A circuit has a section AB as shown in the fi | vedclass

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(D) Let $V_A$ and $V_B$ be the potentials at points $A$ and $B$ respectively. Given $V_A - V_B = V$.Let $q$ be the charge on the capacitors. Since the

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$(V-E)C_2/(C_1+C_2)$

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(D) Let $V_A$ and $V_B$ be the potentials at points $A$ and $B$ respectively. Given $V_A - V_B = V$.Let $q$ be the charge on the capacitors. Since they are in series,the charge $q$ is the same for both.The potential difference across $C_1$ is $V_1 = q/C_1$ and across $C_2$ is $V_2 = q/C_2$.Applying Kirchhoff's Voltage Law from $A$ to $B$:$V_A - V_1 - E - V_2 = V_B$$V_A - V_B = V_1 + V_2 + E$$V = q/C_1 + q/C_2 + E$$V - E = q(1/C_1 + 1/C_2) = q(C_1 + C_2)/(C_1 C_2)$$q = (V - E) (C_1 C_2) / (C_1 + C_2)$The potential difference across $C_1$ is $V_1 = q/C_1 = [(V - E) (C_1 C_2) / (C_1 + C_2)] / C_1 = (V - E) C_2 / (C_1 + C_2)$.

$A$ circuit has a section $AB$ as shown in the figure. If the potential difference between points $A$ and $B$ is $V$ volts,then the potential difference across $C_1$ is

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