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In that below diagram $C_1= 4 \mu F, C_2 =12 \mu F, C_3=8 \mu F$ Energy consumed in charging these capacitors is :

 

$(A)\;22 \times 10^{-6} J \\ (B)\;88 \times 10^{-6} J \\ (C)\; \frac{64}{3} \times 10^{-6} J\\ (D)\;\frac{32}{3} \times 10^{-6} J $

Can you answer this question?
 
 

1 Answer

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Equivalent capacitance $=\large\frac{C_1C_2}{C_1+C_2} +C_3$
$\qquad= \large\frac{4 \times 12}{4+ 12} $$+8$
$\qquad= 11 \mu F$
$Q= CV$
$ \quad= 11 \times 10^{-6} \times 4= 44 \times 10^{-6}V$
$W= \large\frac{1}{2} $$QV=\large\frac{1}{2}$$ \times 44 \times 10^{-6} \times 4$
$\quad= \large\frac{1}{2}$$ \times 44 \times 10^{-6} \times 4$
$\quad= 88 \times 10^{-6} J$
Hence B is the correct answer.

 

answered Jan 8, 2014 by meena.p
edited Jul 31, 2014 by thagee.vedartham
 

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