Chapter 13 Solutions to Exercises

Engineering Circuit Analysis, 7th Edition

1.

v2(t ) = M 21 21

di1 ( t ) dt

Chapter Thirteen Solutions

10 March 2006

= − M 21 (40 (400)(120π )si )sin(120π t )

Taking peak values values and noting sign sign is irrelevant, 100 = M 21 21(400)(120π). Thus, M 21 21 = 663.1 μH

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10 March 2006

2. v1

= M 12

i2

=

1 M 12

di2 dt

∫

v1dt

therefore

=

⎛ 11 115 2 ⎞ o ⎜⎜ ⎟⎟ sin (120π t − 16 ) M 12 ⎝ 120π ⎠ 1

Equating peak values, M 12

=

⎛ 115 2 ⎞ ⎜ ⎟= 45 ⎜⎝ 120π ⎟⎠ 1

9.59 mH




10 March 2006

2. v1

= M 12

i2

=

1 M 12

di2 dt

∫

v1dt

therefore

=

⎛ 11 115 2 ⎞ o ⎜⎜ ⎟⎟ sin (120π t − 16 ) M 12 ⎝ 120π ⎠ 1

Equating peak values, M 12

=

⎛ 115 2 ⎞ ⎜ ⎟= 45 ⎜⎝ 120π ⎟⎠ 1

9.59 mH



3.


10 March 2006

1 and 3, 2 and 4 1 and 4, 2 and 3 3 and 1, 2 and 4



4.

(a) v1

= − L1

di1 dt

+ M


10 March 2006

di2 dt

Substituting in i1 = 30 sin 80 t and i2 = 30 cos 80 t , we find that v1 = –2400 cos 80 t – 1200 sin 80 t

⎛ ⎝

= – 24002 + 1200 2 cos ⎜ 80t − tan −1

1200 ⎞

⎟

2400 ⎠

= –2683 cos (80 t – 26.57o) V

(b) v2

= − L2

di2 dt

+ M

di1 dt

Substituting in i1 = 30 sin 80 t and i2 = 30 cos 80 t , we find that v2 = 7200 sin 80 t + 1200 cos 80 t

=

⎛ ⎝

7200 2 + 1200 2 cos ⎜ 80t − tan −1

7200 ⎞

⎟

2400 ⎠

= 7299 cos (80 t – 80.54o) V

PROPRIETARY MATERIAL. © 2007 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission.


5.

(a) v1


10 March 2006

di di = − ⎛⎜ L1 1 + M 2 ⎞⎟ dt ⎠ ⎝ dt

Substituting in i1 = 3 cos 800 t nA and i2 = 2 cos 800 t nA,we find that v1

= − ⎡⎣ −(22 ×10−6 )(3)(800) ×10−9 sin 800t − (5 ×10 −6 )(2)(800) ×10−9 sin 800t ⎤⎦ = 60.8 sin 800t pV

(b) v2

di di = + ⎛⎜ L2 2 + M 1 ⎞⎟ dt ⎠ ⎝ dt

Substituting in i1 = 3 cos 800 t nA and i2 = 2 cos 800 t nA,we find that v1

= −(15 ×10−6 )(2)(800) ×10−9 sin 800t − (5 ×10 −6 )(3)(800) ×10−9 sin 800t = 36 sin 800t pV



6.

8

di1

+ 0.4

dt

0.4

di1 dt

Let i1

+8

di2

= 5e−t

[1]

= 3e−2t

[2]

dt di2 dt

= ae −t + be −2t

and i2


10 March 2006

= ce−t + de −2t

Then from Eq. [1] we have –8a – 0.4c = 5 [3]

and

–16b – 0.8d = 0

[4]

and

–0.8b – 16d = 3

[6]

And from Eq. [2] we have –0.4a – 8c = 0 [5]

Solving, we find that a = –0.6266, b = 0.0094, c = 0.03133, and d = –0.1880

(a)

(b)

di1 dt di2 dt

(c) i1

= =

d

⎡⎣ −0.6266e −t + 0.0094e−2t ⎤⎦ = 0.6266e −t − 0.0188e −2t A/s dt d

⎡⎣0.0313e−t − 0.1880e−2t ⎤⎦ = −0.0313e −t + 0.376e −2t A/s dt

= −0.6266e −t + 0.0094e−2t A



7.


⎛ −2 di1 + 1.5 di2 ⎞ ×10−3 = 2e−t ⎜ dt ⎟ dt ⎠ ⎝

[1]

⎛ −1.5 di1 + 2 di2 ⎞ = 4e −3t ⎜ ⎟ dt dt ⎠ ⎝

[2]

= ae −t + be −3t

= ce−t + de −3t

Let i1

and i2

10 March 2006

Then from Eq. [1] we have 3

2a – 1.5c = 2×10 [3]

and

6b – 4.5d = 0

[4]

And from Eq. [2] we have 1.5a – 2c = 0 [5]

and

4.5b – 6d = 4×10

3

[6]

Solving, we find that a = 2286, b = -1143, c = 1714, and d = –1524

(a)

(b)

di1 dt di2 dt

(c) i2

= =

d

⎡⎣ 2286e−t − 1143e −3t ⎤⎦ = −2286e −t + 3429e −3t dt d

⎡⎣1714e−t − 1524e−3t ⎤⎦ = −1714e −t + 4572e −3t dt

A/s

A/s

= 1714e−t + 4572e −3t A




10 March 2006

8. (a)

−V 2 = jω 0.4 1∠0 V 2 = − j100π× 0.4 × 1∠0

= 126 ∠ 90 V o

Thus, v(t ) = 126 cos (100 πt + 90o) V (b)

