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Capacitors $C1$ and $C2$ are joined as shown below. If potential at point A i $V_A$ and point B is $V_B$, Potential at point C is

 

$(A)\;\frac{C2 V_A-C1 V_B}{C1+C2} \\ (B)\;\frac{C1 V_A+C2 V_B}{C1+C2} \\ (C)\; \frac{1}{2} (V_A+V_B) \\ (D)\;None $

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Let potential of point C be V.
If q is change on each capacitor.
$q= C1 (V_A-V)$
$q= C2 (V-V_B)$
=> $\large\frac{C1}{C2} =\large\frac{V-V_B}{V_A-V}$
=> $ V(C1+C2) =V_AC1+V_B C2$
=> $ V= \large\frac{C1 V_A+C2 V_B}{C1+C2}$
Hence B is the correct answer.
answered Jan 8, 2014 by meena.p
edited Aug 21, 2014 by meena.p
 

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