In the circuit shown,the energy stored in the | vedclass

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(C) The $5\,\mu F$ and $1\,\mu F$ capacitors are in parallel,so their equivalent capacitance is $C_{p} = 5\,\mu F + 1\,\mu F = 6\,\mu F$.This $C_{p}$

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$32$

Vedclass · Answer

(C) The $5\,\mu F$ and $1\,\mu F$ capacitors are in parallel,so their equivalent capacitance is $C_{p} = 5\,\mu F + 1\,\mu F = 6\,\mu F$.This $C_{p}$ is in series with the $3\,\mu F$ capacitor.The potential difference across the branch containing the $3\,\mu F$ and $6\,\mu F$ capacitors is $24\,V$.Let $V_{AB}$ be the potential difference across the $3\,\mu F$ capacitor and $V_{BC}$ be the potential difference across the parallel combination $(6\,\mu F)$.Since they are in series,the ratio of potential differences is inversely proportional to the ratio of capacitances: $V_{AB} : V_{BC} = C_{p} : C_{3\mu F} = 6 : 3 = 2 : 1$.Thus,$V_{BC} = \frac{1}{2+1} 	imes 24\,V = \frac{1}{3} 	imes 24\,V = 8\,V$.The $1\,\mu F$ capacitor is in parallel with the $5\,\mu F$ capacitor,so the potential difference across the $1\,\mu F$ capacitor is also $8\,V$.The energy stored in the $1\,\mu F$ capacitor is $U = \frac{1}{2} C V^{2} = \frac{1}{2} 	imes (1 	imes 10^{-6}\,F) 	imes (8\,V)^{2} = \frac{1}{2} 	imes 10^{-6} 	imes 64 = 32 	imes 10^{-6}\,J = 32\,\mu J$.

List-$I$	List-$II$
$(i) P$	$(A) (0, 0)$
$(ii) Q$	$(B) (6, 0)$
$(iii) R$	$(C) (-2, 1)$
$(iv) S$	$(D) (-6, 0)$
	$(E) (-6, -3)$
	$(F) (-6, 3)$

In the circuit shown,the energy stored in the $1\,\mu F$ capacitor is ..... $\mu J$.

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