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(D) In steady state,the capacitor acts as an open circuit,so no current flows through it.The circuit consists of two parallel branches connected acros

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$60$

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(D) In steady state,the capacitor acts as an open circuit,so no current flows through it.The circuit consists of two parallel branches connected across the $6 \, V$ battery.Branch $1$ (left side) consists of resistors $3 \, \Omega$ and $7 \, \Omega$ in series. The potential at $D$ relative to $A$ is given by the voltage divider rule: $V_{A} - V_{D} = \frac{3}{3+7} 	imes 6 = 1.8 \, V$.Branch $2$ (right side) consists of resistors $2 \, \Omega$ and $4 \, \Omega$ in series. The potential at $B$ relative to $A$ is: $V_{A} - V_{B} = \frac{2}{2+4} 	imes 6 = 2 \, V$.The potential difference across the capacitor (between points $D$ and $B$) is $V_{DB} = |V_{A} - V_{B} - (V_{A} - V_{D})| = |2 - 1.8| = 0.2 \, V$.The energy stored in the capacitor is $U = \frac{1}{2} CV^{2}$.Given $C = 3 \, \mu F = 3 	imes 10^{-6} \, F$ and $V = 0.2 \, V$.$U = \frac{1}{2} 	imes (3 	imes 10^{-6}) 	imes (0.2)^{2} = \frac{1}{2} 	imes 3 	imes 10^{-6} 	imes 0.04 = 6 	imes 10^{-8} \, J = 60 	imes 10^{-9} \, J = 60 \, nJ$.

In steady state,the energy stored in the capacitor as shown in the figure is ..... $nJ$.

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