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(B) At steady state,the capacitors act as open circuits. Therefore,no current flows through the branches containing the capacitors.The current $i$ flo

Vedclass · Accepted Answer

$\frac{4}{5}$

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(B) At steady state,the capacitors act as open circuits. Therefore,no current flows through the branches containing the capacitors.The current $i$ flows through the series combination of the $6\,\Omega$ resistor,the galvanometer resistance $(2\,\Omega)$,and the $8\,\Omega$ resistor.The total resistance of the circuit is $R_{eq} = 6\,\Omega + 2\,\Omega + 8\,\Omega = 16\,\Omega$.The current in the circuit is $i = \frac{V}{R_{eq}} = \frac{4\,V}{16\,\Omega} = 0.25\,A = \frac{1}{4}\,A$.The potential difference across $C_1$ is the voltage across the branch $AC$. Since $C_1$ is in series with the $6\,\Omega$ resistor and the galvanometer,the voltage across $C_1$ is the potential difference between points $A$ and $C$. $V_1 = V_{AC} = i 	imes (6\,\Omega + 2\,\Omega) = \frac{1}{4} 	imes 8 = 2\,V$.The potential difference across $C_2$ is the voltage across the branch $BD$. Since $C_2$ is in series with the galvanometer and the $8\,\Omega$ resistor,the voltage across $C_2$ is the potential difference between points $B$ and $D$. $V_2 = V_{BD} = i 	imes (2\,\Omega + 8\,\Omega) = \frac{1}{4} 	imes 10 = 2.5\,V$.The ratio of the potential difference across $C_1$ and $C_2$ is $\frac{V_1}{V_2} = \frac{2}{2.5} = \frac{4}{5}$.

In the given figure,the resistance of the coil of the galvanometer $G$ is $2\,\Omega$. The emf of the cell is $4\,V$. The ratio of the potential difference across $C_1$ and $C_2$ is:

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