A capacitor is connected to a 20 V battery through a resistance of 10Ω. It is found that the potential difference across the capacitor rises to 2 V in 1 μs. The capacitance of the capacitor is __________ μF. Given : ln ( 10/9 ) = 0.105

Solution

Physics sir · 4 steps

The capacitor reached only a tenth of the battery voltage, so the exponential left is 0.9. Take its log.

1. Charging law. While a capacitor charges through a resistor, its voltage grows towards the battery voltage V₀ with time constant RC.

   V = V₀(1 - e^(-t/RC))

2. Put in numbers. 2 = 20(1 - e^(-t/RC)) gives e^(-t/RC) = 0.9 = 9/10. Taking logs, t/RC = ln 10/9 = 0.105.
3. Find RC. RC = (1 × 10⁻⁶ s)/0.105 = 9.52 × 10⁻⁶ s.
4. So. C = (9.52 × 10⁻⁶)/10 = 0.952 × 10⁻⁶ F ≈ 0.95 μF, which is option (b).

The idea to keep. For charging, write the fraction still to go, 1 - V/V₀ = e^(-t/RC), then take the log. That gives the time constant, and you divide by R to get C.

Answer from the NTA final answer key. Solution by PrepWiser.