Page 165 - Schaum's Outline of Theory and Problems of Electric Circuits

FIRST-ORDER CIRCUITS
154
di p [CHAP. 7
¼ 500A sin 500t þ 500B cos 500t
dt
Substituting these expressions for i p and di p =dt into (27) and expanding the right-hand side,
500A sin 500t þ 500B cos 500t þ 250A cos 500t þ 250B sin 500t ¼ 530:3 cos 500t þ 530:3 sin 500t
Now equating the coefficients of like terms,
500A þ 250B ¼ 530:3 and 500B þ 250A ¼ 530:3
Solving these simultaneous equations, A ¼ 0:4243 A, B ¼ 1:273 A.
i p ¼ 0:4243 cos 500t þ 1:273 sin 500t ¼ 1:342 sin ð500t 0:322ÞðAÞ
and i ¼ i c þ i p ¼ ke 250t þ 1:342 sin ð500t 0:322Þ ðAÞ

At t ¼ 0, i ¼ 0. Applying this condition, k ¼ 0:425 A, and, finally,
i ¼ 0:425e 250t þ 1:342 sin ð500t 0:322Þ ðAÞ

7.20 For the circuit of Fig. 7-33, obtain the current i L , for all values of t.

Fig. 7-33

For t < 0, the 50-V source results in inductor current 50=20 ¼ 2:5 A. The 5-A current source is applied
for t > 0. As t !1, this current divides equally between the two 10-
resistors, whence i L ð1Þ ¼ 2:5A.
The time constant of the circuit is
3
0:2 10 H 1
¼ ¼ ms
20
100

þ
and so, with t in ms and using i L ð0 Þ¼ i L ð0 Þ¼ 2:5A,
þ
i L ¼½i L ð0 Þ i L ð1ފe t= þ i L ð1Þ ¼ 5:0e 100t 2:5 ðAÞ
Finally, using unit step functions to combine the expressions for t < 0 and t > 0,
i L ¼ 2:5uð tÞþð5:0e 100t 2:5ÞuðtÞ ðAÞ

7.21 The switch in Fig. 7-34 has been in position 1 for a long time; it is moved to 2 at t ¼ 0. Obtain
the expression for i, for t > 0.

þ

With the switch on 1, ið0 Þ¼ 50=40 ¼ 1:25 A. With an inductance in the circuit, ið0 Þ¼ ið0 Þ. Long
after the switch has been moved to 2, ið1Þ ¼ 10=40 ¼ 0:25 A. In the above notation,
þ
B ¼ ið1Þ ¼ 0:25 A A ¼ ið0 Þ B ¼ 1:00 A
and the time constant is ¼ L=R ¼ð1=2000Þ s. Then, for t > 0,
i ¼ 1:00e 2000t þ 0:25 ðAÞ