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If $I_1=2\,\text{A}$, $I_2=3\,\text{A}$, $I_3=5\,\text{A}$, $I_4=4\,\text{A}$, and $I_5=1\,\text{A}$ in the junction of the figure, then $I_6$ is 
A$3\,\text{A}$
B$5\,\text{A}$
C$7\,\text{A}$
D$9\,\text{A}$
Answer & Solution
Correct answer: B. $5\,\text{A}$
Using KCL, $I_1+I_2+I_3=I_4+I_5+I_6$. Substituting gives $2+3+5=4+1+I_6$, so $10=5+I_6$, hence $I_6=5\,\text{A}$.