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Two capacitors of capacitance $C_1 = 6 \mu F$ and $C_2 = 3 \mu F$ are connected in series across a cell of emf $18\; V$. Calculate the potential difference across capacitor $C_2$.

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$C_1 = 6 \mu F, C_2 = 3\mu F, V = 18V$
As $C_1$ and $C_2$ are in series, the effective capacitance is given as:
$\large\frac{1}{C}=\frac{1}{C_1} +\frac{1}{C_2}$
$C=\large\frac{C_1C_2}{C_1+C_2}$
$\quad=\large\frac{6 \times 3}{6+3}$
$\quad= 2 \mu F$
$v_1=\large\frac{C_2}{C_1+C_2}$$ V= \large\frac{3}{6+3} $$\times 18=6 volts$
$v_2=\large\frac{C_1}{C_1+C_2}$$ V= \large\frac{6}{6+3} $$\times 18=12 volts$
answered Jun 17, 2014 by meena.p
 

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