A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential V₀. Another capacitor of capacitance 2C is charged to a potential of 2V₀. Now both capacitors are connected to each other in such a way that positive plate of one is connected to negative plate of the other. The final potential difference across the combination is

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$V_{0}$

$\frac{3 V_{0}}{2}$

$\frac{2 V_{0}}{3}$

$\frac{4 V_{0}}{3}$

answer is A.

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Detailed Solution

Net charge stored in the combination is given by $Q = 2 C \times 2 V_{0} - C V_{0} = 3 C V_{0}$

Capacitance of the combination, $C = C_{0} + 2 C_{0} = 3 C_{0}$

$∴$ Final potential difference $V = \frac{Q}{C} = \frac{3 C V_{0}}{3 C_{0}} = V_{0}$

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