Find the voltage across the capacitor as a function of time for t>0 for the circuit in Fig. 8.72. Assume steady-state conditions exist at t=0 1a 025 H 20 V Figure 8.72 For Prob. 8. 18. Chapter 8, Solution 18. When the switch is off, we have a source-free parallel RLC circuit. 2, JLC V0.25xl 0.5 2RC underdamped case V4 –0.25 = 1.936 a

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Find the voltage across the capacitor as a function of time for t>0 for the circuit in Fig.
8.72. Assume steady-state conditions exist at t= o
0.25 H
20 V
Figure 8.72
For Prob. 8.18.
Chapter 8, Solution 18.
When the switch is off, we have a source-free parallel RLC circuit.
1
-JIc J0.25xl
2,
0.5
2RC
a<@.
underdamped case
4–0.25 = 1.936
L(0) = i(0) = initial inductor current 20/5
= 4A
V0) = v(0) = initial capacitor voltage = 0 V
v(1) =e(4, cosot+ A, sin ot)=es (A, cos 1.936 + A, sin1.9361)
v(0) -0= A,
dv
s"(-0.5X4, cos1.936t + A, sin L92
+1.9364, cos 1.9361)
dr
dv(0)
dt
(V + RI
A =-2.066
--0.54, +1.936A,
RC
= =V-LR -(vo+IR)
R
Thus,
(1) --2.066e
sin 1.936
How did he get the value
inside the red circle
88
i)
Transcribed Image Text:Find the voltage across the capacitor as a function of time for t>0 for the circuit in Fig. 8.72. Assume steady-state conditions exist at t= o 0.25 H 20 V Figure 8.72 For Prob. 8.18. Chapter 8, Solution 18. When the switch is off, we have a source-free parallel RLC circuit. 1 -JIc J0.25xl 2, 0.5 2RC a<@. underdamped case 4–0.25 = 1.936 L(0) = i(0) = initial inductor current 20/5 = 4A V0) = v(0) = initial capacitor voltage = 0 V v(1) =e(4, cosot+ A, sin ot)=es (A, cos 1.936 + A, sin1.9361) v(0) -0= A, dv s"(-0.5X4, cos1.936t + A, sin L92 +1.9364, cos 1.9361) dr dv(0) dt (V + RI A =-2.066 --0.54, +1.936A, RC = =V-LR -(vo+IR) R Thus, (1) --2.066e sin 1.936 How did he get the value inside the red circle 88 i)
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