In the adjoining figure, four capacitors are shown with their respective capacities and the P.D. applied. The charge and the P.D. across the 4μF capacitor will be.

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600μC;150 volts

300μC;75 volts

800μC;200 volts

580μC;145 volts

answer is D.

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Detailed Solution

Total capacitance 1C=120+18+112⇒C=12031μFTotal charge Q=CV=12031×300=1161μCCharge, through 4μF condenser =11612=580μCand potential difference across it =5804=145 V.

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