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 = 485.0 the capacitor has capacitance of C = 81.0 F and the battery maintains the emf of = 10.0 V. The switch is closed at time t = 0.0 s and the capacitor begins to charge.

1.What is the time constant for this circuit?

2.What is the charge on the capacitor after the switch has been closed for t = 1.57×10-2 s?

3. What is the current through the circuit after the switch has been closed for t = 1.57×10-2 s?

4. What is the voltage across the capacitor after the switch has been closed for t = 1.57×10-2 s?

Explanation / Answer

At the moment the switch is closed, the capacitor is modeled as a short circuit.
I(0) = 10 / 485 = 20.61 mA

1.solution

time constant T = RC = 485*81x10-6= 3.93 x 10-2 s

2.solution

Vc = 10 [1 - e(-t/RC)]

Qc = C *Vc

Vc at 1.57*10-2 is Vc = 10 [1 - e(-t/RC)]

Vc = 10 [1 - e(-1.57*10^-2 / 485*81*10^-6)] = 3.294 V
Qc = C*Vc
Qc = 81x10-6 *3.294 = 2.668*10-4 C

3.
I(1.57×10-2) = 20.61*e(-t/RC) = 13.82 mA

4.

Vc(1.57×10-2) = 3.294V

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