What is the energy stored in the capacitor of | vedclass

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(B) To find the energy stored in the capacitor,we first need to determine the potential difference $V_{AB}$ across it.Applying Kirchhoff's Current Law

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$8 	imes 10^{-4} \, J$

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(B) To find the energy stored in the capacitor,we first need to determine the potential difference $V_{AB}$ across it.Applying Kirchhoff's Current Law at junction $A$: The current entering from the left is $2 \, A$ and from the top is $1 \, A$. Thus,the current flowing through the $5 \, \Omega$ resistor towards $D$ is $i_1 = 2 + 1 = 3 \, A$.Applying Kirchhoff's Current Law at junction $C$: The current $i_1 = 3 \, A$ enters $C$ from $D$. The current $i_2$ flows from $C$ to $B$ through the $2 \, \Omega$ resistor. The current $i_3$ flows out of $C$ through the $4 \, \Omega$ resistor. Based on the circuit diagram,the current $i_2$ is $1 \, A$.Now,calculate the potential difference $V_{AB} = V_A - V_B$ by traversing the path $A 	o D 	o C 	o B$:$V_A - V_B = V_A - V_D + V_D - V_C + V_C - V_B$$V_A - V_B = (i_1 	imes 5) + (i_1 	imes 1) + (i_2 	imes 2)$$V_A - V_B = (3 	imes 5) + (3 	imes 1) + (1 	imes 2) = 15 + 3 + 2 = 20 \, V$.The energy stored in the capacitor is given by $U = \frac{1}{2} C V^2$.$U = \frac{1}{2} 	imes (4 	imes 10^{-6} \, F) 	imes (20 \, V)^2$$U = 2 	imes 10^{-6} 	imes 400 = 800 	imes 10^{-6} \, J = 8 	imes 10^{-4} \, J$.

What is the energy stored in the capacitor of the given circuit?

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