problem 13.5,

===

=

105

)10)(10(5

)L)(L(M

k21

5.0

To find the currents, begin by finding an equivalent circuit which takes into account the couplingeffects, i.e., the induced voltages.

Use mesh analysis to find 1I and 2I .Loop 1 : 21 I5jI)10j10(10 ++=

Loop 2 : 21 I)10j10(I5j10- ++=

In matrix form,

=

+

+

10-10

II

10j105j5j10j10

2

1

where )8j1)(25(25200j)5j)(5j()10j10)(10j10( +=+=++= .

R2R1

+

V1 V2+

I1 I2

M

L2L1

10 10

+

10 V 10 V+

I1 I2j10 j10

++

j5I1j5I2

+

+

+

=

10-10

)8j1)(25(10j105j-

5j-10j10

II

2

1

+

+

++

=

+

+

+

++

+

=

)8j1)(25(100j-100-50j-)8j1)(25(50j100j100

10-10

)8j1)(25(10j10

)8j1)(25(5j-

)8j1)(25(5j-

)8j1)(25(10j10

II

2

1

+

+

+

=

+

+

+

=

8j16j-4-8j16j4

)8j1)(25()3j-2-)(50()8j1)(25()3j2)(50(

II

2

1

Thus,

=

=

+

+= 88.820623.8

31.562111.78j16j4

I1 A57.62-8944.0

=

=

+= 88.820623.8

69.123-2111.78j16j-4-

I2 A43.1538944.0

Now, find the power absorbed in the circuit. Look at the power absorbed by each element.

Starting with the primary circuit,W9994.7-))57.26-(0cos()8944.0(10)(-cosIV-p 1V ===

W9995.7)10()8944.0(RIp 21211R ===W0)270cos(9998.3))57.26-(43.243cos()8944.0)(4720.4(p1 ===


Ending with the secondary circuit,W9994.7-)43.1530cos()8944.0)(10(cosIVp 2V ===

W9995.7)10()8944.0(RIp 22222R ===W0)90-cos(9998.3)43.15343.63cos()8944.0)(4720.4(p2 ===





Figure 13.1

=k 5.0

=1I A31.56-5547.0

=2I A69.1235547.0




Figure 13.1

=k 5.0

=1I A57.62-8944.0

=2I A43.1538944.0


R2R1

+

V1 V2+

I1 I2

M

L2L1

R2R1

+

V1 V2+

I1 I2

M

L2L1

LINEAR TRANSFORMERS

Problem 13.10 [13.23] For the circuit in Figure 13.1, find :(a) the T-equivalent circuit,(b) the -equivalent circuit.

Figure 13.1

(a) == MLL 1a H10

== MLL 2b H15

== MLc H5

(b) 27525300MLL 221 ==

==

= 15275

MLMLL

L2

221

A H33.18

=

=

MLMLL

L1

221

B H5.27

==

= 5275

MMLL

L2

21C H55

5 H

20 H15 H

IDEAL TRANSFORMERS

Problem 13.11 Given the ideal transformer circuit in Figure 13.1, find 10V .

Figure 13.1

Carefully DEFINE the problem.Each component is labeled completely. The problem is clear.

PRESENT everything you know about the problem.We know the values of the independent source. We also know the values of all the elements.

Establish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.The three solution techniques that can be used are nodal analysis, mesh analysis, and basiccircuit analysis. Using either nodal or mesh analysis will produce two equations with fourunknowns. In either case, we need to find two constraint equations. So, let's use meshanalysis for the initial attempt to find a solution. Then, nodal analysis will be used to checkthe solution.

ATTEMPT a problem solution.We need to make some assumptions. First, assume that a positive voltage for the inductor inthe primary circuit yields a positive voltage for the inductor in the secondary circuit. Second,assume that a positive (or clockwise) current in the primary circuit yields a positive (orclockwise) current in the secondary circuit. The assumptions are shown in the followingcircuit.

0.9

100 V +

10 +

V10

1 : 10

0.9

100 V +

10 +

V10

1 : 10

I1 I2+

V2

+

V1

Mesh analysis yields,Loop 1 : 11 VI9.010 +=

Loop 2 : 22 I10V =

This is a set of two equations and four unknowns. Two constraint equations are needed.

From the ideal transformer, as shown, we know that 12 nVV = and 21 nII = .

This implies that12 V10V = and 21 I10I =

which are the two constraint equations.

There are many ways to find the values of 1V , 2V , 1I , and 2I . Let's find 1I . To do this,find 1V in terms of 1I and substitute into the equation for loop 1.

12221 I1.0I)I10)(1.0(V1.0V ====

and the equation for loop 1 becomes11 I1.0I9.010 +=

1I10 =Hence,

A10I1 = V1V1 =A1I2 = V10V2 =

Therefore,== 210 VV V10

EVALUATE the solution and check for accuracy.Using nodal analysis,

At node 1, 0I9.010V

11

=+

or 0I9.010V 11 =+ (1)

At node 2, 0)I-(10

0V2

2=+

or 0I10V 22 = (2)

VS +

ZL

1 : n

I1 I2+

V2

+

V1

Again, this is a set of two equations and four unknowns. Two constraint equations areneeded. From the ideal transformer, we have

12 V10V = (3)21 I10I = (4)

From (2) and (4),1122 II)101)(10(I10V === (5)

From (1), (3), and (5),010V)91(V)10)(9.0(10VV9.010V 11121 =+=+=+

V1V10V10 11 ==

Then, V10V10V 12 == A10VI 21 == and A1I)101(I 12 ==

Our check for accuracy was successful.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to ALTERNATIVE solutions and continue through the process again.This problem has been solved satisfactorily.

