Two condensers C₁ & 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

Charge on $C_{1} =$ Charge on $C_{2}$
$\begin{array}{l} \Rightarrow C_{1} (V_{A} - V_{D}) = C_{2} (V_{D} - V_{B}) \\ \Rightarrow C_{1} (V_{1} - V_{D}) = C_{2} (V_{D} - V_{2}) \\ V_{D} = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}} \end{array}$

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