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Byju's Answer

Standard XII

Physics

Kirchoff's Voltage Law

In the circui...

Question

In the circuit, shown in fig. ‘K’ is open. The charge on capacitor C in steady state is $q_{1}$ . Now key is closed and at steady state, the charge on C is $q_{2}$ . The ratio of charges $(\frac{q_{1}}{q_{2}})$ is

\frac{4}{3}

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\frac{2}{3}

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1

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\frac{1}{2}

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Solution

The correct option is A $\frac{4}{3}$
When key is open, charge in steady state will be $q_{1} = C E$
When key is closed, potential difference across capacitor will be $V = \frac{3 R}{R + 3 R} E = \frac{3}{4} R$
Charge in steady state will be $q_{2} = \frac{3}{4} C E \Rightarrow \frac{q_{1}}{q_{2}} = \frac{4}{3}$

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In the circuit, shown in fig. ‘K’ is open. The charge on capacitor C in steady state is q1. Now key is closed and at steady state, the charge on C is q2. The ratio of charges (q1q2) is

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