For the network of resistors shown in the fig | vedclass

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(B) Let the left junction be $A$ and the right junction be $B$. The circuit is a bridge. The left part consists of two parallel branches: one with $2\

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$I_1 = 3A$

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(B) Let the left junction be $A$ and the right junction be $B$. The circuit is a bridge. The left part consists of two parallel branches: one with $2\Omega$ and $1\Omega$ in series (total $3\Omega$) and another with $4\Omega$ in series with the $PQ$ branch. However,looking at the bridge structure,the left side has two parallel branches connected to $P$ and $Q$: one branch has $2\Omega$ and $4\Omega$ resistors. The potential at $P$ and $Q$ can be calculated. The equivalent resistance of the left part is $R_L = (2 || 4) = 8/6 = 4/3 \Omega$. The equivalent resistance of the right part is $R_R = (2 || 4) = 4/3 \Omega$. The central part has $1\Omega$ (between $P$ and $S$) and $1\Omega$ (between $Q$ and $T$) and $4\Omega$ (between $Q$ and $T$). Actually,the bridge is balanced if the ratio of resistances is equal. The left side has $2\Omega$ and $4\Omega$ connected to the source. The potential at $P$ is $V_P = 12 	imes (4/(2+4)) = 8V$ and $V_Q = 12 	imes (2/(2+4)) = 4V$. The right side is symmetric,so $V_S = 4V$ and $V_T = 8V$. Since $V_P = 8V$ and $V_S = 4V$,current flows from $P$ to $S$. The branch $PQ$ has a $1\Omega$ resistor. Since $V_P=8V$ and $V_Q=4V$,current flows through $PQ$. Thus,option $A$ is incorrect. Calculating $I_1$: The total resistance is $R_{eq} = 4\Omega$. $I_1 = V/R_{eq} = 12/4 = 3A$. Option $B$ is correct. Since $V_S = 4V$ and $V_Q = 4V$,the potential at $S$ is equal to the potential at $Q$. Thus,$C$ is incorrect. Therefore,the correct option is $B$.

For the network of resistors shown in the figure,choose the correct option.

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