Page 187 - Schaum's Outline of Theory and Problems of Electric Circuits

HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
[CHAP. 8
176
For sustained oscillations the roots of the characteristic equation in Example 8.14 should be imaginary numbers.
This happens when k ¼ 3or R 2 ¼ 2R 1 , in which case ! ¼ 1=RC.

Solved Problems

8.1 A series RLC circuit, with R ¼ 3k
, L ¼ 10 H, and C ¼ 200 mF, has a constant-voltage source,
V ¼ 50 V, applied at t ¼ 0. (a) Obtain the current transient, if the capacitor has no initial
charge. (b) Sketch the current and find the time at which it is a maximum.
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R 1
2
2
2
ðaÞ ¼ ¼ 150 s 1 ! 0 ¼ ¼ 500 s 2 ¼ ! ¼ 148:3s 1
0
2L LC
The circuit is overdamped ð > ! 0 Þ.
s 1 ¼ þ ¼ 1:70 s 1 s 2 ¼ ¼ 298:3s 1
and i ¼ A 1 e 1:70t þ A 2 e 298:3t
þ

þ
þ

Since the circuit contains an inductance, ið0 Þ¼ ið0 Þ¼ 0; also, Qð0 Þ¼ Qð0 Þ¼ 0. Thus, at t ¼ 0 ,
KVL gives

di di V
0 þ 0 þ L ¼ V or ¼ ¼ 5A=s
dt 0 þ dt 0 þ L
Applying these initial conditions to the expression for i,
0 ¼ A 1 ð1Þþ A 2 ð1Þ
5 ¼ 1:70A 1 ð1Þ 298:3A 2 ð1Þ
from which A 1 ¼ A 2 ¼ 16:9 mA.
i ¼ 16:9ðe 1:70t e 298:3t Þ ðmAÞ

(b) For the time of maximum current,
di 1:70t 298:3t
¼ 0 ¼ 28:73e þ 5041:3e
dt
Solving by logarithms, t ¼ 17:4 ms. See Fig. 8-18.

Fig. 8-18