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(C) According to Kirchhoff's Current Law $(KCL)$,the sum of currents entering a junction must equal the sum of currents leaving the junction.Let us an

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$13$

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(C) According to Kirchhoff's Current Law $(KCL)$,the sum of currents entering a junction must equal the sum of currents leaving the junction.Let us analyze the junctions in the circuit.$1$. At the left junction: $10 \, A$ enters,and it splits into two branches. Let the upper branch current be $I_1$ and the lower branch current be $6 \, A$. Thus,$10 = I_1 + 6$,which gives $I_1 = 4 \, A$.$2$. At the top junction: $1 \, A$ enters from the top and $I_1 = 4 \, A$ enters from the left. These combine to flow into the right branch. Let this current be $I_2$. Thus,$I_2 = 1 + 4 = 5 \, A$.$3$. At the bottom junction: $6 \, A$ leaves from the left branch and $2 \, A$ enters from the bottom. The current flowing into the right branch from this junction is $I_3$. By $KCL$,$6 + I_3 = 2$ is incorrect based on the diagram flow; looking at the arrows,$6 \, A$ enters the bottom junction and $2 \, A$ enters the bottom junction,so the total current leaving towards the right is $I_3 = 6 + 2 = 8 \, A$.$4$. At the right junction: $I_2 = 5 \, A$ enters from the top and $I_3 = 8 \, A$ enters from the bottom. These combine to form the output current $I$. Thus,$I = 5 + 8 = 13 \, A$.

The figure shows a network of currents. The values of the currents are shown in the figure. What is the value of current $I$ in $A$?

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