CHAP. 7]

153

FIRST-ORDER CIRCUITS

Fig. 7-32 di1 ¼ 100 dt 10ði1 þ i2 Þ þ 5i2 ¼ 100

10ði1 þ i2 Þ þ 5i1 þ 0:01

From (25), i2 ¼ ð100  10i1 Þ=15.

ð24Þ ð25Þ

Substituting in (24), di1 þ 833i1 ¼ 3333 dt

ð26Þ

The steady-state solution (particular solution) of (26) is i1 ð1Þ ¼ 3333=833 ¼ 4:0 A; hence i1 ¼ Ae833t þ 4:0 ðAÞ The initial condition i1 ð0 Þ ¼ i1 ð0þ Þ ¼ 0 now gives A ¼ 4:0 A, so that i1 ¼ 4:0ð1  e833t Þ

ðAÞ

and

i2 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 Req ¼ 5 þ

5ð10Þ ¼ 8:33  15

Then 1= ¼ Req =L ¼ 833 s1 . At t ¼ 1, the circuit resistance is RT ¼ 10 þ

5ð5Þ ¼ 12:5  10

so that the total current is iT ¼ 100=12:5 ¼ 8 A. And, at t ¼ 1, this divides equally between the two 5- resistors, yielding a ﬁnal inductor current of 4 A. Consequently, iL ¼ i1 ¼ 4ð1  e833t Þ ðAÞ

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 0:785Þ ðVÞ applied at t ¼ 0.

Obtain the current for t > 0.

The circuit equation for t > 0 is di þ 250i ¼ 750 sin ð500t þ 0:785Þ dt

ð27Þ

The solution is in two parts, the complementary function (ic ) and the particular solution ðip Þ, so that i ¼ ic þ ip . The complementary function is the general solution of (27) when the right-hand side is replaced by zero: ic ¼ ke250t . The method of undetermined coeﬃcients for obtaining ip consists in assuming that ip 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

Mahmood_Nahvi_eBook_Schaum_s_Outlines_Theory_An