Three wires meet at a junction. Wire 1 carries 0.40 A into the junction; Wire 2 carries 0.67 A out of the junction. What is the magnitude of the current in wire 3?

• A. 0.27 A
• B. 0.43 A
• C. 0.67 A
• D. 0.40 A

Answer: (A)

At a junction, current is conserved: the currents flowing into the junction must equal the currents flowing out. Treat into the junction as positive and out as negative, so the sum must be zero.

Wire 1 brings in 0.40 A. Wire 2 takes 0.67 A away, which is like −0.67 A in our sign convention. Let the current in wire 3 be I3 (positive if it goes into the junction). Then 0.40 + I3 − 0.67 = 0. Solving gives I3 = 0.27 A. The magnitude is 0.27 A, and the direction is into the junction (since it came out positive).

So the current in wire 3 is 0.27 A. The other options don’t balance the currents at the junction.