A capacitor of capacitance 2C is charged to a potential 2V and then connected to an uncharged capacitor of capacitance C with a resistance R to a battery of emf 4V as shown.
The key is closed at t = 0. The charge on uncharged capacitor as a function of time will be

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$\frac{4 CV}{3} (1 - e^{\frac{- 3 t}{2 RC}})$

$4 CV (1 - e^{\frac{- t}{2 RC}})$

$4 CV (1 - e^{\frac{- 3 t}{2 RC}})$

$\frac{4 CV}{3} (1 - e^{\frac{- t}{2 RC}})$

answer is D.

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Detailed Solution

Let at any time a charge q flows through the circuit. The circuit is as shown
Time constant $(τ) = \frac{2 RC}{3}$

$∴ q = \frac{4 CV}{3} (1 - e^{\frac{- 3 t}{2 RC}})$

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