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

V=C1V1+C2V2C1+C2 Equivalent circuit

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

V=C1V1+C2V2C1+C2 Equivalent circuit

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

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

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

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

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

answer is C.

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$V = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$ Equivalent circuit

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

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