Chapter 13 - Alexander`s Fundamentals of Electric circuits Ch.13 Solution PDF

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Alexander`s Fundamentals of Electric circuits Ch.13 Solution...

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Chapter 13, Solution 1. For coil 1, L 1 – M 12 + M 13 = 12 – 8 + 4 = 8 For coil 2, L 2 – M 21 – M 23 = 16 – 8 – 10 = – 2 For coil 3, L 3 + M 31 – M 32 = 20 + 4 – 10 = 14 L T = 8 – 2 + 14 = 20H or

L T = L 1 + L 2 + L 3 – 2M 12 – 2M 23 + 2M13

LT = 12 + 16 + 20 – 2x8 – 2x10 + 2x4 = 48 – 16 – 20 + 8 = 20H

Chapter 13, Solution 2. Using Fig. 13.73, design a problem to help other students to better understand mutual inductance. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Determine the inductance of the three series-connected inductors of Fig. 13.73.

Figure 13.73 Solution L = L 1 + L 2 + L 3 + 2M12 – 2M 23 –2M 31 = 10 + 12 +8 + 2x6 – 2x6 –2x4 = 22H

Chapter 13, Solution 3. L 1 + L 2 + 2M = 500 mH

(1)

L1 + L 2 – 2M = 300 mH

(2)

Adding (1) and (2), 2L 1 + 2L 2 = 800 mH But,

L 1 = 3L 2, , or 8L 2 + 400,

and L 2 = 100 mH

L 1 = 3L 2 = 300 mH From (2),

150 + 50 – 2M = 150 leads to M = 50 mH k = M/ L1 L2  50 / 100 x300 = 0.2887

300 mH, 100 mH, 50 mH, 0.2887

Chapter 13, Solution 4. (a) For the series connection shown in Figure (a), the current I enters each coil from its dotted terminal. Therefore, the mutually induced voltages have the same sign as the self-induced voltages. Thus, L eq = L 1 + L 2 + 2M M

Is

L1

I

I1 Vs

M

I2

+ –

L2

L1

L2

L eq (a) (b)

(b)

For the parallel coil, consider Figure (b). Is = I1 + I2

and

Z eq = V s /I s

Applying KVL to each branch gives, Vs = jL 1 I 1 + jMI2

(1)

Vs = jMI 1 + j L 2 I 2

(2)

 Vs   jL1  V    jM  s 

or

jM   I 1  jL 2   I 2 

2 2 2  = – L 1 L 2 +  M ,  1 = jV s (L2 – M),  2 = jV s (L 1 – M)

I 1 =  1 /, and I 2 =  2 / I s = I 1 + I 2 = ( 1 +  2 )/ = j(L 1 + L 2 – 2M)V s /( –2(L 1 L 2 – M2)) = (L 1 + L 2 – 2M)V s /( j(L 1 L 2 – M2)) Z eq = V s /I s = j(L1 L 2 – M2)/(L 1 + L 2 – 2M) = jL eq i.e.,

Leq = (L1 L 2 – M2)/(L 1 + L 2 – 2M)

Chapter 13, Solution 5. (a) If the coils are connected in series,

L  L1  L2  2 M  50 120  2(0.5) 50 x120  247.4 mH (b) If they are connected in parallel, L

L1 L2  M 2 50x120  38.72 2 mH  48.62 mH  L1  L2  2M 50  120  2x 38.72

(a) 247.4 mH, (b) 48.62 mH

Chapter 13, Solution 6. M  k L1 L2  0.6 40 x5  8.4853 mH 40 mH 5mH

 

3

j L  j2000 x40 x10  j80 3

j L  j 2000 x5 x10  j10

 

3

  8.4853mH j M  j2000 x8.4853 x10  j16.97 We analyze the circuit below.

