Page 146 - Schaum's Outline of Theory and Problems of Electric Circuits

135
FIRST-ORDER CIRCUITS
CHAP. 7]
EXAMPLE 7.6 In Fig. 7-12 the 9-mF capacitor is connected to the circuit at t ¼ 0. At this time, capacitor voltage
is v 0 ¼ 17 V. Find v A , v B , v C , i AB , i AC , and i BC for t > 0:

Fig. 7-12
Apply KCL at nodes A, B, and C for t > 0 to find voltages in term of i:

1 1 1 1 1
Node A: þ þ v A v B v C ¼ 0 or 6v A 3v B v C ¼ 0 (11)
2 3 6 2 6

1 1 1 1
3
3
Node B: v A þ þ v B 10 i v C ¼ 0 or 2v A þ 3v B v C ¼ð4 10 Þi (12)
2 2 4 4

1 1 1 1 1
Node C: v A v B þ þ þ v C ¼ 0 or 2v A 3v B þ 6v C ¼ 0 (13)
6 4 4 6 12
Solving (11), (12), and (13) simultaneously,
3
3
3
8
7
v A ¼ ð10 Þi v B ¼ 34 ð10 Þi v C ¼ ð10 Þi
3 9 3
The circuit as seen by the capacitor is equivalent to a resistor R ¼ v B =i ¼ 34=9k
. The capacitor discharges
3
6
its initial voltage V 0 in an exponential form with a time constant ¼ RC ¼ 34 ð10 Þð9 10 Þ¼ 0:034 s. For t > 0,
9
the voltages and currents are
v B ¼ V 0 e t= ¼ 17e 1000t=34 ðVÞ
dv B 3 1000t=34 3 1000t=34
i ¼ C ¼ð9 17 10 =34Þe ¼ð4:5 10 Þe ðAÞ
dt
3
3
7
8
v A ¼ ð10 Þi ¼ 10:5e 1000t=34 ðVÞ v C ¼ ð10 Þi ¼ 12e 1000t=34 ðVÞ
3 3
3
v AB ¼ v A v B ¼ 6:5e 1000t=34 ðVÞ i AB ¼ v AB =2000 ¼ð 3:25 10 Þe 1000t=34 ðAÞ
3
v AC ¼ v A v C ¼ 1:5e 1000t=34 ðVÞ i AC ¼ v AC =6000 ¼ð 0:25 10 Þe 1000t=34 ðAÞ
3
v BC ¼ v B v C ¼ 5e 1000t=34 ðVÞ i BC ¼ v BC =4000 ¼ð1:25 10 Þe 1000t=34 ðAÞ
All voltages and currents are exponential functions and have the same time constant. For simplicity, it is custom-
ary to use units of V, mA, k
, and ms for voltage, current, resistance, and time, respectively, so that the multipliers
1000 and 10 3 can be omitted from the equations as summarized below.
v A ¼ 10:5e t=34 ðVÞ v AB ¼ 6:5e t=34 ðVÞ i AB ¼ 3:25e t=34 ðmAÞ
v B ¼ 17e t=34 ðVÞ v AC ¼ 1:5e t=34 ðVÞ i AC ¼ 0:25e t=34 ðmAÞ
v C ¼ 12e t=34 ðVÞ v BC ¼ 5e t=34 ðVÞ i BC ¼ 1:25e t=34 ðmAÞ
i ¼ 4:5e t=34 ðmAÞ