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 $μ J$ ) in one of the resistors is ?(Take $C = 1 μ F, V = 3 V$ )

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answer is 0.67.

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Detailed Solution

Charge on each plate (polarity shown) shown capacitors are charged $Q = \frac{C V}{3}$ .

Total charge available for distribution amongst the capacitors = $Q + Q - Q = Q$
When charge flow stops, charge on each capacitor $\frac{Q}{3} = \frac{C V}{9}$
Heat dissipated = loss in energy stored in capacitor system $= \frac{1}{2} . \frac{C}{3} V^{2} - \frac{1}{2 C} {(\frac{C V}{9})}^{2} \times 3 = \frac{4}{27} C V^{2}$

Heat dissipated in each resistor $= \frac{2}{27} C V^{2}$ .

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