16.77 

I1 +

I2 +

 j80 

j10 

V1

V2 

_

_

V1  j80 I1  j16.97 I 2 V2  16.97 I1  j10I2

(1) (2)

But, V 1 = 200˚ V and I 2 = 4–90˚ A. Substituting these into (1) produces I 1 = [(V 1 +j16.97I 2 )/j80] = [(20+j16.97(–j4))/j80] = 1.0986–90˚ A or i 1 = 1.0986sin(ωt) A From (2), V 2 = –16.97x(–j1.0986) + j10(–j4) = 40 + j18.643 = 44.1325˚ or v 2 = 44.13cos(ωt+25˚) V.

Chapter 13, Solution 7. We apply mesh analysis to the circuit as shown below. j1 

1

2

–j1 

 12

+ _

j6 

I1

+ j4 

I2

1



For mesh 1, (2+j6)I 1 + jI 2 = 24 For mesh 2, jI 1 + (2–j+j4)I 2 = jI 1 + (2+j3)I 2 = 0 or I 1 = (–3+j2)I 2 Substituting into the first equation results in I 2 = (–0.8762+j0.6328) A. V o = I 2 x1 = 1.081 144.16˚ V.

Vo _

Chapter 13, Solution 8. 2H 1H

   

j L  j 4 x2  j8 j L  j 4 x1  j 4

Consider the circuit below.

j4

4

2 0o

+ _

I1





j8

j4

+ I2

2  (4  j 8) I1  j 4 I2 0   j 4I1  (2 j 4)I 2

2

v(t) _

(1) (2)

In matrix form, these equations become 2  4  j8  j4   I 1  0    j 4 2 j 4  I    2     Solving this leads to I 2 = 0.2353 – j0.0588 V = 2I2 = 0.4851 –14.04o Thus, v(t) = 485.1cos(4t–14.04°) mV

Chapter 13, Solution 9. Consider the circuit below.

2

8 30o

+ –

I1

2

j4

j4

I2

-j1

-j2V

+ –

For loop 1, 830 = (2 + j4)I 1 – jI 2 For loop 2,

(j4 + 2 – j)I 2 – jI 1 + (–j2) = 0 or

Substituting (2) into (1),

(1)

I1 = (3 – j2)i2 – 2 830 + (2 + j4)2 = (14 + j7)I 2 I 2 = (10.928 + j12)/(14 + j7) = 1.03721.12 V x = 2I 2 = 2.074 21.12 V

(2)

Chapter 13, Solution 10. 2H 0.5H 1 F 2

     

j L  j 2 x 2  j 4 j L  j 2 x0.5  j 1 1 jC  j2 x1/ 2   j

Consider the circuit below. j 

 +

24  0

+ _

j4

I1

j4

I2

Vo

–j

_

24  j4 I1  jI 2 0   jI1  ( j 4  j) I 2 In matrix form,

  0   I1  3 I 2

 24  j 4  j   I 1        0    1 3   I2 

Solving this, I 2   j 2.1818,

V o   jI 2   2.1818 v o (t) = –2.1818cos2t V

(1) (2)

Chapter 13, Solution 11. 800mH 600 mH

   

j L  j 600 x800 x10

3

 j 480

j L  j600 x600 x10

3

 j360 3

j L  j600 x1200 x10  j720 1 j 12F   = –j138.89 jC 600x12x10  6  

1200 mH

After transforming the current source to a voltage source, we get the circuit shown below. 200



-j138.89

j480

150

Ix 800  0

+ _

I1

j360

j720

I2

+ _

110  30



For mesh 1, 800  (200  j480  j 720)I 1  j 360I 2  j 720I 2 or 800  (200  j1200) I1  j360 I 2

(1)

For mesh 2, 11030˚ + 150–j138.89+j720)I 2 + j360I 1 = 0 or  95.2628  j 55   j 360I1  (150  j 581.1)I 2

(2)