Define V2 across the 2-H inductor with + reference at the dot, and a clockwise currents I1 and I2, respectively, in each mesh. Then, V = -V 2

and we may also write V2 = jω L2 I2 + jω MI1

-V = jω L2

or

V

+ jω M

10

Solving for V, V=

− ( j100π )(0.4) 1 + ( j100π )(2)

=

125.7∠ − 90

o

=

1 + j 62.83

125.7∠ - 90o 62.84∠89.09

o

=

2.000 ∠ - 179.1o

Thus, v(t ) = 2 cos (100πt – 179.1 ) V. o

(c)

Define V1 across the left inductor, and V2 across the right inductor, with the “+” reference at the respective dot; also define two clockwise mesh currents I1 and I2. Then, V1 = jω L1 I 1 + jω M I 2 V2

Now

=

1I

1∠0 − V 1 4

= jω L2 I 2 + jω M I 1

and

V out

and I 2

=

=−

2V

V out

10

V out 1∠0 − V 1 ⎤ j M EQN 1 ⇒ V 1 = jω L1 ⎡⎢ + ω 10 ⎣ 4 ⎥⎦ V 1∠0 − V 1 ⎤ EQN 2 −V out = jω L2 out + jω M ⎡⎢ 10 ⎣ 4 ⎥⎦ − jω M ⎤ ⎡1 − jω L1 ⎡ jωL1 1∠0 ⎤ ⎢ ⎥ ⎥ ⎡ V ⎤ ⎢ 4 10 4 ⎢ ⎥ ⎢ 1 ⎥=⎢ ⎥ V ω ω ω ∠ j M j L j M 1 0 2 ⎥ ⎣ out ⎦ ⎢ ⎢ ⎥ −1 + ⎢⎣ 4 ⎥ ⎢ ⎥⎦ ⎣ 10 ⎦ 4 ⎡1 − j 39 − j12.6 ⎤ ⎡ V 1 ⎤ ⎡39.3 j ⎤ ⎢ j 31.4 −1 + j 62.8⎥ ⎢V ⎥ = ⎢31.4 j ⎥ ⎦ ⎣ ⎦ ⎣ out ⎦ ⎣

Solving, we find that Vout (= V) = 1.20 ∠ -2.108 V and hence o

v(t ) = 1.2 cos (100πt – 2.108 ) V. o



9. (a)


10 March 2006

j j I2 , (2000 + 500) j j I1 = 0 100 = (50 + 200) I1 + 300 I2 + 300

⎛ − j 3 900 ⎞ , 100 = ⎜ 50 + j 200 + ⎟ I 1 j 20 + j 5 20 5 + ⎝ ⎠ 900 + j 4250 ∴100 = I1 ∴ I1 = 0.47451 ∠ − 64.01° A 20 + j 5 ∴ I2 =

1

∴ PS ,abS = − ×100 × 0.4745cos 64.01° = −10.399 W 2

1

1

= × 50 × 0.4745 = 5.630 W, P2000 = × 2000 × 0.4745

(b)

P50

(c)

0 each

(d)

0

2

2

2

2

2

− j3 × = 4.769 W 20 + j 5



= 4t

A, iS 2


10 March 2006

= 10t A

10.

iS 1

(a)

v AG

= 20 × 4 + 4 × 10 = 120 V

(b)

vCG

= −4 × 6 = −24 V

(c)

v BG

= 3 ×10 + 4 × 4 − 6 × 4 = 30 + 16 − 24 = 22 V




10 March 2006

11. (a)

Vab,oc

=

100 50 + j 200

( − j 300) = 145.52∠ − 165.96° V

j j I2SC , 500 j I2SC 100 = (50 + 200) I1 + 300

j I1 300

=0

⎤ + j 300⎥ I2 SC ∴ I2 SC = 1.1142 ∠158.199° A ⎟ 3 3⎠ ⎦ 145.52∠ − 165.96° ∴ Zth = Vab,bc / I2 SC = = 130.60∠35.84° = 105.88 + j 76.47 Ω 1.1142∠158.199° ∴ I1 = −

(b)

Z L

5

⎡ ⎣

+

⎛ ⎝

I2 SC , 100 = ⎢ (50 + j 200) ⎜ −

= 105.88 − j 76.47 Ω ∴

IL

=

5⎞

145.52 2 × 105.88

= 0.6872 A

1

∴ P L max = × 0.6872 2 × 105.88 = 25.00 W 2




10 March 2006

12. KVL Loop 1

100 ∠0 = 2(I1 – I2) + jω 3 (I1 – I3) + jω 2 (I2 – I3)

KVL Loop 2

2(I2 – I1) + 10I2 + jω 4 (I2 – I3) + jω 2 (I1 – I3) = 0

KVL Loop 3

5I3 + jω 3 (I3 – I1) + jω 2 (I3 – I2) + jω 4 (I3 – I2) + jω 2 (I3 – I1) = 0

∴LINEAR EQUATIONS ⎡ 2 + jω 3 − 2 + jω 2 − jω 5 ⎤ ⎡ I1 ⎤ ⎡100∠0⎤ ⎢− 2 + jω 2 12 + jω 4 − jω 6 ⎥ ⎢I ⎥ = ⎢ 0 ⎥ ⎢ ⎥ ⎢ 2⎥ ⎢ ⎥ ⎢⎣ − jω 5 ⎥ ⎢ ⎥ ⎢ 5 + j11⎦ ⎣I 3 ⎦ jω 2 ⎣ 0 ⎥⎦ Since ω = 2π f = 2π(50) = 314.2 rad/s, the matrix becomes

⎡ 2 + j 942.6 − 2 + j 628.4 ⎢− 2 + j 628.4 12 + j1257 ⎢ ⎢⎣ − j1571 j 628.4

− j1571 ⎤ ⎡ I1 ⎤ ⎡100∠0⎤ − j1885 ⎥⎥ ⎢⎢I 2 ⎥⎥ = ⎢⎢ 0 ⎥⎥ ⎢⎣ 0 ⎥⎦ 5 + j 3456 ⎥⎦ ⎢⎣I 3 ⎥⎦

