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Potential difference between points A and B is 5V, then voltage across $2 \mu F$ capacitor is :


$(A)\;15 v \\ (B)\;8.5 v \\ (C)\; 10 v \\ (D)\;0V $

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If Q is the charge flowing.
By Kirchhoff 's law,
$V_A-\large\frac{Q}{C_1} $$+E-\large\frac{Q}{C_2}$$=V_B$
=> $V_A-V_B+E= Q \bigg( \large\frac{1}{C_1}+\frac{1}{C_2}\bigg)$
=> $5+10= Q \bigg( \large\frac{C_1+C_2}{C_1C_2}\bigg)$
=> $\qquad= Q \bigg( \large\frac{2+2}{2 \times 2}\bigg)$
=>$Q= 15 \mu c$
V across $C_2 =\large\frac{15 \mu C}{2 \mu F}$
$\qquad =7.5 V$
Hence B is the correct answer.
answered Jan 10, 2014 by meena.p

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