In matrix form, 800  j 360   I 1     200  j 1200  95.2628  j55    j360 150  j581.1  I 2     Solving this using MATLAB leads to: >> Z = [(200+1200i),-360i;-360i,(150+581.1i)] Z= 1.0e+003 * 0.2000 + 1.2000i 0 - 0.3600i

0 - 0.3600i 0.1500 + 0.5811i >> V = [800;(-95.26-55i)] V= 1.0e+002 * 8.0000 -0.9526 - 0.5500i >> I = inv(Z)*V I= 0.1390 - 0.7242i 0.0609 - 0.2690i I x = I 1 – I 2 = 0.0781 – j0.4552 = 0.4619–80.26˚. Hence, i x (t) = 461.9cos(600t–80.26˚) mA.

Chapter 13, Solution 12. Let   1.

j8 j4  +

j12

1V -

j16

I1

j20 I2



Applying KVL to the loops, 1  j16 I1  j8 I 2 0  j8I 1  j 36 I 2

(1) (2)

Solving (1) and (2) gives I 1 = –j0.0703. Thus

Z

1  jL eq I1

 

L eq 

1  14.225 H. jI1

We can also use the equivalent T-section for the transform to find the equivalent inductance.

Chapter 13, Solution 13.

4

I1

80  0

4

j5

–j Ω

j5 I2

+ _ j2I 2

+ –

– +

j2I 1

j2 

–80 + (4+j5)I 1 + j2I 2 = 0 or (4+j5)I 1 + j2I 2 = 80

j2I 1 +(4+j6)I 2 = 0 or I 2 = [–j2/(7.211156.31°)]I 1 = (0.27735–146.31°)J 1

[4+j5 + j2(–0.230769–j0.153846)]I 1 = [4+j5+0.307692–j0.461538]I 1 = 80

[4.307692+j4.538462]I 1 = 80 or I 1 = 80/(6.257346.494°)

= 12.78507–46.494° A.

Z in = 80/I 1 = 6.257346.494° Ω = (4.308 + j4.538) Ω

An alternate approach would be to use the equation, Z in  4  j(5) 

4 4  4  j5  j5  4  j  j2 7.211156.31

= 4+j5+0.5547–56.31° = 4+0.30769+j(5–0.46154)

= [4.308+j4.538] Ω.

Chapter 13, Solution 14. To obtain V Th , convert the current source to a voltage source as shown below.

j2 5

j6 

j8 

- j3 

2

a j50 V

+

+ –

I

40 V

V Th

+ –

– b

Note that the two coils are connected series aiding. L = L 1 + L 2 – 2M jL = j6 + j8 – j4 = j10 Thus,

–j50 + (5 + j10 – j3 + 2)I + 40 = 0 I = (– 40 + j50)/ (7 + j7)

But,

–j50 + (5 + j6)I – j2I + V Th = 0 V Th = j50 – (5 + j4)I = j50 – (5 + j4)(–40 + j50)/(7 + j7) V Th = 26.74 34.11 V

To obtain Z Th , we set all the sources to zero and insert a 1-A current source at the terminals a–b as shown below.

j2 5

j6 

I1

a

j8 

+ Vo

1A

–

b

-j3 

2

I2

Clearly, we now have only a super mesh to analyze. (5 + j6)I 1 – j2I 2 + (2 + j8 – j3)I 2 – j2I 1 = 0 (5 + j4)I1 + (2 + j3)I 2 = 0

(1)

But,

I 2 – I 1 = 1 or I 2 = I 1 – 1

(2)

Substituting (2) into (1),

(5 + j4)I 1 +(2 + j3)(1 + I 1 ) = 0 I 1 = –(2 + j3)/(7 + j7)

Now,

((5 + j6)I 1 – j2I 1 + V o = 0 V o = –(5 + j4)I 1 = (5 + j4)(2 + j3)/(7 + j7) = (–2 + j23)/(7 + j7) = 2.33250 Z Th = V o /1 = 2.332 50 Ω.

