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Question

A Circuit contains a resistor of resistance R and a capacitor C1 of capacitance C, a second Capacitor C2 of capacitance 2C, and a battery that provides an emf of V, as shown above.
Initially, both switch 1 is closed, switch 2 is open and the circuit is at steady state. At time t=0, both switch 1 is closed and switch 2 is closed. After the circuit reaches steady state, what is the charge Q2 on capacitor C2?
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A
Q2=13VC
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B
Q2=12VC
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C
Q2=23VC
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D
Q2=VC
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E
Q2=32VC
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Solution

The correct option is D Q2=VC
Capacitor C2 has a capacitance 2C. Emf is V and we know that
Q=CV
Initially when switch was opened the potential difference C1 is V but across C2 it will become V2.
So charge on C2isQ2=2C×V2
Q2=CV

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