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Suppose at a node $7\,\text{A}$ and $4\,\text{A}$ enter the junction, while $5\,\text{A}$ leaves through one branch. How much current must leave through the remaining branch?
A$2\,\text{A}$
B$6\,\text{A}$
C$11\,\text{A}$
D$16\,\text{A}$
Answer & Solution
Correct answer: B. $6\,\text{A}$
Using KCL, total entering current equals total leaving current. The entering current is $7+4=11\,\text{A}$. If $5\,\text{A}$ already leaves through one branch, the remaining branch must carry $11-5=6\,\text{A}$.