A capacitor of capacitance 4 μ F is charged to a potential difference of 6 V with a battery. The battery is then removed and in its place another capacitor of capacitance 8 μ F is introduced and the circuit is closed. The potential difference attained by each of the capacitors in V is

Question

A capacitor of capacitance 4 μ F is charged to a potential difference of 6 V with a battery. The battery is then removed and in its place another capacitor of capacitance 8 μ F is introduced and the circuit is closed. The potential difference attained by each of the capacitors in V is, 2, 4, 6, 8

MARKS App · Accepted Answer

When two capacitors are connected in series then their equivalent capacitance is, C e q = C 1 C 2 C 1 + C 2 = 4 × 8 4 + 8 = 8 3 μ F Q = C eq V = 8 3 × 6 = 16 μ C Now V 1 = Q C 1 = 16 4 = 4 V & V 2 = Q C 2 = 16 8 = 2 V ∴ ∆ V = 4 - 2 = 2 V

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