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(C) The square loop is made of a $16 \ \Omega$ wire,so each side has a resistance of $4 \ \Omega$. Let the vertices of the square be $P, Q, R, S$ in o

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(C) The square loop is made of a $16 \ \Omega$ wire,so each side has a resistance of $4 \ \Omega$. Let the vertices of the square be $P, Q, R, S$ in order. Let the battery be connected across side $PQ$. The capacitor is connected across diagonal $PR$. Looking at the circuit,the current $I$ from the battery flows into the node $P$. One path from $P$ to $Q$ is the direct side $PQ$ with resistance $4 \ \Omega$. The other path from $P$ to $Q$ is through the rest of the loop: $P ightarrow S ightarrow R ightarrow Q$,which has a resistance of $4 \ \Omega + 4 \ \Omega + 4 \ \Omega = 12 \ \Omega$. These two paths are in parallel. The equivalent resistance of the loop is $R_{loop} = \frac{4 	imes 12}{4 + 12} = \frac{48}{16} = 3 \ \Omega$. Including the internal resistance $r = 1 \ \Omega$,the total resistance of the circuit is $R_{total} = 3 \ \Omega + 1 \ \Omega = 4 \ \Omega$. The total current from the battery is $I = \frac{V}{R_{total}} = \frac{9 \ V}{4 \ \Omega} = 2.25 \ A$. The potential difference across the parallel combination (nodes $P$ and $Q$) is $V_{PQ} = I 	imes R_{loop} = 2.25 	imes 3 = 6.75 \ V$. The current through the path $P ightarrow S ightarrow R ightarrow Q$ is $I_2 = \frac{V_{PQ}}{12 \ \Omega} = \frac{6.75}{12} = 0.5625 \ A$. The potential at $P$ is $V_P = 6.75 \ V$ and at $Q$ is $0 \ V$ (taking $Q$ as reference). The potential at $R$ is $V_R = V_P - I_2 	imes (4 + 4) = 6.75 - 0.5625 	imes 8 = 6.75 - 4.5 = 2.25 \ V$. The potential difference across the capacitor connected between $P$ and $R$ is $V_{PR} = V_P - V_R = 6.75 - 2.25 = 4.5 \ V$. The energy stored in the capacitor is $U = \frac{1}{2} C V_{PR}^2 = \frac{1}{2} 	imes 4 \ \mu F 	imes (4.5 \ V)^2 = 2 	imes 20.25 = 40.5 \ \mu J$. Given $U = \frac{x}{2} \ \mu J$,we have $40.5 = \frac{x}{2}$,which gives $x = 81$.

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