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Q.

Three identical capacitors, each of capacitance C are connected in series. The capacitors are charged by connecting a battery of emf V to the terminals (a and d) of the circuit. Now the battery is removed and two resistors of resistance R each are connected as shown. The heat dissipated(in μJ )  in one of the resistors is ?(Take C=1μF,V=3V )
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answer is 0.67.

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Detailed Solution

Charge on each plate (polarity shown) shown capacitors are charged Q=CV3.

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Total charge available for distribution amongst the capacitors = Q+QQ=Q
When charge flow stops, charge on each capacitor  Q3=CV9
Heat dissipated = loss in energy stored in capacitor system  =12.C3V212C(CV9)2×3=427CV2

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Heat dissipated in each resistor =227CV2.

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