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Chapter 13, Problem 1.
For the three coupled coils in Fig. 13.72, calculate the total inductance.
Figure 13. For Prob. 13.1.
Chapter 13, Solution 1.
For coil 1, L 1 – M 12 + M 13 = 6 – 4 + 2 = 4
For coil 2, L 2 – M 21 – M 23 = 8 – 4 – 5 = – 1
For coil 3, L 3 + M 31 – M 32 = 10 + 2 – 5 = 7
L (^) T = 4 – 1 + 7 = 10H
or L (^) T = L 1 + L 2 + L 3 – 2M 12 – 2M 23 + 2M 12
L (^) T = 6 + 8 + 10 = 10H
Chapter 13, Problem 2.
Determine the inductance of the three series-connected inductors of Fig. 13.73.
Figure 13. For Prob. 13.2.
Chapter 13, Solution 2.
L = L 1 + L 2 + L 3 + 2M 12 – 2M 23 –2M (^31)
= 10 + 12 +8 + 2x6 – 2x6 –2x
= 22H
Chapter 13, Problem 3.
Two coils connected in series-aiding fashion have a total inductance of 250 mH. When connected in a series-opposing configuration, the coils have a total inductance of 150 mH. If the inductance of one coil ( L 1 ) is three times the other, find L 1 , L 2 , and M. What is the coupling coefficient?
Chapter 13, Solution 3.
L 1 + L 2 + 2M = 250 mH (1)
L 1 + L 2 – 2M = 150 mH (2)
Adding (1) and (2),
2L 1 + 2L 2 = 400 mH
But, L 1 = 3L2,, or 8L 2 + 400, and L 2 = 50 mH
L 1 = 3L 2 = 150 mH
From (2), 150 + 50 – 2M = 150 leads to M = 25 mH
k = M/ L 1 L 2 = 25 / 50 x 150 = 0.
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, Leq = L 1 + L 2 + 2M
(b) For the parallel coil, consider Figure (b).
I (^) s = I 1 + I 2 and Zeq = Vs /I (^) s
Applying KVL to each branch gives,
Vs = jωL 1 I 1 + jωMI 2 (1)
Vs = jωMI 1 + jω L 2 I 2 (2)
or (^) ⎥ ⎦
ω ω
ω ω ⎥= ⎦
2
1 2
1 s
s I
I
j M j L
j L j M V
V
∆ = –ω^2 L 1 L 2 + ω^2 M 2 , ∆ 1 = jωVs (L 2 – M), ∆ 2 = jωVs (L 1 – M)
I 1 = ∆ 1 /∆, and I 2 = ∆ 2 /∆
I (^) s = I 1 + I 2 = (∆ 1 + ∆ 2 )/∆ = jω(L 1 + L 2 – 2M)Vs /( –ω^2 (L 1 L 2 – M 2 )) = (L 1 + L 2 – 2M)Vs /( jω(L 1 L 2 – M 2 ))
Zeq = Vs /I (^) s = jω(L 1 L 2 – M^2 )/(L 1 + L 2 – 2M) = jωLeq
i.e., Leq = (L 1 L 2 – M 2 )/(L 1 + L 2 – 2M)
(a) (b)
I (^) s
I 1 I (^2)
L 1
L 2
L 1
Leq
L 2
L = L 1 +L 2 + 2 M= 25 + 60 + 2 ( 0. 5 ) 25 x 60 = 123.7 mH
(b) If they are connected in parallel,
= mH 25 60 2 x 19. 36
25 x 60 19. 36 L L 2 M
LL M
L
2
1 2
2 (^1 2) 24.31 mH
Chapter 13, Solution 6.
M = k L L 1 2 (^) = 0.6 40 5 x =8.4853 mH
3
40 mH j ω L j 2000 40 10 x x j 80
− ⎯⎯→ = = 3
5 mH j ω L j 2000 5 10 x x j 10
− ⎯⎯→ = = 3
8.4853 mH j ω M j 2000 8.4853 10 x x j 16.
− ⎯⎯→ = =
We analyze the circuit below.
V 1 (^) = j 80 I 1 (^) − j 16.97 I 2 (1) V 2 (^) = −16.97 I 1 (^) + j 10 I 2 (2)
But V 1 (^) = 10 < 0 o^ and I (^) 2 = 2 < − 90 o = − j 2. Substituting these in eq.(1) gives
1 2 1
I V^ j^ I^ j^ x^ j^ o j j
i 1 ( ) t =0.5493sin ω t A
From (2),
2 16.97 ( 0. 5493)^ 10 (^ 2)^20 9.3216^ 22.0656^ 24. V = − x − j + j x − j = + j = < o 2 ( )^ 22.065cos(^ 25 ) V
v t = ω t + o
_
V 2
I 1 I 2
_
V 1
j80 Ω j10 Ω
For the circuit in Fig. 13.76, find Vo.
Figure 13. For Prob. 13.7.
Chapter 13, Solution 7.
We apply mesh analysis to the circuit as shown below.
For mesh 1, 12 = I 1 (^) (2 + j 6) + jI 2 (1)
For mesh 2, 0 = jI 1 (^) + (2 − j 1 + j 4) I 2
or 0 = jI 1 (^) + (2 + j 3) I 2 (2)
In matrix form,
1 2
j j I I
_
j6 Ω^1 Ω
_
j4 Ω Vo
1 Ω –j1 Ω
j1 Ω
I 1 I 2
Chapter 13, Problem 9.
Find V x in the network shown in Fig. 13.78.
Figure 13. For Prob. 13.9.
Chapter 13, Solution 9.
Consider the circuit below.
For loop 1,
8 ∠ 30 ° = (2 + j4)I 1 – jI 2 (1)
For loop 2, ((j4 + 2 – j)I 2 – jI 1 + (–j2) = 0
or I 1 = (3 – j2)i 2 – 2 (2)
Substituting (2) into (1), 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 °
o –
j 4 j 4
-j
Find v (^) o in the circuit of Fig. 13.79.
Figure 13. For Prob. 13.10.
Chapter 13, Solution 10.
2 H ⎯⎯→ j ω L = j 2 2 x = j 4
0.5 H ⎯⎯→ j ω L = j 2 0.5 x = j
F j
j ω C j x
Consider the circuit below.
24 = j 4 I 1 (^) − jI 2 (1) 0 = − jI 1 (^) + ( 4 j − j I ) (^) 2 ⎯⎯→ 0 = − I 1 (^) + 3 I 2 (2)
In matrix form,
1 2
j j I I
⎡ ⎤ ⎡ −^ ⎤⎡ ⎤
⎢ ⎥ =^ ⎢ ⎥⎢^ ⎥
Solving this, I 2 (^) = − j 2.1818, Vo = − jI 2 = −2.
vo = –2.1818cos2t V
24 ∠ 0 ° _
_
Vo
j
j
I 1 I 2 –j
j
