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 = 910 Ohm and an emf source with epsilon = 26.0 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 75.0 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. ________________ C Compute the voltage drops across the resistor and across the capacitor at the instant described in part (a). Across the resistor. _______________ V Across the capacitor. ______________ V 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. ___________________ V Across the capacitor. ______________ V Compute the charge on the capacitor 75.0 ms after the switch is thrown from position 2 back to position 1. ________________ C

Explanation / Answer

a)

T = time constant = RC = 910 (3 x 10-5) = 0.0273

t = given time after which is moved from position 2 to position 1 = 0.075 sec

Qo = maximum charge = CE = (3 x 10-5) (26) = 0.00078

Q = charge after time "t"

Q = Qo (1 - e-t/T)

Q = (0.00078 ) (1 - e-0.075/0.0273) = 0.00073 C

b)

Vr = E = 26 volts

Vc = voltage across the capacitor = 0      since there is no charge on capacitor when at the start of position 1

c)

Voltage across the capacitor = Vc = Q/C = 0.00073/(3 x 10-5) = 24.33 volts

Vr = voltage across resistor = 0

d)

Q = Qo e-t/T

Q = ( 0.00073) e-0.075/0.0273 = 0.000047 C

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