=10V V10

Problem 13.12 [13.33] For the circuit in Figure 13.1, find oV . Switch the dot on thesecondary side and find oV again.

Figure 13.1

F02.0C = becomes 10j-)02.0)(5(j1

Cj1

==

2

20 mF

10

10 cos(5t) V +

+

Vo

3 : 1

We apply mesh analysis to the circuit shown below.

Using mesh analysis,

For mesh 1, 131 VI10I1010 += (1)For mesh 2, o22 VI2V == (2)For mesh 3, 1213 VVI10I)10j10(0 += (3)

At the terminals, 3VVnV 112 == (4)3IInI 221 == (5)

From (2) and (4),21 I6V = (6)

Substituting this into (1),31 I10I1010 = (7)

Substituting (4) and (6) into (3) yields321 I)j1)(10(I4I10-0 += (8)

From (5), (7), and (8),

=

0100

III

j10104-10-10-61000.333-1

3

2

1

A9.32482.193.33j20-100j100

I 22 =

=

=

== 2o I2V V9.32963.2

Switching the dot on the secondary side affects only equations (4) and (5).

3V-V 12 = (9)

2

j10

10

+

+

Vo

3 : 1

100 V I1 I2

I3

+

V2

+

V1

3I-I 21 = (10)

From (2) and (9),21 I-6V =

Substituting this into (1),1231 I)5j23(I6I10I1010 == (11)

Substituting (9) and (10) into (3),321 I)j1)(10(I4I10-0 ++= (12)

From (10) to (12), we get

=

0100

III

10j10410-10-6-100333.01

3

2

1

A147.1-482.193.33j20-100j100

I 22 =+

=

=

== 2o I2V V147.1-963.2

THREE-PHASE TRANSFORMERS

Problem 13.13 [13.53] In order to meet an emergency, three single-phase transformerswith rmsV7200470,12 are connected in Y to form a three-phase transformer which isfed by a 12,470-V transmission line. If the transformer supplies 60 MVA to a load, find :

(a) the turns ratio for each transformer,(b) the currents in the primary and secondary windings of the transformer,(c) the incoming and outgoing transmission line currents.

(a) Consider just one phase at a time.

===

3124707200

V3V

nLp

L

31

(b) The load carried by each transformer is MVA20360 =

Hence, == kV47.12MVA20

ILp A604,1

== kV2.7MVA20

ILs A778,2

(c) The current in each incoming line a, b, c is

== 85.16033I3 Lp A778,2

The current in each outgoing line A, B, C is

=

3n2778

A812,4

1:n

c

b

a

C

20MVALoad

B

A

PSPICE ANALYSIS OF MAGNETICALLY COUPLED CIRCUITS

Problem 13.14 [13.63] Use PSpice to find 1V , 2V , and oI in the circuit in Figure 13.1.

Figure 13.1

The schematic is shown below.

In the AC Sweep box, we type Total Pts = 1, Start Freq = 0.1592, and End Freq = 0.1592.After simulation, we obtain the output file which includes

FREQ IM(V_PRINT1) IP(V_PRINT1)1.592 E01 1.955 E+01 8.332 E+01

FREQ IM(V_PRINT2) IP(V_PRINT2)1.592 E01 6.847 E+01 4.640 E+01

FREQ IM(V_PRINT3) IP(V_PRINT3)1.592 E01 4.434 E01 9.260 E+01

j12 2

+

1 : 2

4060 V+

V2

+

V1

16

j4 20

j8

300 V+

Io

Thus,=1V V32.8355.19 =2V V4.4647.68

=oI mA92.6-4.443

APPLICATIONS

Problem 13.15 [13.73] A 4800-V rms transmission line feeds a distribution transformerwith 1200 turns on the primary and 28 turns on the secondary. When a 10- load is connectedacross the secondary, find :

(a) the secondary voltage,(b) the primary and secondary currents,(c) the power supplied to the load.

(a) nNN

VV

1

2

1

2==

=

== )4800(

120028

VNN

V 11

22 V112

(b) === 10112

RV

I 22 A2.11

21 InI = , where 120028n =

=

= )2.11(1200

28I1 mA3.261

(c) === )10()2.11(RIp 222 W1254

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PrefaceChapter 1 Basic ConceptsChapter 2 Basic LawsChapter 3 Methods of AnalysisChapter 4 Circuit TheoremsChapter 5 Operational AmplifierChapter 6 Capacitors and InductorsChapter 7 First-Order CircuitsChapter 8 Second-Order CircuitsChapter 9 Sinusoids and PhasorsChapter 10 Sinusoidal Steady-State AnalysisChapter 11 AC Power AnalysisChapter 12 Three-Phase CircuitsChapter 13 Magnetically Coupled CircuitsMutual InductanceEnergy in a Coupled CircuitLinear TransformersIdeal TransformersThree-Phase TransformersPSpice Analysis of Magnetically Coupled CircuitsApplications

Chapter 14 Frequency ResponseChapter 15 Laplace TransformChapter 16 Fourier SeriesChapter 17 Fourier TransformChapter 18 Two-Port Networks

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