Two condensers C₁ and C₂ in a circuit are joined as shown in the figure. The potential of point A is V₁ and that of B is V₂. The potential of point D will be

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$\frac{1}{2} (V_{1} + V_{2})$

$\frac{C_{2} V_{1} + C_{1} V_{2}}{C_{1} + C_{2}}$

$\frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$

$\frac{C_{2} V_{1} - C_{1} V_{2}}{C_{1} + C_{2}}$

answer is C.

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Detailed Solution

$V_{1} - V_{2}$ will be divided between C₁ and C₂ and V_D is potential at point D in series.

Potential difference across C₁ is

$\begin{aligned} V_{1} - V_{D} = \frac{C_{2} (V_{1} - V_{2})}{C_{1} + C_{2}} \\ \Rightarrow V_{D} = V_{1} - \frac{C_{2} (V_{1} - V_{2})}{C_{1} + C_{2}} \\ \Rightarrow V_{D} = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} \end{aligned}$

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