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`        A capacitor of capacitance 10 μF is connected to a battery of emf 2 V. It is found that it takes 50 ms for the charge on the capacitor to become 12.6 μC. Find the resistance of the circuit.`
3 years ago

Kevin Nash
332 Points
```										Sol. C = 10 μF = 10^–5 F, emf = 2 V
t = 50 ms = 5 * 10^–2 s, q = Q(1 – e^–t/RC)
Q = CV = 10^–5 * 2
q = 12.6 * 10^–6 F
⇒ 12.6 * 10–6 = 2 * 10^–5 (1 e^-5 10 2 /R 10 5 )
⇒12.6 * 10^-6/2 * 10^-5 = 1 – e^-5*10^-2 /R*10^-5
⇒1 – 0.63 = -^-5*10^3 /R
⇒ -5000/R = In 0.37
⇒ R = 5000/0.9942 = 5028 Ω = 5.028 * 10^3Ω = 5 KΩ.

```
3 years ago
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