Page 357

346

MUTUAL INDUCTANCE AND TRANSFORMERS

1 ¼ 11 þ 12 ¼ 0:6 mWb M¼

L1 ¼

N1 1 500ð0:6Þ ¼ 60 mH ¼ 5:0 I1

N2 12 1500ð0:4Þ ¼ 120 mH ¼ 5:0 I1

pffiffiffiffiffiffiffiffiffiffiffi Then, from M ¼ k L1 L2 , L2 ¼ 540 mH.

14.2

[CHAP. 14

12 ¼ 0:667 1

Two coupled coils have self-inductances L1 ¼ 50 mH and L2 ¼ 200 mH, and a coefficient of coupling k ¼ 0:50. If coil 2 has 1000 turns, and i1 ¼ 5:0 sin 400t (A), find the voltage at coil 2 and the flux 1 . pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffi M ¼ k L1 L2 ¼ 0:50 ð50Þð200Þ ¼ 50 mH di d ð5:0 sin 400tÞ ¼ 100 cos 400t v2 ¼ M 1 ¼ 0:05 dt dt

ðVÞ

Assuming, as always, a linear magnetic circuit, M¼

14.3

N2 12 N2 ðk1 Þ ¼ i1 i1

 or

1 ¼

 M i ¼ 5:0  104 sin 400t N2 k 1

ðWbÞ

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. ð di di 1 di di i dt þ L2  M ¼ v Ri þ L1  M þ dt dt C dt dt ð di 1 or i dt ¼ v Ri þ L 0 þ dt C where L 0  L1 þ L2  2M.

Because M

pffiffiffiffiffiffiffiffiffiffiffi L1 þ L2 L1 L2  2

L 0 is nonnegative.

14.4

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 þ L2 þ 2M ¼ LA which give

L1 þ L2  2M ¼ LB

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An  
Advertisement