Four capacitors,each of capacity 4,mu F,are c | vedclass

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(B) The circuit consists of one $4\,\mu F$ capacitor in series with a parallel combination of two $4\,\mu F$ capacitors,which is then in series with a

Vedclass · Accepted Answer

$1800$

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(B) The circuit consists of one $4\,\mu F$ capacitor in series with a parallel combination of two $4\,\mu F$ capacitors,which is then in series with another $4\,\mu F$ capacitor.First,the two $4\,\mu F$ capacitors in parallel have an equivalent capacitance of $C_p = 4\,\mu F + 4\,\mu F = 8\,\mu F$.Now,the total equivalent capacitance $C_{eq}$ is given by the series combination of three capacitors ($4\,\mu F$,$8\,\mu F$,and $4\,\mu F$):$\frac{1}{C_{eq}} = \frac{1}{4} + \frac{1}{8} + \frac{1}{4} = \frac{2+1+2}{8} = \frac{5}{8}\,\mu F^{-1}$.Thus,$C_{eq} = \frac{8}{5}\,\mu F = 1.6 	imes 10^{-6}\,F$.The energy stored $U$ is given by $U = \frac{1}{2} C_{eq} V^2$.$U = \frac{1}{2} 	imes (1.6 	imes 10^{-6}) 	imes (15)^2 = 0.8 	imes 10^{-6} 	imes 225 = 180 	imes 10^{-6}\,J$.Since $1\,J = 10^7\,ergs$,the energy in $ergs$ is $U = 180 	imes 10^{-6} 	imes 10^7 = 1800\,ergs$.

Four capacitors,each of capacity $4\,\mu F$,are connected as shown in the figure. If $V_P - V_Q = 15\,V$,the energy stored in the system is . . . . . . $ergs$.

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