Solving using a scientific calculator or MATLAB, we find that I1 = 278.5 ∠ -89.65o mA, I2 = 39.78 ∠ -89.43o mA, I3 = 119.4 ∠ -89.58o mA. Returning to the time domain, we thus find that i1(t ) = 278.5 cos (100 πt – 89.65 ) mA, i2(t ) = 39.78 cos (100 πt – 89.43 ) mA, and o i3(t ) = 119.4 cos (100 πt – 89.58 ) mA. o

o




10 March 2006

13. vs v

=

10t 2u (t ) t 2 + 0.01

= 0.01i′S ∴ i′S =

=x 0.015i′ =S

∴ iC = 100 × 10

15t 2 t 2 + 0.01

−6

∴ iC = 15 ×10−4

v′x = 10

1000t 2 t 2 + 0.01

u (t ), 100v

−4

=x

u (t )

1500t 2 t 2 + 0.01

u (t )

2 2 ⎛ 15t 2 ⎞ −4 (t + 0.01)2t − t × 2t ( ) 15 10 ( ) = × u t u t ⎜ ⎟ (t 2 + 0.01) dt ⎝ t 2 + 0.01 2 ⎠

d

0.02t (t 2 + 0.01) 2

∴ iC (t ) =

30t (t 2 + 0.01) 2

μA,

t > 0



14.


(a)

v A (t ) = L1i′1 − Mi′2, v B (t ) = L1i ′1 − Mi′2 + L 2i′2 − Mi′1

(b)

V1( jω ) = jω L1 IA + jω M(IB + IA)

10 March 2006

V2( jω ) = jω L2 (IB + IA) + jω MIA




10 March 2006

15. (a)

100 = j5ω (I1 – I2) + j3ω I2 + 6(I1 – I3)

[1]

(4 + j4ω )I2 + j3ω (I1 – I2) + j2ω (I3 – I2) + j6ω (I2 – I3) – j2ω I2 + j5ω (I2 – I1) – j3ω I2 = 0 [2] 6 (I3 – I1) + j6ω (I3 – I2) + j2ω I2 + 5 I3 = 0

[3]

Collecting terms,

(b)

(6 + j5ω ) I1 – j2ω I2 – 6 I3 = 100

[1]

- j2ω I1 + (4 + j5ω ) I2 – j4ω I3 = 0

[2]

-6 I1 - j4ω I2 + (11 + j6ω ) I3 = 0

[3]

For ω = 2 rad/s, we find (6 + j10) I1 – j4 I2 – 6 I3 = 100 - j4 I1 + (4 + j10) I2 – j8 I3 = 0 -6 I1 – j8 I2 + (11 + j12) I3 = 0 o Solving, I3 = 4.32 ∠ -54.30 A




10 March 2006

16. (a)

= jω L1 I a + jω M I b Vb = jω L2 I b + jω M I a

= I 1 I b = − I 2

Va

V1

V2

(b)

Ia

= I1 R1 + V a = 1IR1 + jω L1 aI+ jω M bI = 1IR1 + jω L1 1I− jω M 2I = I 2 R2 − V b = 2IR2 − jω L2 = 2I R2 + jω L2

− jω M 2I− jω M

bI

aI

1I

Assuming that the systems connecting the transformer are fully isolated.

= jω L1 I a + jω MI b Vb = jω L2 I b + jω MI a

= − I 1 I b = − I 2

Va

V1

V2

= I1 R − V a = 1IR − jω L1 aI− jω M bI = 1IR + jω L1 1I+ jω M 2I

Ia

= Vb + I b R2 = − 2I R2 + jω L2 bI+ jω M aI = − 2IR2 − jω L2 2I− jω M 1I




10 March 2006

17. (a) Z

ω2 (0.2)2 2 + jω0.1 + 5 + jω 0.5 jω0.5 ω2 (0.2) 2 5ω2 (0.2) 2 − 2 2 + jω 0.1 + 2 5 + (ω0.5) 2 5 + (ω0.5) 2 ⎡ 0.2ω2 0.02ω2 ⎤ + jω ⎢0.1 − 2+ ⎥ 25 + 0.25ω2 25 + 0.25ω2 ⎦ ⎣

(b)

(c)

Zin( j jω ) at ω = 50 is equal to 2 + 0.769 + j(50)(0.023) = 2.77 + j1.15 Ω.




10 March 2006

18. Z in

= Z11 +

ω2 M 2 Z 22

ω2 M 2 = jω50 × 10 + − 8 + jω10 ×10 10 3 2 ω2 M ωj10 ×10−3 ω2 M 8 −3 ⇒ Z in = jω50 × 10 + 2 − 2 8 + (ω10 1 0 ×10 10−3 ) 2 8 + (ω10 ×10 1 0−3 ) 2 ⎡ 10 ×10 −3 ω2 M 2 ⎤ ω2 M 2 8 −3 = 2 + jω ⎢50 × 10 − 2 ⎥ 8 + (ω10 × 10−3 ) 2 8 + (ω10 × 10 −3 ) 2 ⎦ ⎣ −3

In this circuit the real power delivered by the source is all consumed at the speaker, so 1 2 2 V ⎛ 20 ⎞ ω2 M 2 8 P = rms ⇒ 3.2 = ⎜ × ⎟ −3 2 2 R ⎝ 2 ⎠ 8 (ω10 ×10 )

ω2 M 2 8 202 ⇒ 2 = 8 + (ω10 10 ×10 −3 ) 2 2 × 3.2

= 62.5 W



= 2cos1 2cos10 0t A, iS 2 = 1.2co 1.2cos1 s10 0t A


19.

