In the circuit shown below, C_1 = 10 , mu F, | vedclass

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(B) Let the potential at the junction between the two capacitor blocks be $V_m$. Let the potential at $X$ be $V_X$ and at $Y$ be $V_Y$. Given $C_1 = 1

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$3$

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(B) Let the potential at the junction between the two capacitor blocks be $V_m$. Let the potential at $X$ be $V_X$ and at $Y$ be $V_Y$. Given $C_1 = 10 \, \mu F$, $C_2 = 20 \, \mu F$, $C_3 = 20 \, \mu F$, and $C_4 = 40 \, \mu F$. Charge on $C_1$ is $q_1 = 20 \, \mu C$. Potential difference across $C_1$ is $V_1 = \frac{q_1}{C_1} = \frac{20 \, \mu C}{10 \, \mu F} = 2 \, V$. Since $C_1$ and $C_2$ are in parallel, the potential difference across $C_2$ is also $2 \, V$. Charge on $C_2$ is $q_2 = C_2 V_1 = 20 \, \mu F 	imes 2 \, V = 40 \, \mu C$. The total charge entering the junction from the left side is $q_1 + q_2 = 20 \, \mu C + 40 \, \mu C = 60 \, \mu C$. This total charge must be distributed among $C_3$ and $C_4$ in series with the first block. Since $C_3$ and $C_4$ are in parallel, the total charge $60 \, \mu C$ is shared between them. The equivalent capacitance of the second block is $C_{34} = C_3 + C_4 = 20 \, \mu F + 40 \, \mu F = 60 \, \mu F$. The potential difference across the second block is $V_2 = \frac{q_{total}}{C_{34}} = \frac{60 \, \mu C}{60 \, \mu F} = 1 \, V$. The total potential difference between $X$ and $Y$ is $V_{XY} = V_1 + V_2 = 2 \, V + 1 \, V = 3 \, V$.

In the circuit shown below, $C_1 = 10 \, \mu F$, $C_2 = C_3 = 20 \, \mu F$, and $C_4 = 40 \, \mu F$. If the charge on $C_1$ is $20 \, \mu C$, then the potential difference between $X$ and $Y$ is ......... $V$.

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