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(D) To find the current in the portion $PQ$,we apply Kirchhoff's current law $(KCL)$ at the junctions.$1$. At the junction where the $2 \, A$ and $8 \

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$6 \, A$ from $Q$ to $P$

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(D) To find the current in the portion $PQ$,we apply Kirchhoff's current law $(KCL)$ at the junctions.$1$. At the junction where the $2 \, A$ and $8 \, A$ currents meet,the total incoming current is $2 \, A + 8 \, A = 10 \, A$. This $10 \, A$ current flows towards the next junction.$2$. At the junction below $Q$,the total outgoing current is $4 \, A + 2 \, A = 6 \, A$. Since $10 \, A$ enters this junction and $6 \, A$ leaves downwards,the remaining $10 \, A - 6 \, A = 4 \, A$ must flow upwards towards $Q$.$3$. At junction $Q$,the incoming currents are $3 \, A$ (from the top branch) and $4 \, A$ (from the lower branch). The outgoing current is $1 \, A$ (to the right). Let the current in $PQ$ be $I_{PQ}$ flowing from $Q$ to $P$.$4$. Applying $KCL$ at junction $Q$: $I_{incoming} = I_{outgoing}$.$3 \, A + 4 \, A = 1 \, A + I_{PQ}$$7 \, A = 1 \, A + I_{PQ}$$I_{PQ} = 6 \, A$.Since the result is positive,the current $6 \, A$ flows in the assumed direction,i.e.,from $Q$ to $P$.

The figure below shows a portion of an electric circuit with the currents in amperes and their directions. The magnitude and direction of the current in the portion $PQ$ is

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