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 = 252.0 the capacitor has capacitance of C = 13.0 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 time constant for this circuit?

What is the charge on the capacitor after the switch has been closed for t = 3.24×10-3 s?

What is the current through the circuit after the switch has been closed for t = 3.24×10-3 s?

What is the voltage across the capacitor after the switch has been closed for t = 3.24×10-3 s?

Explanation / Answer

formula for time constant is

= RC

= (252)(13 X 10-6) = 3.27 ms

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The formula for the charge, q, on a capacitor is

q = C(1-e-t/RC)

= (13 X 10-6)(49)(1 - e-(3.24 X 10-3/3.27 X 10-3)

=4.00 * 10^-4 C

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I = (/R)e-t/RC

I = (49/252)e-(3.24 X 10-3/3.27 X 10-3

=0.072 A

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V = Q/C=4.00 * 10^-4 C/13 * 10^-6 F = 30.76 V

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