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(B) Let $i$ be the total current flowing from $A$ to $B$. The $2\,\Omega$ resistor is in parallel with the series combination of the $9\,V$ battery an

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$3.5$

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(B) Let $i$ be the total current flowing from $A$ to $B$. The $2\,\Omega$ resistor is in parallel with the series combination of the $9\,V$ battery and $1\,\Omega$ resistor. However,applying Kirchhoff's laws to the whole circuit:Let $i_1$ be the current through the $2\,\Omega$ resistor and $i_2$ be the current through the $1\,\Omega$ resistor and $9\,V$ battery branch.At the junction,the total current $i = i_1 + i_2$.Applying Kirchhoff's voltage law from $A$ to $B$:$V_A - 4i - 9 - 1(i_2) + 3 - 4i = V_B$Given $V_A - V_B = 16\,V$,so $16 - 8i - 1(i_2) - 6 = 0 \implies 8i + i_2 = 10$ (Eq. $1$).Also,the potential difference across the parallel branches is equal:$9 - 1(i_2) = 2(i_1) \implies 9 - i_2 = 2i_1$ (Eq. $2$).Substituting $i = i_1 + i_2$ into Eq. $1$: $8(i_1 + i_2) + i_2 = 10 \implies 8i_1 + 9i_2 = 10$.Solving the system:$2i_1 + i_2 = 9 \implies i_2 = 9 - 2i_1$.$8i_1 + 9(9 - 2i_1) = 10 \implies 8i_1 + 81 - 18i_1 = 10 \implies 10i_1 = 71 \implies i_1 = 7.1\,A$ (Wait,re-evaluating the circuit path).Correct approach: The potential difference across the parallel part is $V_p$. $V_A - 4i - V_p - 3 - 4i = V_B \implies 16 - 8i - V_p - 3 = 0 \implies 8i + V_p = 13$.$V_p = 9 - 1(i_2) = 2(i_1)$. Since $i = i_1 + i_2$,$i_2 = V_p$ and $i_1 = V_p/2$.$i = 1.5 V_p$. Substituting: $8(1.5 V_p) + V_p = 13 \implies 12 V_p + V_p = 13 \implies 13 V_p = 13 \implies V_p = 1\,V$.Current through $2\,\Omega$ resistor $i_1 = V_p / 2 = 1 / 2 = 0.5\,A$. Given the options,there might be a typo in the question's provided solution. Based on standard circuit analysis,the current is $0.5\,A$.

In the circuit shown in the figure,the potential difference between points $A$ and $B$ is $16\,V$. The current passing through the $2\,\Omega$ resistance will be $...........\,A$.

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