Page 159 - Schaum's Outline of Theory and Problems of Electric Circuits

FIRST-ORDER CIRCUITS
148

Fig. 7-23 [CHAP. 7

þ

iðtÞ¼ ið0 Þ¼ ið0 Þ¼ 1A
For t > 0, the current source is replaced by an open circuit and the 50-V source acts in the RL series
circuit ðR ¼ 20
Þ. Consequently, as t !1, i ! 50=20 ¼ 2:5 A. Then, by Sections 6.10 and 7.3,
þ
iðtÞ¼ ½ðið0 Þ ið1ފe Rt=L þ ið1Þ ¼ 3:5e 100t 2:5 ðAÞ
By means of unit step functions, the two formulas may be combined into a single formula valid for all t:
iðtÞ¼ uð tÞþð3:5e 100t 2:5ÞuðtÞ ðAÞ

7.9 In Fig. 7-24(a), the switch is closed at t ¼ 0. The capacitor has no charge for t < 0. Find i , i ,
C
R
v C , and v s for all times if i s ¼ 2 mA.
For t < 0, i R ¼ 2 mA, i C ¼ v C ¼ 0, and v s ¼ð2mAÞð5000
Þ¼ 10 V.
For t > 0, the time constant is ¼ RC ¼ 10 ms and
þ
i R ð0 Þ¼ 0; i R ð1Þ ¼ 2 mA, and i R ¼ 2ð1 e 100t ÞðmAÞ [See Fig. 7-24ðbÞ:Š
þ
v C ð0 Þ¼ 0; v C ð1Þ ¼ ð2mAÞð5k
Þ¼ 10 V, and v C ¼ 10ð1 e 100t ÞðVÞ [See Fig. 7-24ðcÞ:Š
þ
i C ð0 Þ¼ 2mA; i C ð1Þ ¼ 0, and i C ¼ 2e 100t ðmAÞ [See Fig. 7-24ðdÞ:Š
þ
v s ð0 Þ¼ 0; v s ð1Þ ¼ ð2mAÞð5k
Þ¼ 10 V, and v s ¼ 10ð1 e 100t Þ ðVÞ [See Fig. 7-24ðeÞ:Š

7.10 In Fig. 7-25, the switch is opened at t ¼ 0. Find i , i , v , and v .
C
s
R
C
For t < 0, the circuit is at steady state with i R ¼ 6ð4Þ=ð4 þ 2Þ¼ 4 mA, i C ¼ 0, and v C ¼ v s ¼ 4ð2Þ¼ 8V.
During the switching at t ¼ 0, the capacitor voltage remains the same. After the switch is opened, at

þ
þ
t ¼ 0 , the capacitor has the same voltage v C ð0 Þ¼ v C ð0 Þ¼ 8V.
For t > 0, the capacitor discharges in the 5-k
resistor, produced from the series combination of the
3 6
3-k
and 2-k
resistors. The time constant of the circuit is ¼ð2 þ 3Þð10 Þð2 10 Þ¼ 0:01 s. The
currents and voltages are
100t
v C ¼ 8e ðVÞ
i R ¼ i C ¼ v C =5000 ¼ð8=5000Þe 100t ¼ 1:6e 100t ðmAÞ
v s ¼ð6mAÞð4k
Þ¼ 24 V
since, for t > 0, all of the 6 mA goes through the 4-k
resistor.
7.11 The switch in the circuit of Fig. 7-26 is closed on position 1 at t ¼ 0 and then moved to 2 after one
time constant, at t ¼ ¼ 250 ms. Obtain the current for t > 0.
It is simplest first to find the charge on the capacitor, since it is known to be continuous (at t ¼ 0 and at
t ¼ ), and then to differentiate it to obtain the current.
For 0 t , q must have the form
q ¼ Ae t= þ B