Question

A capacitor of capacitance 100 μF is connected across a battery of emf 6 V through a resistance of 20 kΩ for 4 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4 s after the battery is disconnected?

Answer 1

The charge on the capacitor 4s is 70 μC after the battery is disconnected.

Explanation:

Given:

Capacitor’s capacitance, C = 100 μF

Resistance, R = 20 kΩ  

Battery’s Emf, E = 6V

Charging time, t1 = 4 s

Discharging time, t2 = 4 s

The charge growth across it during the capacitor’s charging,

Q = CE 1-e-t1RCt1RC = 420 $\times$ 103 $\times$ 100 $\times$ 10-6 = 2

⇒Q = 5.187 $\times$ 10-4 C

On the capacitor, this is the charge amount developed after 4s.

The charge decay across it during the capacitor’s discharging is

Q’ = Qe - tRC = 0.7 $\times$ 10-4 C = 70 μC