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Amit Saxena Grade: upto college level
`        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?`
3 years ago

Navjyot Kalra
654 Points
```										Sol. C = 100 μF, emf = 6 V, R = 20 KΩ, t = 4 S.
Charging : Q = CV(1 – e^–t/RC) [-t/RC = 4/2 * 10^4 *10^-4]
= 6 * 10^–4(1 – e^–2) = 5.187 * 10^–4 C = Q
Discharging : q = Q(e^–t/RC) = 5.184 * 10^–4 * e^–2
= 0.7 * 10^–4 C = 70 μc.

```
3 years ago
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