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Standard XII

Physics

Displacement Vector

A Circuit con...

Question

A Circuit contains a resistor of resistance $R$ and a capacitor $C_{1}$ of capacitance $C$ , a second Capacitor $C_{2}$ of capacitance $2 C$ , 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 $Q_{2}$ on capacitor $C_{2}$ ?

Q_{2} = \frac{1}{3} V C

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Q_{2} = \frac{1}{2} V C

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Q_{2} = \frac{2}{3} V C

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Q_{2} = V C

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Q_{2} = \frac{3}{2} V C

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Solution

The correct option is D $Q_{2} = V C$
Capacitor $C_{2}$ has a capacitance 2C. Emf is V and we know that
$Q = C V$
Initially when switch was opened the potential difference $C_{1}$ is V but across $C_{2}$ it will become $\frac{V}{2}$ .
So charge on $C_{2} i s Q_{2} = 2 C \times \frac{V}{2}$
$Q_{2} = C V$

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The correct option is D Q2=VCCapacitor C2 has a capacitance 2C. Emf is V and we know that Q=CVInitially when switch was opened the potential difference C1 is V but across C2 it will become V2.So charge on C2isQ2=2C×V2Q2=CV