Chapter 13, Solution 15. The first step is to replace the mutually coupled circuits with the equivalent circuits using dependent sources. To obtain I N , short-circuit a–b as shown in Figure (a) and solve for I sc .

j20 Ω

20 

a

+ 

+ –

j5(I 1 –I 2 )

I1

I sc

j10 Ω I2

100 30o + –

j5I 2

b (a) Now all we need to do is to write our two mesh equations. Loop 1.

–10060˚ + 20I 1 + j10(I 1 –I2 ) + j5I 2 = 0 or (20+j10)I 1 – j5I 2 = 10060˚ or (4+j2)I 1 – jI 2 = 2030˚

Loop 2.

–j5I 2 + j10(I 2 –I 1 ) + j20I 2 + j5(I 1 –I 2 ) = 0 or –j5I 1 + j20I 2 = 0 or I 1 = 4I2

Substituting back into the first equation, we get, (4+j2)4I 2 – jI 2 = 2030˚ or (16+j7)I 2 = 2030˚. Now to solve for I 2 = I sc = I N = (2030˚)/(16+j7) = (2030˚)/(17.46423.63˚) = 1.1452 6.37˚ A.

j20 Ω

20 

+  j5(I 1 –I 2 )

I1

j10 Ω

+ –

I2

100 30o + –

j5I2

a + V oc – b

(b) To solve for Z N = Z eq = V oc /I sc , all we need to do is to solve for V oc . In circuit (b) we note that I 2 = 0 and we get the mesh equation, –10030˚ + (20+j10)I 1 = 0 or I 1 = (10030˚)/(22.3626.57˚) = 4.4723.43˚ A. V oc = j10I 1 – j5I 1 (induced voltage due to the mutual coupling) = j5I 1 = 22.3693.43˚ V. Z eq = Z N = (22.3693.43˚)/(1.14526.37˚) = 19.525 87.06˚ Ω. or [1.0014+j19.498] Ω.

Chapter 13, Solution 16. To find I N , we short-circuit a-b. 8

j -j2  a

  j4 

+ 800 V o

j6 

I2

IN

I1 b

 80  (8  j2  j4) I1  jI 2  0 j6I 2  jI1  0

 

 

(8  j2)I1  jI 2  80

(1)

I1  6I 2

(2)

Solving (1) and (2) leads to IN  I2 

80  1.584  j0.362  1.6246 –12.91° A 48  j11

To find Z N , insert a 1-A current source at terminals a-b. Transforming the current source to voltage source gives the circuit below. j 8 -j2  2 a

  j4 

+ j6 

2V

I2

I1 b

0  (8  j2)I1  jI 2

 

I1 

jI 2 8  j2

= [j/(8.2462114.036°)]I 2 = 0.12126875.964°I 2 = (0.0294113+j0.117647)I 2

(3)

2  (2  j 6) I 2  jI 1  0

(4)

Solving (3) and (4) leads to (2+j6)I 2 – j(0.0294113+j0.117647)I 2 = –2 or (2.117647+j5.882353)I 2 = –2 or I 2 = –2/(6.2519270.201°) = 0.319902109.8°.

V ab = 2(1+ I 2 ) = 2(1–0.1083629+j0.30099) = (1.78327+j0.601979) V = 1Z eq or

Z eq = (1.78327+j0.601979) = 1.8821 18.65° Ω

An alternate approach would be to calculate the open circuit voltage. –80 + (8+j2)I 1 – jI 2 = 0 or (8+j2)I 1 – jI 2 = 80 (2+j6)I 2 – jI 1 = 0 or I 1 = (2+j6)I 2 /j = (6–j2)I 2

(5) (6)

Substituting (6) into (5) we get,

(8.2462114.036°)(6.32456–18.435°)I 2 – jI 2 = 80 or

[(52.1536–4.399°)–j]I 2 = [52–j5]I 2 = (52.2398–5.492°)I 2 = 80 or

I 2 = 1.53145.492° A and V oc = 2I 2 = 3.06285.492° V which leads to,

Z eq = V oc /I sc = (3.06285.492°)/(1.6246–12.91°) = 1.8853 18.4° Ω

This is in good agreement with what we determined before.