iS 1

(a)

= 0.6(−20 si sin 10t ) − 0.2( −12 si sin 10t ) + 0.5( −32 si sin 10t ) + 9.6 co cos10t ∴ v1 = 9.6 co cos10t − 25.6 si sin 10t = 27.34 co cos (1 (10t + 69.44 ° ) V

(b)

= 0.8( −12 si sin 10t ) − 0.2( −20 si sin 10t ) − 16 si sin 10t + 9.6 co cos 10t ∴ v2 = 9.6 co cos10t − 21.6 si sin 10t = 23.64 co cos (1 (10t + 66.04 °) V

(c)

PS1

10 March 2006

v1

v2

1

1

2

2

= × 27.34 × 2 cos 69.44 ° = 9.601 W, PS 2 = × 23.64 ×1.2 cos 66.04 ° = 5.760 W




10 March 2006

20.

= jω8 I a + jω4 I b Vb = ω j 10 I b + ω j 4I a = ω j 10 I b + ω j 5Ic Vc = jω6 I c + j ω5 I b Va

*

Also

I

= − Ia = − Ib =

Ic

Now examine equation *. j 10 I− ω j 4 I= − ω j 10 I+ ω j5 − ω

Ic

∴ the only solution to this circuit is I = and hence v(t ) = 120 cos ωt V.




10 March 2006

21. 100 = j10 I1 − j15 I2 0 = 200 j I2 − 15 j I1 − 15 j I L 0 = (5 + j10) I L − j15 I2

∴ I2 =

5 + 10j

∴ 0 = ⎜⎛ ⎝

15j

I

1+ 2 j I 3j

=L

⎛ 1+ 2 j 0 200 ∴ = j L ⎜ 3j− ⎝

400 200 ⎞ j − 15 j + ⎟ I L − 15j I1 3 3 ⎠

2 ∴100 = ⎡⎢ (66.67 + j118.33) − 5 − ⎣3 ∴ I L = 1.2597∠ − 60.21° A

∴ I1 = ⎤ ⎦

j10 ⎥ I L

⎞ − L j15 I1 ⎠ j118.33 + 66.67 j15 ⎟ I

j15

IL

= (39.44 + j 68.89) I L




10 March 2006

= 2 cos10t A, t = 0

22.

is

(a)

a − b O.C.

(b)

a − b S.C.

1

1

2

2

= 2∠0° A,

M=

∴ w(0) = × 5 × 22 + × 4 × 22 = 10 + 8 = 18 J ω = 10,

IS

( 30j+ 5) I2 − 10j 3 × 2,

∴ I2 =

1 2

j 20 3

5 + j30

∴ i2 (0) = 1.1235− ∴ w(0) = 10 + 8 −

12

=

3H

= 1.1390∠9.462 °A ∴ 2 =i1.1390 cos (10 + 9.462 °) A t

3 × 2 ×1.1235 +

1 2

× 3 ×1.12352 = 16.001 J




10 March 2006

23. Vs = 12∠0° V rms, ω = 100 rad/s j I1 + 100(0.4K) j j I2 + 40K j I1 = 0 12 = (6 + 20) I2 , (24 + 80)

⎡ ⎤ 3 + j10 ∴12 = ⎢(6 + j 20) + j 40K ⎥ I2 − j 5K − j5K ⎣ ⎦ − 60K j 18 − 200 + 60j+ 60j+ 200K 2 ∴12 = I2 ∴ I2 = − j 5K −182 + 200K 2 + ∴ I1 =

3 + j10

∴ P24 =

I2

602 K 2 24 (200K 2 − 182) 2 + 1202

=

86, 400 K 2 40, 000K 4 − 72,800K 2 + 47, 524

j120

=

2.16K 2 K 4 − 1.82K 2 + 1.1881

W



24.


M

•

•

2Ω

k =

→

Zin

L1 L2

= 2 × 80 × 10−6 = 12.6μH ω2 M2 R22 − jM2 ω2 X22 = Z11 + 2 + 2 2 2 R22 + X22 R22 + X22

M=

Z11

M

ω = 250k rad / s

j10 Ω

Zin

10 March 2006

L1 L2

= j × 250 ×103 × 2 ×10 −6 = j 0.5

= 2Ω 3 −6 X 22 = (250 ×10 ) (80 ×10 ) = 20 R22

Thus, Zin = j0.5 + 19.8/404 – j198/ 404 = 0.049 + j0.010 Ω.



25.

ω = 100 rad/s

(a)

K1 → j 50Ω, K 2

→


10 March 2006

j 20Ω, 1H → j100 Ω

100 = j200 I1 − j50 I2 − j 20 I3 0 = (10 + j100) I2 − j 50 I1 0 = (20 + j100) I3 − j 20 I1

⎡ 5j 2j ⎤ j − 5j ∴10 = ⎢ 20 − 2j ⎥ I1 j j + + 2 + 10j 1 + 10j 1 10 2 10 ⎣ ⎦ ⎛ 25 4 ⎞ ∴10 = ⎜ j 20 + + ⎟ I1 ∴ I1 = 0.5833 ∠ − 88.92° A, I2 = 0.2902∠ − 83.20° A, + + 1 10 2 10 j j ⎝ ⎠ I3 = 0.11440 ∠ − 77.61° A ∴ P10Ω = 0.2902 2 ×10 = 0.8422 W ∴ I3 =

(b)

P20

(c)

Pgen

2j

I1 , I2

=

5j

I1

= 0.1144 2 × 20 = 0.2617 W = 100 × 0.5833cos 88.92° = 1.1039 W



26. (a)

k =


10 March 2006

M L1 L2

⇒ M = 0.4 5 ×1.8 = 1.2H (b)

