A condenser of capacity $0.2 μ F$ is charged to a potential of 600 V. The battery is now disconnected and the condenser of capacity $1 μ F$ is connected across it. The potential of the condenser will reduce to

600 V

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300 V

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100 V

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120 V

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Solution

The correct option is B $100 V$
For (1), charge on condenser , $Q = C_{1} V_{1} = 0.2 \times 10^{- 6} \times 600 = 12 \times 10^{- 5} C$
For (2), now the equivalent capacitance of the circuit, $C = C_{1} + C_{2} = 0.2 + 1 = 1.2 μ F$
potential drop of the condenser is $V_{2} = \frac{Q}{C} = \frac{12 \times 10^{- 5}}{1.2 \times 10^{- 6}} = 100 V$

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