Page 164 - Schaum's Outline of Theory and Problems of Electric Circuits

FIRST-ORDER CIRCUITS
CHAP. 7]

Fig. 7-32 153

di 1
10ði 1 þ i 2 Þþ 5i 1 þ 0:01 ¼ 100 ð24Þ
dt
10ði 1 þ i 2 Þþ 5i 2 ¼ 100 ð25Þ
From (25), i 2 ¼ð100 10i 1 Þ=15. Substituting in (24),
di 1
þ 833i 1 ¼ 3333 ð26Þ
dt
The steady-state solution (particular solution) of (26)is i 1 ð1Þ ¼ 3333=833 ¼ 4:0 A; hence

i 1 ¼ Ae 833t þ 4:0 ðAÞ

þ
The initial condition i 1 ð0 Þ¼ i 1 ð0 Þ¼ 0 now gives A ¼ 4:0 A, so that
i 1 ¼ 4:0ð1 e 833t Þ ðAÞ and i 2 ¼ 4:0 þ 2:67e 833t ðAÞ
Alternate Method
When the rest of the circuit is viewed from the terminals of the inductance, there is equivalent resistance

5ð10Þ
R eq ¼ 5 þ ¼ 8:33
15
1
Then 1= ¼ R eq =L ¼ 833 s .At t ¼1, the circuit resistance is
5ð5Þ
R T ¼ 10 þ ¼ 12:5
10
so that the total current is i T ¼ 100=12:5 ¼ 8 A. And, at t ¼1, this divides equally between the two 5-
resistors, yielding a final inductor current of 4 A. Consequently,
i L ¼ i 1 ¼ 4ð1 e 833t ÞðAÞ

7.19 A series RL circuit, with R ¼ 50
and L ¼ 0:2 H, has a sinusoidal voltage
v ¼ 150 sin ð500t þ 0:785ÞðVÞ
applied at t ¼ 0. Obtain the current for t > 0.
The circuit equation for t > 0is
di
þ 250i ¼ 750 sin ð500t þ 0:785Þ ð27Þ
dt
The solution is in two parts, the complementary function (i c ) and the particular solution ði p Þ, so that
i ¼ i c þ i p . The complementary function is the general solution of (27) when the right-hand side is replaced
by zero: i c ¼ ke 250t . The method of undetermined coefficients for obtaining i p consists in assuming that
i p ¼ A cos 500t þ B sin 500t
since the right-hand side of (27) can also be expressed as a linear combination of these two functions. Then