Chapter 13, Solution 17.

j L  j40

40 40   2667 rad/s L 15 x10 3

  

M  k L1L2  0.6 12 x103 x30x103  11.384 mH If Then

15 mH

40 Ω

12 mH 30 mH 11.384 mH

32 Ω 80 Ω 30.36 Ω

The circuit becomes that shown below. j30.36 

22 

60 



j32 

j80 

Z L =j40



Z in  22  j 32 

2 M 2

 22  j32 

(30.36) 2 60  j120

j80  60  j 40 921.7  22  j32   22  j32  6.87  63.43  22  j32  3.073  j6.144 134.16 63.43

= [25.07 + j25.86] Ω.

Chapter 13, Solution 18. Replacing the mutually coupled circuit with the dependent source equivalent we get, –j4 Ω

j2 

j5Ω

j5I 2 120V

j20Ω

– +

+ –

j5I 1

+ 

I2 I1

4

j6 Ω

Now all we need to do is to find V oc and I sc . To calculate the open circuit voltage, we note that I 2 is equal to zero. Thus,

–120 + (4 + j(–4+5+6))I 1 = 0 or I 1 = 120/(4+j7) = 120/(8.0622660.255°)

= 14.8842–60.255°.

V oc = V Thev = j5I 1 + (4+j6)I 1 = (4+j11)I 1 = (11.704770.017°)(14.8842–60.255°) = 174.22 9.76° V

To find the short circuit current (I sc = I 2 ), we need to solve the following mesh equations,

Mesh 1 –120 + (–j4+j5)I 1 – j5I 2 + (4+j6)(I 1 –I 2 ) = 0 or (4+j7)I 1 – (4+j11)I 2 = 120

(1)

Mesh 2 (4+j6)(I 2 –I 1 ) – j5I 1 + j22I 2 = 0 or –(4+j11)I 1 + (4+j28)I 2 = 0 or I 1 = (28.284381.87°)I 2 /(11.704770.0169°) = (2.416511.853°)I 2

Substituting this into equation (1) we get,

(8.0622660.255°)(2.416511.853°)I 2 – (4+j11)I 2 = 120 or

[(19.482572.108°) – 4 –j11]I 2 = 120 and [5.9855+j18.5403–4–j11]I 2 = (1.9855+j7.5403)I 2 = 120 or

I 2 = I sc = 120/(7.7973375.248°) = 15.3899─75.248° A

Checking using MATLAB we get, >> Z = [(4+7j) (-4-11j);(-4-11j) (4+28j)] Z= 4.0000 + 7.0000i -4.0000 -11.0000i -4.0000 -11.0000i 4.0000 +28.0000i >> V = [120;0] V= 120 0

>> I = inv(Z)*V I= 16.6551 -33.2525i

(I 1 )

3.9188 -14.8829i

(I 2 = I sc ) = 15.3902─75.248° (answer checks)

Finally, Z eq = V Thev /I sc = (174.229.76°)/(15.3899─75.248°)

= (11.32 85.01°) Ω

Chapter 13, Solution 19. X La = X L1 – (–X M ) = 40 + 25 = 65 Ω and X Lb = X L2 – (–X M ) = 40 + 25 = 55 Ω. Finally, X C = X M thus, the T-section is as shown below. j65 

j55 

–j25 

Chapter 13, Solution 20. Transform the current source to a voltage source as shown below.

