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If at a node $I_1=2\,\text{A}$, $I_2=3\,\text{A}$, $I_3=1\,\text{A}$ enter the junction and $I_4=4\,\text{A}$, $I_5=1\,\text{A}$ leave it, then the current $I_6$ leaving the junction is
A$1\,\text{A}$
B$2\,\text{A}$
C$3\,\text{A}$
D$6\,\text{A}$
Answer & Solution
Correct answer: A. $1\,\text{A}$
By KCL, total entering current equals total leaving current. So $2+3+1=4+1+I_6$, giving $6=5+I_6$ and hence $I_6=1\,\text{A}$.
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