A capacitor of capacitance 100 μF is connected across a battery of emf 6.0 V through a resistance of 20 kΩ for 4.0 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4.0 s after the battery is disconnected?
Concepts/Formulas used:
Charging a capacitor:
A capacitor of capacitance C is connected in series with a resistor of resistance R, a switch, and battery of emf ϵ . It is uncharged at first. The switch is closed at t = 0, then at time any time t the charge stored on the capacitor is given by
Discharging a capacitor:
A capacitor of capacitance C is connected in series with a resistor of resistance R and a switch. Before the switch is closed, it has charge Qi . If the switch is closed at t = 0, then at any time t, the charge on the capacitor is given by:
Given,
Emf = 6V
Capacitance,
Resistance,
Time for charging = time for discharging =
When charging,
When discharging,
Now,
Hence,