+ I 2 = I 3 ⇒ I2 = I3 − I1 I1

−t

− t

= 5 × 10 − 4 × 10

10

5

(c)

The total energy stored at t = 0. I1

W total

= 4 A =

1 2 1

L1

= 1A

I 2 2

I1

+

1 2

L2

2

I2

+ M 12 I1 I2

1

= × 5 × 16 + ×1.8 × 1 − 1.2 × 4 × 1 2 2 = 40 + 0.9 − 4.8

= 36.1J




10 March 2006

27. K → j1000K L1L 2 , L1 → j1000L1, L 2

∴ Vs = (2 + j1000L1 ) I1 −

j1000K

→ j1000L 2

L1L 2 I2

0 = − j1000K L1L 2 I1 + (40 + j1000L 2 ) I2

ω = 1000 rad/s ∴ I1 =

∴

V2 Vs

j1000K L1L 2

I2

j1000K L1L 2 6 2 80 + j 40, 000L1 + j 2000L 2 − 10 L1L 2 (1 − K )

=

j 40, 000K L1L 2

80 − 106 L1 L2 (1 − K 2 ) + j (40, 000L1 + 2000L 2 )

(a)

L1 = 10−3 , L2

(b)

L1

= 1,

∴

V2

(c)

I2

j1000K L1L 2

6 2 (2 + j1000L1 )(40 + j1000L 2 ) + 10 K L1L 2

∴ Vs = ∴ I2 =

40 + j1000L 2

L1

VS

=

= 1,

L2

= 25 × 10−3 , K = 1 ∴

= 25,

K

= 0.99 ∴

V2 V3

j198, 000

80 − 497,500 + j90, 000

L2

= 25,

K

=1 ∴

V2 Vs

=

=

V2 Vs

=

j 40 × 5

80 − 0 + j (40 + 50)

=

j 200

80 + j 90

= 1.6609∠ 41.63°

j 40, 000 × 0.99 × 5

80 − 25 × 10 (1 − 0.99 ) + j (40, 000 + 50, 000) 6

2

= 0.3917∠ − 79.74° j 40, 000 × 5

80 − 0 + j 90, 000

=

j 200, 000

80 + j90, 000

= 2.222∠0.05093 °



28. (a)


10 March 2006

= 10 mH, L CD = 5 mH , ABOC L AB,CDSC = 8 mH ∴ L1 = 10 mH, L 2 = 5 mH, 8 = 10 − M + M (5 − M) (mH) M(5 − M) 2 ∴ 8 = 10 − M + , ∴ 5M = (10 − 8)5 + 5M − M ∴ M = 3.162 mH (= L

AB , CDOC

5

∴K = (b)

3.162 50

10)

∴ K = 0.4472

Dots at A and D, i1

= 5 A, wtot = 100 mJ

1

1

2

2

∴100 ×10−3 = ×10 ×10 −3 × 25 + × 5 ×10 −3 i22 − 100 = 125 + 2.5i22

−5

10 i2

∴i22 − 2

10 × 5i2 ×10 −3

10 i2 +10 = 0, i2

=

2 10 ± 40 − 40 2

=

10

∴ i2 = 3.162 A



29.


10 March 2006

Define coil voltages v1 and v2 with the “+” reference at the respective dot. Also define two clockwise mesh currents i1 and i2. We may then write: v1

= L1

v2

= L2

d I1 dt d I 2

+ M

dt

+M

d I 2 dt d I 1 dt

M

=k

L1 L2

ω = 2π60 rad / s

or, using phasor notation,

= jω L1 I 1 + jω M I 2 V2 = jω L2 I 2 + jω M I 1 V1

100∠0 = 50

−25

+ jω L1 1I+ jω M 2I 2I = jω L2 2I + jω M 1I

Rearrange:

or

1I

[50 + jω L1 ] I1 + jω MI 2

= 100∠0 jω M I1 [−25 + jω L2 ] I 2 = 0

jω M ⎤ ⎡ I 1 ⎤ ⎡100∠0⎤ ⎡50 + jω L1 ⎢ jω M ⎥ ⎢ I ⎥ = ⎢ 0 ⎥ j L − + ω 25 ⎣ ⎦ ⎣ 2⎦⎣ 2⎦

We can solve for I2 and V2 = −25I2: V2 =

−

j1.658 k L1L 2

+

1




10 March 2006

30. i1

= 2 cos 500t A 1

Wmax at t = 0 1

1

∴ wmax = × 4 × 22 + × 6 × 22 + × 5 × 2 2 + 3× 2 2 2 2 2 = 8 + 12 + 10 + 12 = 42 J




2

31.

(a) Reflected impedance = Z 22

= 2 + 7∠320 + jω 10 −2

ω

M 2

10 March 2006

Z 22

.

where ω = 100π

Thus, the reflected impedance is 4.56 – j3.94 nΩ (essentially zero).

–2

–9

(b) Zin = Z11 + reflected impedance = 10 + jω (20×10 ) + (4.56 – j3.94)×10

= 10 + j62.84 Ω (essentially Z11 due to small reflected impedance)



32.

Reflected impedance =

ω

2

M 2 Z22

(


=

2

ω

10 March 2006

M 2

. 3.5 + j(ω L2 + X L )

We therefore require 1 + jω 3 ×10

−3

2

ω

10−6

) = 3.5 + j (10−

3

ω

+ X L )

.

Thus, 2 −6 ⎡ ω 10 ⎢ − 3.5 − j LX= − ⎢⎣1 + jω ( 3 ×10−3 )

10j

−3

⎤ ω ⎥ = −0.448 + ⎥⎦

j 3.438 .

This is physically impossible; to be built, X L must be a real number.



33.


10 March 2006

M = 5 H. L1 – M = 4 H, therefore L2 – M = 6 H, therefore

L1 = 9 H L2 = 11 H.



