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Question

Three capacitors of capacitances 4 μF, 12 μF, and 8 μF are connected in series to a battery of 110 V. The charge on 12 μF capacitor is

A
520 μC
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B
120 μC
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C
240 μC
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D
280 μC
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Solution

The correct option is C 240 μCIn series combination, the equivalent capacitance is given by, 1Ceq=1C1+1C2+1C3 Substituting the values, we get ⇒1Ceq=14+112+18 ⇒1Ceq=30+10+15120 ⇒Ceq=2411 μF Moreover, in a series combination, the charge on each capacitor is the same, which is the same as the charge on the equivalent capacitor. Using, Q=CV we get, Q=2411 μF×110 V=240 μC Hence, option (c) is the correct answer. Key concept: Charge on a capacitor in a series arrangement in a circuit.

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