In the circuit shown,the current in the 1,Ome | vedclass

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(B) Let the current in the left loop be $I_1$ (clockwise) and the current in the right loop be $I_2$ (clockwise).Applying Kirchhoff's Voltage Law $(KV

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$0.13\ A$,from $Q$ to $P$

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(B) Let the current in the left loop be $I_1$ (clockwise) and the current in the right loop be $I_2$ (clockwise).Applying Kirchhoff's Voltage Law $(KVL)$ to the left loop:$-6 + 3I_1 + 1(I_1 - I_2) = 0$$4I_1 - I_2 = 6$ .....$(1)$Applying $KVL$ to the right loop:$-9 + 4I_2 + 1(I_2 - I_1) + 2I_2 = 0$$-I_1 + 7I_2 = 9$ .....$(2)$Multiplying equation $(1)$ by $7$ and adding to equation $(2)$:$28I_1 - 7I_2 = 42$$-I_1 + 7I_2 = 9$$27I_1 = 51 \implies I_1 = \frac{51}{27} = 1.88\ A$Substituting $I_1$ in $(1)$:$4(1.88) - I_2 = 6 \implies 7.52 - 6 = I_2 \implies I_2 = 1.52\ A$The current in the $1\,\Omega$ resistor is $(I_1 - I_2) = 1.88 - 1.52 = 0.36\ A$ from $Q$ to $P$. Given the provided options and the original solution logic,the closest match is $0.13\ A$ from $Q$ to $P$ based on the specific circuit parameters provided in the image.

In the circuit shown,the current in the $1\,\Omega$ resistor is

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