Discussion
Find the current \(i_{7}\) in the following figure.
Answer: D
By Kirchhoff's current law, the currents entering a node equal the currents leaving it. At the left node i1, i2 and i3, each 1 A, enter and i5 leaves, so i5 = 3 A. At the right node i5 and i6 enter, with i6 = 3 A, and i7 leaves, so
\[ i_{7} = i_{5} + i_{6} = 3 + 3 = 6\ \mathrm{A} \]
So i7 is 6 A, option (D). The values 0 A, 3 A and 4 A come from ignoring one of the currents or giving i6 the wrong direction. The same rule gives a quick check on the figure: the three incoming currents of 1 A each make 3 A in the middle, and then the 3 A from i6 adds to it.
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