A capacitor with C = 3.00 times 10^-5 F is connected as shown in the figure belo

Question

A capacitor with C = 3.00 times 10^-5 F is connected as shown in the figure below with a resistor with R = 980 Ohm and an emf source with epsilon = 25.0 V and negligible internal resistance. Initially, the capacitor is uncharged and the switch S is in position 1. The switch is then moved to position 2, so that the capacitor begins to charge. After the switch has been in position 2 for 60.0 ms, the switch is moved back to position 1 so that the capacitor begins to discharge. Compute the charge on the capacitor just before the switch is thrown from position 2 back to position 1. Compute the voltage drops across the resistor and across the capacitor at the instant described in part (a). Across the resistor. Across the capacitor. Compute the voltage drops across the resistor and across the capacitor just after the switch is thrown from position 2 back to position 1. Across the resistor. Across the capacitor. Compute the charge on the capacitor 60.0 ms after the switch is thrown from position 2 back to position 1.

Explanation / Answer

a)

Maximum Charge

Qo=CVo=(5*10-5)*21=1.05*10-3 C

Time Constant

T=RC =910*5*10-5=0.0455 sec

So in RC circuit Charge as a function of time is given by

Q=Qo[1-e-t/T]

Q=(1.05*10-3)[1-e-0.06/0.0455]

Q=7.7*10-4 C

b)

Voltage across capacitor isr is

Vc=Voe-t/T=21*e-0.06/0.0455

Vc=5.62Volts

Voltage drop across resistor is

VR=21-5.62=15.38 Volts

c)

Voltage drop across capacitor and resistor is

VR=VC=15.38 Volts

d)

Q=Qoe-t/T =(7.7*10-4)e-0.6/0.045

Q=2.03*10-4 C

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