(C) Since $C_1$ and $C_2$ are connected in series,the charge on $C_1$ must be equal to the charge on $C_2$ because the junction $D$ is isolated.Let $V

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$\frac{C_1V_1 + C_2V_2}{C_1 + C_2}$

(C) Since $C_1$ and $C_2$ are connected in series,the charge on $C_1$ must be equal to the charge on $C_2$ because the junction $D$ is isolated.Let $V_D$ be the potential at point $D$.The charge on capacitor $C_1$ is $Q_1 = C_1(V_1 - V_D)$.The charge on capacitor $C_2$ is $Q_2 = C_2(V_D - V_2)$.Since $Q_1 = Q_2$,we have:$C_1(V_1 - V_D) = C_2(V_D - V_2)$$C_1V_1 - C_1V_D = C_2V_D - C_2V_2$$C_1V_1 + C_2V_2 = V_D(C_1 + C_2)$$V_D = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$

Class 11 Chemistry Data Entry - Chemistry Level 0 of all question bank

Two capacitors $C_1$ and $C_2$ in a circuit are joined as shown in the figure. The potential of point $A$ is $V_1$ and that of $B$ is $V_2$. The potential of point $D$ will be

MHT CET 2022 All MHT CET

A
$\frac{1}{2}(V_1 + V_2)$
B
$\frac{C_2V_1 + C_1V_2}{C_1 + C_2}$
C
$\frac{C_1V_1 + C_2V_2}{C_1 + C_2}$
D
$\frac{C_2V_1 - C_1V_2}{C_1 + C_2}$

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Two capacitors $C_1$ and $C_2$ in a circuit are joined as shown in the figure. The potential of point $A$ is $V_1$ and that of $B$ is $V_2$. The potential of point $D$ will be

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