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(B) Let the currents in the left and right branches be $I_1$ and $I_2$ respectively. The middle branch carries current $I = I_1 + I_2$.Applying Kirchh

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(B) Let the currents in the left and right branches be $I_1$ and $I_2$ respectively. The middle branch carries current $I = I_1 + I_2$.Applying Kirchhoff's Voltage Law $(KVL)$ to the left loop:$2 - 2I_1 - 2I = 0 \implies 2 - 2I_1 - 2(I_1 + I_2) = 0 \implies 2 - 4I_1 - 2I_2 = 0 \implies 2I_1 + I_2 = 1$  ---$(i)$Applying $KVL$ to the right loop:$-2 - 2I_2 - 2I = 0 \implies -2 - 2I_2 - 2(I_1 + I_2) = 0 \implies -2 - 2I_1 - 4I_2 = 0 \implies I_1 + 2I_2 = -1$  ---(ii)Solving equations $(i)$ and (ii):From $(i)$, $I_2 = 1 - 2I_1$. Substituting into (ii):$I_1 + 2(1 - 2I_1) = -1$$I_1 + 2 - 4I_1 = -1$$-3I_1 = -3 \implies I_1 = 1 A$Then $I_2 = 1 - 2(1) = -1 A$.The current $I$ in the middle branch is $I = I_1 + I_2 = 1 + (-1) = 0 A$.

The current $I$ in the circuit shown is:

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