34.


10 March 2006

Lz = L1 – M = 300 – 200 = 100 mH Ly = L2 – M = 500 – 200 = 300 mH Lx = M = 200 mH



35. (a)

All DC: L1− 2

(b)

AB SC: L1− 2

= −1 + 2

(c)

BC SC: L1− 2

= 2 + (−1)

(d)

AC SC: L1− 2

= (2 − 1)


10 March 2006

= 2 −1 = 1 H 8 = 0.6 H 9 = 2 − 9 / 8 = 0.875 H

(1 + 2) = 1 3 = 0.750 H




10 March 2006

36. (a)

IL VS

=

=

(b)

⎛ j 2ω ⎞ 1 ⎜ ⎟ j 2ω (20 + jω ) ⎜⎝ 20 + j 3ω ⎠⎟ 15 + j 3ω + 20 + j 3ω j 2ω

300 − 11ω 2

+ j145ω

vs (t ) = 100u (t ), is (0) = 0, iL (0) = 0, s1,2

=

−145 ±

1452 − 13, 200 22

= − 2.570, − 10.612

= 0, ∴ i L= Ae −2.57t + Be −10.61t , ∴ 0 = A + B 100 = 15is + 5i′s − 2i′L, 0 = 20iL + 3i′L − 2i′s At t = 0 + : 100 = 0 + 5i′s (0 + ) − 2i′L (0 + ) and 0 = 0 + 3i L′ (0+ ) − 2i′s (0+ ) ∴ i′s (0+ ) = 1.5i′L (0+ ) ∴100 = 7.5i′L (0+ ) − 2i′L (0+ ) = 5.5i′L (0 + ) ∴ i L′ (0+ ) = 18.182 A/s ∴18.182 = −2.57A − 10.61B = −2.57A + 10.61A = 8.042A ∴ A = 2.261, B = −2.261, i L (t ) = 2.261(e−2.57t − e −10.612t ) A, t > 0 i

= i Lf+ i

L

,i

Ln

Lf




10 March 2006

37.

(a)

Open-Circuit T×A Z oc

= jω4 M Ω T ×B Z oc = jω4 M Ω (b)

Short-Circuit T×A Z SS

(c)

= Z SST ×B = − jω4 M Ω + jω8 jω10 M Ω

If the secondary is connected in parallel with the primary T×A

= − jω4 − jω10 + jω8 MΩ

T ×B

= jω26 jω12 − jω8 MΩ

Z in Z in



38.


10 March 2006

Define three clockwise mesh currents I1, I2, and I3 beginning with the left-most mesh. Vs = j8ω I1 – j4ω I2 0 = -4 jω I1 + (5 + j6ω ) I2 – j2ω I3 0 = - j2ω I2 + (3 + jω ) I3

Solving, I3 = jω / (15 + j17ω ). Since Vo = 3 I3, Vo VS

=

j 3ω

15 + j17ω



39.


10 March 2006

Leq = 2/ 3 + 1 + 2 + 6/5 = 4.867 H Z( jω ) = 10 jω (4.867)/ (10 + jω 4.867) = j4.867ω / (1 + j0.4867ω ) Ω.




10 March 2006

40.

ω = 100 rad/s Vs = 100∠0° V rms

(a)

Zina −b

= 20 + j 600 +

j j 400(10 − 200)

10 + j 200

= 20 +

j 600 +

j 80, 000 + 4, 000

10 + j 200

= 210.7∠73.48o V and Voc = 0. (b)

VOC ,cd = Zincd , VS

100( j 400) 20 + j1000

= 39.99∠1.146° V rms

= 0 = − j 200 +

j j + 600) 400(20

20 + j1000

= − j 200 +

j −240, 000 + 8, 000 = 40.19 ∠85.44 °Ω 20 + j1, 000



41.

(a)

L1 Z


10 March 2006

= 1 H, L 2 = 4 H, K = 1, ω = 1000 rad/s = 1000 Ω ∴ Z = jin1000 + L

106 ×1× 4 j 4000 + 100

= 24.98 +

4 × 10

j0.6246 Ω

6

(b)

(c)

Z

ZL

= j1000 × 0.1 ∴ Z = L = − j100 ∴ Zin =

j1000 + in

j1000 +

j 4000 + j100

4 ×10

= j 24.39 Ω

6

j 4000 − j100

= − j 25.46 Ω




10 March 2006

42. L1

= 6 H, L2 = 12 H, M = 5 H

#1, L inAB,CDOC = 6 H #2, L inCD , ABOC = 12 H #3, L inAB,CDSC = 1 + 7 5 = 3.917 H #4, L inCD , ABSC = 7 + 5 1 = 7.833 H #5, LinAC ,BDSC = 7 + 1 = 8 H #6, L inAB, ACSC ,BDSC = 7 1 + 5 = 5.875 H #7, L inAD,BCSC = 11 + 17 = 28 H #8, L inA B, ADSC = −5 + 11/17 = 1.6786 H




10 March 2006

43.

ω2 M 2 Z in = Z11 + R jX22 22 + 1

ωC

1

= 31.83 ⇒ ω =

= 314 rad / s 31.83 × C ie. a 50Hz system

ω2 k 2 L1 L2 Z in = 20 + jω100 × 10 + 2 − j 31.83 ω2 k 2 L1L2 2 jω2k 2 L1L2 31.83 −3 Z in = 20 + jω100 × 10 + − 22 + 31.832 22 + 31.832 ⎡ 493 − 7840 ⎤ 2k = 20 + 31.4 j +⎢ j ⎣1020 1020 ⎥⎦ = 20 + j31.4 + [0.483 − j 7.69]k 2 Z in (k = 0) = 20 + j 31.4 Ω Z in (k = 0.5) = 20.2 + j 27.6 Ω Z in (k = 0.9) = 20.4 + j 24.5 Ω Z in (k = 1.0) = 20.5 + j 23.7 Ω −3

(a) (b) (c) (d)




44.

