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At a junction, currents of 3 A and 5 A flow in, and 2 A flows out through one branch. By Kirchhoff's junction rule, the current leaving through the remaining branch is:
A4 A
B2 A
C10 A
D6 A
Answer & Solution
Correct answer: D. 6 A
Kirchhoff's junction rule: current in = current out. 3 + 5 = 2 + x ⟹ x = 6 A.
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