A galvanometer (G) of 2 Omega resistance is | vedclass

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(D) In steady state,the capacitors act as open circuits,so no current flows through them.The circuit consists of three resistors in series: $4 \ \Omeg

Vedclass · Accepted Answer

$\frac{1}{2}$

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(D) In steady state,the capacitors act as open circuits,so no current flows through them.The circuit consists of three resistors in series: $4 \ \Omega$,$2 \ \Omega$ (galvanometer),and $6 \ \Omega$.Total resistance $R_{eq} = 4 + 2 + 6 = 12 \ \Omega$.The current in the circuit is $I = \frac{V}{R_{eq}} = \frac{6 \ V}{12 \ \Omega} = 0.5 \ A$.Let the nodes be $A$ (left),$B$ (top),$C$ (bottom),and $D$ (right). The potential difference across $C_1$ is the potential difference between nodes $A$ and $C$. The potential difference across $C_2$ is the potential difference between nodes $B$ and $D$.Potential at $A = 6 \ V$,potential at $D = 0 \ V$.Potential at $B = V_A - I 	imes 4 \ \Omega = 6 - 0.5 	imes 4 = 4 \ V$.Potential at $C = V_B - I 	imes 2 \ \Omega = 4 - 0.5 	imes 2 = 3 \ V$.Potential difference across $C_1$ $(V_{C1})$ = $V_A - V_C = 6 - 3 = 3 \ V$.Potential difference across $C_2$ $(V_{C2})$ = $V_B - V_D = 4 - 0 = 4 \ V$.Charge $q_1 = C_1 	imes V_{C1} = 4 \ \mu F 	imes 3 \ V = 12 \ \mu C$.Charge $q_2 = C_2 	imes V_{C2} = 6 \ \mu F 	imes 4 \ V = 24 \ \mu C$.The ratio $\frac{q_1}{q_2} = \frac{12 \ \mu C}{24 \ \mu C} = \frac{1}{2}$.

$A$ galvanometer $(G)$ of $2 \ \Omega$ resistance is connected in the given circuit. The ratio of charge in $C_1$ $(4 \ \mu F)$ and $C_2$ $(6 \ \mu F)$ is:

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