↑ L1 → 125 H, L 2 → 20 H, K = 1, ∴ M =

(a)

Zina −b

= 20 + j 7500 +

= 20 + j 7500 +

2500

= 50 H, jωM =

10 March 2006

j5000 Ω

j 5000(10 − j 3000)

10 + j 2000

15 × 106 + j50, 000 10 + j 2000

= 82.499∠0.2170°Ω

= 82.498 + j0.3125− Ω VOC = 0 (b)

VOC ,cd = Zincd , VS

100( j 5000) 20 + j12,500

= 39.99995∠0.09167° V rms

= 0 = − j3000 +

j 5000(20 + j 7500)

20 + j12, 500

= 3.19999 +

j 0.00512 Ω




10 March 2006

45.

∴

I a

=

I b

=

280 1280 1000 1280

× 2 = 0.438A × 2 = 1.56A

∴ I 1 = 1.56A ⇒ I 2 = 5 ×1.56 = 7.8A ⇒ I 3 = 1.5 × 7.8 A = 11.7A ⇒ P(1k ) = I a2 R = 0.4382 ×1×103 = 192W ⇒ P(30Ω) = I12 R = (1.56) 2 × 30 = 73W ⇒ P(1Ω) = I 22 R = 7.82 × 1 = 60.8W ⇒ P(4Ω) = I 32 R = 11.7 2 × 4 = 548W



46. (a)

R L sees 10 × 42


10 March 2006

= 160 Ω ∴ use R L = 160 Ω

2

⎛ 100 ⎞ ×10 = 250 W P L max = ⎜ ⎟ ⎝ 20 ⎠ (b)

R L I2

= 100 Ω

= I1 / 4,

V2

= 4 V1 ∴ I X =

3V ∴100 = 10 ⎛⎜ I1 1 ⎞⎟ + V1 , ⎝ 40 ⎠

I1 4

V2 − V1

=

40 3V1 40

+

=

3V1 40

4V1 100

∴ I1 = 0.46V1 ∴100 = 10(0.46V1 − 0.075V1) + V1 = 4.85 V1 ∴ V1 = ∴ V2 = 4V1 =

400 4.85

= 82.47 V ∴ P L =

82.47 100

2

100 4.85

= 68.02 W




10 March 2006

47. I2

=

V2 8

∴ I1 =

V2 40

, V1

= 5V2

∴100 = 300(C + 0.025) V2 + 5V2 ∴ V2 = (a)

(b)

(c)

100 12.5 + 300C C=0

∴ V2 = 8 V ∴ P L =

C = 0.04

∴ V2 =

C = −0.04

100 24.5

∴ V2 =

100 0.5

82 8

=8 W

100 ⎞ ∴ P L = ⎛⎜ ⎟ ⎝ 24.5 ⎠

2

1 8

= 200 V ∴ P L =

= 2.082 W (neg. fdbk )

200 2 8

= 5000 W (pos. fdbk )




10 March 2006

48.

= 1 V ∴ Ix = 0.05 A, V2 = 4 V 4 −1 = 0.05 A I2 + 20 × 0.05 ∴ I2 =

Apply Vab

∴ 4 = 60

60

∴ I1 = 0.2 A ∴ Iin = 0.25 A ∴ R th = 4 Ω,

Vth

=0




10 March 2006

49. Pgen

= 1000 W, P100 = 500 W

∴ I L =

500

= 5 A, VL = 100 5 V 100 1000 = 10 A ∴ V1 = 100 − 40 = 60 V IS = 100 Now, P25

= 1000 − 500 − 10 2 × 4 = 100W ∴ I X =

I x = b 5

= 2, b =

2 5

100 25

= 2 A; also

= 0.8944

Around center mesh: 60a = 2 × 25 + 100 5

1 0.8944

∴a =

300 60

=5



50. (a)

2

⎛ 4 ⎞ 16 Ω, 16 + 2 = 22 Ω, 3× ⎜ ⎟ = 3 3 3 ⎝ 3⎠ 66 + 25 = 91Ω

100 91


22 3

(3) 2

10 March 2006

= 66 Ω

= 1.0989∠0° A = I1

(b)

I2

= 3I1 = 3.297∠0° A

(c)

I3

= − × 3.297 = 4.396∠180° A

(d)

P25

(e)

P2

= 3.297 2 × 2 = 21.74 W

(f)

P3

= 4.3962 × 3 = 57.96 W

4 3

= 25 ×1.0989 2 = 30.19 W



51. V1 = 2.5 V2 , I1 = 0.4 I2 , I50

=

∴ 60 = 40(0.4

∴ I2 =

I2 ) − 2.5V2


10 March 2006

I2 + 0.1 V2

Also, 60 = 50 ( I2 + 0.1 V2 ) + V2

60 + 2.5 V2 16

= 50 I2 + 6V2

60 + 2.5 V2 ⎞ ∴ 60 = 50 ⎛⎜ ⎟ + 6 V2 = 187.5 + (7.8125 + 6) V2 16 ⎝ ⎠ 60 − 187.5 ∴ V2 = = −9.231 V 13.8125




10 March 2006

52. 400 52 16 22 2 2

= 16 Ω, 16

= 4 Ω ∴ Is =

48 = 12Ω, 12 + 4 = 16 Ω 10 4 +1

= 2 A ∴ P1 = 4 W

= 1 A ∴ P4 = 4 W, 10 − 2 × 1 = 8 V

8 × 2 = 16 V, 16 − 4 × 1 = 12 V, 12 2 / 48 = 3 W P400

=

602 400

= P48 , 12 × 5 = 60 V

=9W




10 March 2006

53. I1

= 2I2 , 2I2 = Is + I x ∴ I x + Is − 2I2 = 0

100 = 3I s

+

1 2

(4I 2 + 20I 2 − 20I x )

