Question

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

A 4/3
B 2/3
C 1
D 1/2

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Answer

The correct option is A $\frac{4}{3}$.

  1. When the key is open:

    • The charge on the capacitor $C$ in the steady state is given by: $$ q_1 = C \cdot E $$

  2. When the key is closed:

    • The potential difference across the capacitor $V$ can be determined using:

      $$ V = \frac{3R}{R + 3R} \cdot E = \frac{3}{4}E $$

    • Therefore, in the steady state, the charge on the capacitor $C$ will be: $$ q_2 = C \cdot \frac{3}{4}E $$

  3. Ratio of charges $\frac{q_1}{q_2}$:

    • Using the expressions for $q_1$ and $q_2$: $$ \frac{q_1}{q_2} = \frac{C \cdot E}{C \cdot \frac{3}{4}E} = \frac{1}{\frac{3}{4}} = \frac{4}{3} $$

Thus, the ratio of charges $\left( \frac{q_1}{q_2} \right)$ is $\frac{4}{3}$.


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