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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 the current $I_6$ in the figure 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 for the shown junction, $I_1+I_2+I_3=I_4+I_5+I_6$. Substituting gives $2+3+1=4+1+I_6$, so $6=5+I_6$. Therefore $I_6=1\,\text{A}$.