∴10I x − 3I s − 12I 2 = −100 100 = 3 I s − 5I x + 20I 2 − 20I x ∴ 25I x − 3I s − 20I2 = −100 −2 0 1 −100 −3 −12 ∴ I X =

−100 −3 −20 −800 0 + 100( −26) − 100(−18) = = = 4.819 A −2 1 1 1(60 − 36) − 10(− 20 − 6) + 25(− 12 − 6) −166 10

−3 −12

25

−3 −20




10 March 2006

54. (a)

50 10 =

25 3

100 ⎞ ∴ P10 AB = ⎛⎜ ⎟ ⎝ 3 ⎠

= 1× 3 ×

VCD (b)

25

Ω ∴ V AB = 1× 4 ×

25 3

2

1 10

=

1000 9

3

=

100 3

V

= 111.11 W 252

= 25 V, P10CD =

10

= 62.5 W

Specify 3 A and 4 A in secondaries I

= I f+ 4

AB

ICD

= − Ib − 3 ∴

25

(I f

+ 4) =

25

3 ∴ 2I f = −7, I f = −3.5 A

∴ V AB= V CD= ∴ P10 AB= P10

25 3

(−3.5 + 4) =

25 = ⎛⎜ ⎞⎟ CD ⎝6⎠

2

1 10

3

(− I f

25 6

− 3)

V

= 1.7361 W



55.


10 March 2006

Corrections required to the problem text : both speakers that comprise the load are 4-Ω devices. We desire a circuit that will connect the signal generator (whose Thévenin resistance is 4 Ω) to the individual speakers such that one speaker receives twice the power delivered to the other. One possible solution of many:

We can see from analysing the above circuit that the voltage across the right-most 1.732 or 2 times that across the left speaker. Since power is speaker will be 1.225 proportional to voltage squared , twice as much power is delivered to the right speaker.



56.


10 March 2006

(a) We assume Vsecondary = 230∠0 V as a phasor reference. Then, o

Iunity PF load =

I0.8 PF load =

8000 230

15000 230

Thus, Iprimary =

∠0 o =

34.8∠0o A

∠(− cos−1 0.8) =

230 2300

(34.8∠0

+

o

and

65.2∠ − 36.9o A 65.2∠ - 36.9o

)

= 0.1 (86.9 – j39.1) = 9.5 ∠-24.3 A o

(b) The magnitude of the secondary current is limited to 25×10 /230 = 109 A. If we include a new load operating at 0.95 PF lagging, whose current is 3

I0.95 PF load = | I0.95 PF load | ∠ (-cos-1 0.95) = | I0.95 PF load | ∠ -18.2o A,

then the new total secondary current is 86.9 – j39.1 + | I0.95 PF load | cos 18.2

o

o

– j | I0.95 PF load | sin 18.2 A.

Thus, we may equate this to the maximum rated current of the secondary: 109

=

(86.9 + | I

0.95 PF load

| cos 18.2o

)

2

+

(39.1 + | I

0.95 PF load

| sin 18.2o

)

2

Solving, we find | I 0.95 PF load |

2

=

- 189 ± 1892

+ (4)(2800)

2

So, |I0.95 PF load | = 13.8 A (or –203 A, which is nonsense). This transformer, then, can deliver to the additional load a power of 13.8×0.95×230 = 3 kW.



57.


10 March 2006

After careful examination of the circuit diagram, we (fortunately or unfortunately) determine that the meter determines individual IQ based on age alone. A simplified version of the circuit then, is simply a 120 V ac source, a 28.8-k Ω resistor and a 2 (24 )RA resistor all connected in series. The IQ result is equal to the power (W) dissipated in resistor RA divided by 1000. 2

P=

Thus, IQ =

⎛ ⎞ 120 ⎜⎜ ⎟⎟ × 576R A 3 × + 28 . 8 10 576R A ⎠ ⎝ 2

⎛ ⎞ 120 ⎜⎜ ⎟ × 576 × Age 3 1000 ⎝ 28.8 × 10 + 576 × Age ⎠⎟ 1

(a) Implementation of the above equation with a given age will yield the “measured” IQ. (b) The maximum IQ is achieved when maximum power is delivered to resistor R A, 3 which will occur when 576RA = 28.8×10 , or the person’s age is 50 years. (c) Well, now, this arguably depends on your answer to part (a), and your own sense of ethics. Hopefully you’ll do the right thing, and simply write to the Better Business Bureau. And watch less television.



58.


10 March 2006

We require a transformer that converts 240 V ac to 120 V ac, so that a turns ratio of 2:1 is needed. We attach a male european plug to the primary coil, and a female US plug to the secondary coil. Unfortunately, we are not given the current requirements of the CD writer, so that we will have to over-rate the transformer to ensure that it doesn’t overheat. Checking specifications on the web for an example CD writer, we find that the power supply provides a dual DC output: 1.2 A at 5 V, and 0.8 A at 12 V. This corresponds to a total DC power delivery of 15.6 W. Assuming a moderately efficient ac to DC converter is being used ( e.g. 80% efficient), the unit will draw approximately 15.6/0.8 or 20 W from the wall socket. Thus, the secondary coil should be rated for at least that (let’s go for 40 W, corresponding to a peak current draw of about 333 mA). Thus, we include a 300-mA fuse in series with the secondary coil and the US plug for safety.



59.


10 March 2006

You need to purchase (and wire in) a three-phase transformer rated at

)

3 (208)(10) 1.923.

=

3.6 kVA. The turns ratio for each phase needs to be 400:208 or


Chapter 13 Solutions to Exercises

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