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 = 118.0 the capacitor has capacitance of C = 41.0 F and the battery maintains the emf of = 49.0 V. The switch is closed at time t = 0.0 s and the capacitor begins to charge.

What is the charge on the capacitor after the switch has been closed for t = 5.32×10-3 s?

What is the current through the circuit after the switch has been closed for t = 5.32×10-3 s?
What is the voltage across the capacitor after the switch has been closed for t = 5.32×10-3 s?

Explanation / Answer

while charging the charege grows as

q = C*E*(1-e^-(t/RC)

part(a)

at t = 5.32*10^-2 s

q = 41*10^-6*49*(1-e^-(5.32*10^-3/(118*41*10^-6))

q = 1.3*10^-3 C

------------------

part(b)

i = (E/R)*e^-(t/RC)

i = (49/118)*e^-(5.32*10^-3/(118*41*10^-6)) = 0.138 A

part(c)

v = i*R = 0.138*118

v = 16.284 V

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