A capacitor is connected to a 20 V battery through a resistance of 10 Ω . 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 In 10 9 = 0 . 105

Step-by-step solution

V = V 0 1 - e - t / R C 2 = 20 1 - e - 1 μ s 10 C 1 10 = 1 - e - 1 × 10 - 6 10 C e - 1 × 10 - 6 10 C = 9 10 1 × 10 - 6 10 C = ln 10 9 C = 1 × 10 - 6 10 × ln 10 9 = 10 - 7 1 . 05 = 1 1 . 05 μF = 100 105 μF = 0 . 95 μF