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

Question

An RC circuit, hooked up to a battery as shown in the figure below, starts with an uncharged capacitor. The resistance in the circuit is R= 584 ohm the capacitor has capacitance C= 58.0 microFarads and the battery maintains the emf of E= 31.0 V. The switch is closed at time t=0s and the capacitor begins to charge.

a.) What is the time constant for this circuit?

b.) What is the charge on the capacitor after the switch has been closed for t=8.81×10-3 seconds?

c.) What is the current through the circuit after the switch has been closed for t=8.81×10-3 seconds?

d.) What is the voltage across the capacitor after the switch has been closed for t= 8.81×10-3 seconds ?

Explanation / Answer

At the moment the switch is closed, the capacitor is modeled as a short circuit.
I(0) = 31/584 = 58.219ma
a)

time constant = RC = 584*58x10-6= 0.033872 seconds

b) We know that function of voltage across capacitor as a function of time is
Vc = 31[1-e(-t/RC)]

Qc = C *Vc

So after 8.81*10-3 seconds
Vc= 7.0997V
Qc = C*Vc
Qc = 58x10-6 *7.0997 = 41.17 mC

c)
I =63.655e(-t/RC)

Now put t=8.81*10-3 seconds.

=49.0765 mA

d)

Vc = 31[1-e(-t/RC)]

=7.0997 Volt

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