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(C) The capacitors $C_1$ and $C_2$ are connected in series between points $A$ and $B$.Let $V_D$ be the potential at point $D$.The charge $q$ on both c

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$\frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$

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(C) The capacitors $C_1$ and $C_2$ are connected in series between points $A$ and $B$.Let $V_D$ be the potential at point $D$.The charge $q$ on both capacitors must be the same because they are in series.The potential difference across $C_1$ is $V_1 - V_D = \frac{q}{C_1}$.The potential difference across $C_2$ is $V_D - V_2 = \frac{q}{C_2}$.From these,we have $q = C_1(V_1 - V_D) = C_2(V_D - V_2)$.Expanding this,$C_1 V_1 - C_1 V_D = C_2 V_D - C_2 V_2$.Rearranging to solve for $V_D$,we get $C_1 V_1 + C_2 V_2 = V_D(C_1 + C_2)$.Therefore,$V_D = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.

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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