Two capacitors of capacity 6μF and 3μF are charged to
100V and 50V separately and connected as shown in figure. Now
all the three switches S1,S2 and S3 are closed. Charge on the 3μF capacitor in steady state will be :
A
400μC
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B
700μC
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C
800μC
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D
250μC
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Solution
The correct option is D250μC Total charge on plate 1 and 2 is q=CV=600μC and total charge on plate 3 and 4 is 50×3=150μC Now charge xμC is flowing in the circuit (150−x)3+(600−x)6−200=0 x=−100μC Hence, final charge on 6μF capacitor is q2=150−x=150+100=250μC