Page 178 - Schaum's Outline of Theory and Problems of Electric Circuits

HIGHER-ORDER CIRCUITS AND COMPLEX FREQUENCY
CHAP. 8]
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 167
6
2
5
2
! d ¼ ! ¼ 10 ð4:9 10 Þ ¼ 714
0
At t ¼ 0, V 0 ¼ 50:0; hence in (3) A 1 ¼ V 0 ¼ 50:0. From the nodal equation
ð
1 t dv
V 0
þ vdt þ C ¼ 0
R L 0 dt

dv
¼ V 0

dt t ¼ 0 RC
at t ¼ 0,
Differentiating the expression for v and setting t ¼ 0 yields

dv
or ! d A 2 A 1 ¼ V 0
dt t¼0 ¼ ! d A 2 A 1 RC
Since A 1 ¼ 50:0,
ðV 0 =RCÞþ V 0
A 2 ¼ ¼ 49:0
! d
and so v ¼ e 700t ð50:0 cos 714t 49:0 sin 714tÞ ðVÞ
The critically damped case will not be examined for the parallel RLC circuit, since it has little or no
real value in circuit design. In fact, it is merely a curiosity, since it is a set of circuit constants whose
response, while damped, is on the verge of oscillation.

8.4 TWO-MESH CIRCUIT
The analysis of the response for a two-mesh circuit which contains two storage elements results in
simultaneous differential equation as shown in the following.

Fig. 8-9

For the circuit of Fig. 8-9, choose mesh currents i 1 and i 2 , as indicated. KVL yields the two first-
order differential equations

di 1
R 1 i 1 þ L 1 þ R 1 i 2 ¼ V ð4Þ
dt
di 2
R 1 i 1 þðR 1 þ R 2 Þi 2 þ L 2 ¼ V ð5Þ
dt
which must be solved simultaneously. To accomplish this, take the time derivative of (4),
2
di 1 d i 1 di 2
R 1 þ L 1 2 þ R 1 ¼ 0 ð6Þ
dt dt dt
and then eliminate i and di =dt between (4), (5), and (6). The following result is a second-order
2
2
equation for i 1 , of the types treated in Sections 8.2 and 8.3, except for the constant term on the right: