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CHAP. 14]

MUTUAL INDUCTANCE AND TRANSFORMERS

349

Fig. 14-24 pffiffiffiffiffiffiffiffiffiffiffi (a) XM ¼ ð0:8Þ 5ð10Þ ¼ 5:66 , and so the Z-matrix is   3 þ j1 3  j1:66 ½Z ¼ 3  j1:66 8 þ j6    3 þ j1 50    3  j1:66 0  ¼ 8:62 24:798 A I2 ¼ Z

Then,

and V ¼ I2 ð5Þ ¼ 43:1 24:798 V.  ½Z ¼ ðbÞ

3 þ j1 3 þ j9:66 3 þ j9:66 8 þ j6



   3 þ j1 50    3 þ j9:66 0  ¼ 3:82 112:128 A I2 ¼ Z and V ¼ 12 ð5Þ ¼ 19:1 112:128 V.

14.9

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 ¼ V1 =I1 :   j!0:3 j!0:043 ½Z ¼ j!0:043 j!0:414 and

Zin ¼

Z ð j!0:3Þð j!0:414Þ  ð j!0:043Þ2 ¼ ¼ j!0:296 11 j!0:414

or Leq is 0.296 H.

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.

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An