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Kirchhoff's Junction Law

Three uncharg...

Question

Three uncharged capacitors of capacities $C_{1}, C_{2}$ and $C_{3}$ are connected to one another as shown in the figure. Points A, B and D are at potentials $V_{1}, V_{2}$ and $V_{3}$ . then the potential at O will be :

\frac{V_{1} C_{1} + V_{2} C_{2} + V_{3} C_{3}}{C_{1} + C_{2} + C_{3}}

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\frac{V_{1} + V_{2} + V_{3}}{C_{1} + C_{2} + C_{3}}

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\frac{V_{1} (V_{2} + V_{3}}{C_{1} (C_{2} + C_{3})}

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\frac{V_{1} V_{2} V_{3}}{C_{1} C_{2} C_{3}}

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Solution

The correct option is A $\frac{V_{1} C_{1} + V_{2} C_{2} + V_{3} C_{3}}{C_{1} + C_{2} + C_{3}}$
From the circuit, the total charge on inner plates of $C_{1}$ , $C_{2}$ and $C_{3}$ is zero, $Q_{1} + Q_{2} + Q_{3} = 0$ . Let, $V_{o}$ is the potential at O
$V_{1}, V_{2}$ and $V_{3} > V_{\circ}$
therefore, $C_{1} (V_{1} - V_{\circ}) + C_{2} (V_{2} - V_{\circ}) + C_{3} (V_{3} - V_{\circ}) = 0$
$C_{1} V_{1} + C_{2} V_{2} + C_{3} V_{3} = V_{0} (C_{1} + C_{2} + C_{3})$
$V_{0} = \frac{C_{1} V_{1} + C_{2} V_{2} + C_{3} V_{3}}{C_{1} + C_{2} + C_{3}}$

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