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(A) Since the capacitors $C_1$ and $C_2$ are in series,the charge $Q$ on each capacitor is the same.The potential difference across the combination is

Vedclass · Accepted Answer

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

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(A) Since the capacitors $C_1$ and $C_2$ are in series,the charge $Q$ on each capacitor is the same.The potential difference across the combination is $V_2 - V_1$ (assuming $V_2 > V_1$ for the direction of flow).Let $V_D$ be the potential at point $D$.The potential drop across $C_1$ is $V_D - V_1 = \frac{Q}{C_1}$ and across $C_2$ is $V_2 - V_D = \frac{Q}{C_2}$.Adding these,we get $(V_D - V_1) + (V_2 - V_D) = \frac{Q}{C_1} + \frac{Q}{C_2} = Q \left( \frac{C_1 + C_2}{C_1 C_2} \right)$.Thus,$V_2 - V_1 = Q \left( \frac{C_1 + C_2}{C_1 C_2} \right)$,which gives $Q = \frac{C_1 C_2 (V_2 - V_1)}{C_1 + C_2}$.Now,substituting $Q$ back into the expression for $V_D$:$V_D - V_1 = \frac{Q}{C_1} = \frac{1}{C_1} \cdot \frac{C_1 C_2 (V_2 - V_1)}{C_1 + C_2} = \frac{C_2 (V_2 - V_1)}{C_1 + C_2}$.$V_D = V_1 + \frac{C_2 V_2 - C_2 V_1}{C_1 + C_2} = \frac{V_1 C_1 + V_1 C_2 + C_2 V_2 - C_2 V_1}{C_1 + C_2} = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.

Two capacitors $C_1$ and $C_2$ in a circuit are joined as shown below. 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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