An RC circuit, hooked up to a battery as shown in the figure, starts with an unc

Question

An RC circuit, hooked up to a battery as shown in the figure, starts with an uncharged capacitor. The resistance in the circuit is R = 768.0 the capacitor has capacitance of C = 63.0 F and the battery maintains the emf of = 41.0 V. The switch is closed at time t = 0.0 s and the capacitor begins to charge.

B) What is the charge on the capacitor after the switch has been closed for t = 2.18×10-2 s?

C) What is the current through the circuit after the switch has been closed for t = 2.18×10-2 s?

D) What is the voltage across the capacitor after the switch has been closed for t = 2.18×10-2 s?

Explanation / Answer

R = 768 , C =63 F, = 41.0 V

(A) Time constant T =RC =768*63*10^-6

T = 0.0484 s

(B) t =2.18*10^-2 s.

Qo = VoC = 41*63 = 2583 C

for charging RC circuit

Q =Qo(1-e^-t/RC)

Q =(0.002583)(1-e^(-0.0218/0.0484))

Q = 9.37*10^-4 C

(C) Q =Qo(1-e^-t/RC)

i =dQ/dt

i =(Qo/RC)(e^(-t/RC))

i =(0.002583/0.0484)(e^(-0.0218/0.0484))

i =0.034 A

(D) V =Vo((1-e^-t/RC)

V = 41(1-e^(-0.0218/0.0484))

V =14.87 V

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