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V1 M 2 s2 L1 s I1 Z2 L2 s M 2 !2 Z2 j!L2
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For the ac steady state where s j!, we have Z1 j!L1 The re ected impedance is Zreflected M 2 !2 Z2 j!L2 19 18
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The load Z2 is seen by the source as M 2 !2 = Z2 j!L2 . The technique is often used to change an impedance to a certain value; for example, to match a load to a source.
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EXAMPLE 14.8 Given L1 0:2 H, L2 0:1 H, M 0:1 H, and R 10  in the circuit of Fig. 14-17. for v1 142:3 sin 100t. Find i1
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Fig. 14-17 The input impedance Z1 at ! 100 is [see (18)] Z1 Then, or p V1 M 2 !2 0:01 10 000 j!L1 j20 5 j15 5 10 71:68 I1 Z2 j!L2 10 j10 I1 V1 =Z1 9 71:68 A i1 9 sin 100t 71:68 A
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EXAMPLE 14.9 Referring to Example 14.8, let v1 u t . The input impedance is [see (17)] Z1 s L1 s Substituting the given values for the elements, we get Z1 s s s 200 10 s 100 or
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Find i1;f , the forced response. M 2 s2 R L2 s
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Y1 s
10 s 100 s s 200
For t > 0, the input v1 1 V is an exponential est whose exponent s 0 is a pole of Y1 s . Therefore, i1;f kt with k 1=L1 5. This result may also be obtained directly by dc analysis of the circuit in Fig. 14-17.
Solved Problems
14.1 When one coil of a magnetically coupled pair has a current 5.0 A the resulting uxes 11 and 12 are 0.2 mWb and 0.4 mWb, respectively. If the turns are N1 500 and N2 1500, nd L1 , L2 , M, and the coe cient of coupling k.
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[CHAP. 14
1 11 12 0:6 mWb M
L1
N1 1 500 0:6 60 mH 5:0 I1 k 12 0:667 1
p Then, from M k L1 L2 , L2 540 mH.
N2 12 1500 0:4 120 mH 5:0 I1
Two coupled coils have self-inductances L1 50 mH and L2 200 mH, and a coe cient of coupling k 0:50. If coil 2 has 1000 turns, and i1 5:0 sin 400t (A), nd the voltage at coil 2 and the ux 1 .
p p M k L1 L2 0:50 50 200 50 mH di d 5:0 sin 400t 100 cos 400t v2 M 1 0:05 dt dt Assuming, as always, a linear magnetic circuit, M N2 12 N2 k1 i1 i1  or 1  M i 5:0 10 4 sin 400t N2 k 1 Wb
Apply KVL to the series circuit of Fig. 14-18.
Fig. 14-18 Examination of the winding sense shows that the signs of the M-terms are opposite to the signs on the L-terms. di di 1 di di i dt L2 M v Ri L1 M dt dt C dt dt di 1 or i dt v Ri L 0 dt C where L 0  L1 L2 2M. Because M L 0 is nonnegative. p L1 L2 L1 L2 2
In a series aiding connection, two coupled coils have an equivalent inductance LA ; in a series opposing connection, LB . Obtain an expression for M in terms of LA and LB .
As in Problem 14.3, L1 L2 2M LA which give L1 L2 2M LB
CHAP. 14]
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1 M LA LB 4 This problem suggests a method by which M can be determined experimentally.
(a) Write the mesh current equations for the coupled coils with currents i1 and i2 shown in Fig. 14-19. (b) Repeat for i2 as indicated by the dashed arrow.
Fig. 14-19 (a) The winding sense and selected directions result in signs on the M-terms as follows: di2 v dt di1 v dt d i i2 M R1 i1 i2 L1 dt 1 di d d i i1 L1 i i1 M R1 i2 i1 R2 i2 L2 2 M dt 2 dt 2 dt R1 i1 L1 di1 M dt di R2 i2 L2 2 M dt
di2 v dt di2 0 dt
Obtain the dotted equivalent circuit for the coupled circuit shown in Fig. 14-20, and use it to nd the voltage V across the 10- capacitive reactance.
Fig. 14-20 To place the dots on the circuit, consider only the coils and their winding sense. Drive a current into the top of the left coil and place a dot at this terminal. The corresponding ux is upward. By Lenz s law, the ux at the right coil must be upward-directed to oppose the rst ux. Then the natural current leaves this winding by the upper terminal, which is marked with a dot. See Fig. 14-21 for the complete dotted equivalent circuit, with currents I1 and I2 chosen for calculation of V.
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