Question

In the circuit below, the switch was open for a long time and then closed at t=0 s. The values of the emf, resistors, and capacitors are ε = 11.5V, R1 = 2.4 Ω, R2 = 7.4 Ω, R3 = 0.3 Ω, CA = 7.1 μF, CB = 5.0 μF.

(a) Immediately after the switch is closed, what is the current through resistor R1?

A long time after the switch was closed, what are the charges stored on the two capacitors?

(b) on CA: (μC)

(c) on CB: (μC)

Answer 1

a) Initially the capacitors are a short circuit , so I = 11.5/2.4 = 4.79 amp

b) long time later, the capacitors do not conduct, so you have a voltage divider. Total R = 2.4+7.4+0.3 = 10.1

Voltage across R2 = 11.5(7.4/10.1) = 8.426 volts, and that voltage is also across Ca.
Q = CV = 7.1µF x 8.426 volts = 59.82 µC

same for Cb
Voltage across R3 = 11.5(0.3/10.1)=0.342 V
and Q = CV=5µF x 0.342 volts = 1.71 µC