Three capacitors of capacitances 25 mu F, 30 | vedclass

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(B) In parallel combination,the potential difference $V$ is the same across all capacitors. The equivalent capacitance is $C_p = C_1 + C_2 + C_3 = (25

Vedclass · Accepted Answer

$86$

Vedclass · Answer

(B) In parallel combination,the potential difference $V$ is the same across all capacitors. The equivalent capacitance is $C_p = C_1 + C_2 + C_3 = (25 + 30 + 45) \mu F = 100 \mu F$.The energy stored is $E = \frac{1}{2} C_p V^2 = \frac{1}{2} 	imes 100 	imes 10^{-6} 	imes (100)^2 = 0.5 \ J$.In series combination,the equivalent capacitance $C_s$ is given by $\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} = \frac{1}{25} + \frac{1}{30} + \frac{1}{45} = \frac{18 + 15 + 10}{450} = \frac{43}{450} \mu F^{-1}$.Thus,$C_s = \frac{450}{43} \mu F$.The energy stored in series is $E' = \frac{1}{2} C_s V^2 = \frac{1}{2} 	imes \frac{450}{43} 	imes 10^{-6} 	imes (100)^2 = \frac{1}{2} 	imes \frac{450}{43} 	imes 10^{-2} = \frac{4.5}{86} \ J$.Given $E' = \frac{9}{x} E$,we have $\frac{4.5}{86} = \frac{9}{x} 	imes 0.5$.$\frac{4.5}{86} = \frac{4.5}{x} \implies x = 86$.

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