In the circuit shown at right, the ideal battery has emf epsilon = 9.0 V, C = 1.
ID: 1420503 • Letter: I
Question
In the circuit shown at right, the ideal battery has emf epsilon = 9.0 V, C = 1.0 mu F, and R_1 = R_2 =R_3 = 10 k Ohm. The capacitor is initially uncharged. The switch is suddenly closed at time t = 0. What is the initial current (at time t = 0) through each of the resistors? After a long time (much greater than the time constant), what is the current through each resistor, and what is the charge on the capacitor? After the capacitor is fully charged, the switch is opened again. How long will it take for the charge on the capacitor to drop to 5% of its peak value?Explanation / Answer
(a)
At the time switch is suddenly closed, Capacitor acts as a Short circut.
Req = R1 + (R2*R2)/(R2+R3)
Req = 10 + (10*10)/(10+10) k ohm
Req = 15 k ohm
V = I*Req
I = 9/15 mA
I = 0.6 mA
Initial Current through each of the resisitor is ,
Current through R1, I1 = 0.6 mA
Current through R2, I2 = 0.3 mA
Current through R3, I3 = 0.3 mA
(b)
After a Long time , Capacitor acts as an open circut and no current will pass through it.
Req = R1 + R2
Req = 10 + 10 k ohm
Req = 20 k ohm
V = I*Req
I = 9/20 mA
I = 0.45 mA
Current through R1, I1 = 0.45 mA
Current through R2, I2 = 0.45 mA
Current through R3, I3 = 0 A
Charge on the Capacitor, Q = C * V
Q = 1.0 uF * (0.45*10)
Q = 4.5 uC
Charge on the Capacitor, Q = 4.5 uC
(c)
Time it will take for capacitor to discharge,
q = Q*e^(-t/RC)
0.05 * Q = Q * e^(-t/RC)
0.05 = e^(-t/RC)
0.05 = e^(-t/ (15*10^3 * 1.0 * 10^-6))
t = 0.045 s
Time, t = 0.045 s
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