k=0.5 4

j10

8

j10 I3

+ –

j12

-j5

I1

I2

20 0o

+ –

k = M/ L1L 2 or M = k L 1L 2 M = k L1L 2 = 0.5(10) = 5 For mesh 1,

j12 = (4 + j10 – j5)I 1 + j5I 2 + j5I 2 = (4 + j5)I 1 + j10I 2

For mesh 2,

0 = 20 + (8 + j10 – j5)I 2 + j5I 1 + j5I 1 –20 = +j10I 1 + (8 + j5)I 2

From (1) and (2),

(1)

 j12 4  j5  j10  I 1   20    j10 8 j5  I    2      = 107 + j60,  1 = –60 –j296,  2 = 40 – j100 I 1 =  1 / = 2.462 72.18 A I 2 =  2 / = 878 –97.48 mA I 3 = I 1 – I 2 = 3.329 74.89 A i 1 = 2.462 cos(1000t + 72.18) A i 2 = 0.878 cos(1000t – 97.48) A

At t = 2 ms, 1000t = 2 rad = 114.6 i 1 (0.002) = 2.462cos(114.6 + 72.18) = –2.445A

(2)

–2.445 i 2 = 0.878cos(114.6 – 97.48) = –0.8391 The total energy stored in the coupled coils is w = 0.5L 1 i 1 2 + 0.5L 2 i 2 2 + Mi 1 i 2 Since L 1 = 10 and  = 1000, L1 = L 2 = 10 mH, M = 0.5L 1 = 5mH w = 0.5(0.01)(–2.445)2 + 0.5(0.01)(–0.8391)2 + 0.05(–2.445)(–0.8391) w = 43.67 mJ

Chapter 13, Solution 21. Using Fig. 13.90, design a problem to help other students to better understand energy in a coupled circuit. Although there are many ways to solve this problem, this is an example based on the same kind of problem asked in the third edition. Problem Find I 1 and I 2 in the circuit of Fig. 13.90. Calculate the power absorbed by the 4- resistor.

Figure 13.90 Solution For mesh 1, 3630 = (7 + j6)I 1 – (2 + j)I 2 For mesh 2,

0 = (6 + j3 – j4)I2 – 2I 1 – jI 1 = –(2 + j)I 1 + (6 – j)I 2

(1) (2)

36 30   7  j6  2  j  I1  Placing (1) and (2) into matrix form,        0   2  j 6  j   I2 

 = 45 + j25 = 51.4829.05,  1 = (6 – j)3630 = 21920.54  2 = (2 + j)3630 = 80.556.57, I1 =  1 / = 4.254 –8.51 A , I 2 =  2 / = 1.563727.52 A Power absorbed by the 4-ohm resistor, = 0.5(I 2 )24 = 2(1.5637)2 = 4.89 watts

Chapter 13, Solution 22. With more complex mutually coupled circuits, it may be easier to show the effects of the coupling as sources in terms of currents that enter or leave the dot side of the coil. Figure 13.85 then becomes, -j50 Io I3 j20Ic

j40

+ 

j10I b

j60

+  Ia j30I c

 +  +

 + Ix

j30I b

j20I a

500 V + 

j80

I1

I2

100 

Ib  +

j10I a

Note the following, Ia = I1 – I3 Ib = I2 – I1 Ic = I3 – I2 and

Io = I3

Now all we need to do is to write the mesh equations and to solve for Io . Loop # 1, -50 + j20(I 3 – I 2 ) j 40(I 1 – I 3 ) + j10(I 2 – I 1 ) – j30(I 3 – I 2 ) + j80(I 1 – I 2 ) – j10(I 1 – I 3 ) =0 j100I 1 – j60I 2 – j40I 3 = 50 Multiplying everything by (1/j10) yields 10I 1 – 6I 2 – 4I 3 = - j5

(1)

Loop # 2, j10 (I 1 – I 3 ) + j80(I 2 –I 1 ) + j30(I 3 –I 2 ) – j30(I 2 – I 1 ) + j60(I 2 – I 3 ) – j20(I 1 – I 3 ) + 100I 2 =0 -j60...