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[CHAP. 14
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Fig. 14-21  5 j5 5 j3 5 j3 10 j6  I1 I2  " 10 08 10 j10 #
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I1 and V I1 j10 10:15 23:968 V.
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10 10 j10 Z
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5 j3 10 j6
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1:015 113:968 A
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Obtain the dotted equivalent for the circuit shown in Fig. 14-22 and use the equivalent to nd the equivalent inductive reactance.
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Fig. 14-22 Drive a current into the rst coil and place a dot where this current enters. The natural current in both of the other windings establishes an opposing ux to that set up by the driven current. Place dots where the natural current leaves the windings. (Some confusion is eliminated if the series connections are ignored while determining the locations of the dots.) The result is Fig. 14-23. Z j3 j5 j6 2 j2 2 j4 2 j3 j12  that is, an inductive reactance of 12 .
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Fig. 14-23
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(a) Compute the voltage V for the coupled circuit shown in Fig. 14-24. polarity of one coil reversed.
(b) Repeat with the
CHAP. 14]
MUTUAL INDUCTANCE AND TRANSFORMERS
Fig. 14-24 p (a) XM 0:8 5 10 5:66 , and so the Z-matrix is   3 j1 3 j1:66 Z 3 j1:66 8 j6 3 j1 50 3 j1:66 0 8:62 24:798 A I2 Z 
Then,
and V I2 5 43:1 24:798 V.  Z b
3 j1 3 j9:66 3 j9:66 8 j6
3 j1 50 3 j9:66 0 3:82 112:128 A I2 Z and V 12 5 19:1 112:128 V.
Obtain the equivalent inductance of the parallel-connected, coupled coils shown in Fig. 14-25.
Currents I1 and I2 are selected as shown on the diagram; then Zin V1 =I1 :   j!0:3 j!0:043 Z j!0:043 j!0:414 and or Leq is 0.296 H. Zin Z j!0:3 j!0:414 j!0:043 2 j!0:296 11 j!0:414
Fig. 14-25
14.10 For the coupled circuit shown in Fig. 14-26, show that dots are not needed so long as the second loop is passive.
MUTUAL INDUCTANCE AND TRANSFORMERS
[CHAP. 14
Fig. 14-26 Currents I1 and I2 are selected as shown. 50 j4 0 5 j10 250 j500 10:96 54:648 A I1 2 j5 j4 24 j45 j4 5 j10 2 j5 50 j4 0 3:92 118:07 908 A I2 Z The value of Z is una ected by the sign on M. Since the numerator determinant for I1 does not involve the coupling impedance, I1 is also una ected. The expression for I2 shows that a change in the coupling polarity results in a 1808 phase shift. With no other phasor voltage present in the second loop, this change in phase is of no consequence.
14.11 For the coupled circuit shown in Fig. 14-27, nd the ratio V2 =V1 which results in zero current I1 .
V1 V2 j2 2 j2 Z
I1 0
Then, V1 2 j2 V2 j2 0, from which V2 =V1 1 j1.
Fig. 14-27
14.12 In the circuit of Fig. 14-28, nd the voltage across the 5  reactance with the polarity shown.
For the choice of mesh currents shown on the diagram, 50 458 j8 0 j3 150 458 I1 3 j15 j8 109 j9 1:37 40:288 A j8 j3 Similarly, I2 3:66 40:288 A.
CHAP. 14]
MUTUAL INDUCTANCE AND TRANSFORMERS
Fig. 14-28 The voltage across the j5 is partly conductive, from the currents I1 and I2 , and partly mutual, from current I1 in the 4  reactance. V I1 I2 j5 I1 j3 29:27 49:728 V Of course, the same voltage must exist across the capacitor: V I2 j8 29:27 49:728 V
14.13 Obtain Thevenin and Norton equivalent circuits at terminals ab for the coupled circuit shown in Fig. 14-29.
In open circuit, a single clockwise loop current I is driven by the voltage source. I 10 08 1:17 20:568 A 8 j3
Fig. 14-29 Then V 0 I j5 4 I j6 4:82 34:608 V. To nd the short-circuit current I 0 , two clockwise mesh currents are assumed, with I2 I 0 . 8 j3 10 4 j1 0 0:559 83:398 A I0 8 j3 4 j1 4 j1 7 j5 V0 4:82 34:608 8:62 48:798  I0 0:559 83:398
Z0
The equivalent circuits are pictured in Fig. 14-30.
14.14 Obtain a conductively coupled equivalent circuit for the magnetically coupled circuit shown in Fig. 14-31.
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