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If at a junction $I_1=2\,\text{A}$, $I_2=3\,\text{A}$, $I_3=1\,\text{A}$, $I_4=4\,\text{A}$, and $I_5=1\,\text{A}$, then according to Kirchhoff's Current Law, $I_6$ must be:
A$1\,\text{A}$
B$2\,\text{A}$
C$3\,\text{A}$
D$6\,\text{A}$
Answer & Solution
Correct answer: A. $1\,\text{A}$
Using KCL, $I_1 + I_2 + I_3 = I_4 + I_5 + I_6$. So $2+3+1 = 4+1+I_6$, giving $6=5+I_6$, hence $I_6=1\,\text{A}$.