Question

Currents I1, I2 and I3 meet at a junction in a circuit. All currents are marked as entering the node. If I1 = -6sin(ωt) mA and I2 = 8 cos(ωt) mA, the I3 is ________

(a) 10 cos(ωt + 36.87) mA

(b) 14 cos(ωt + 36.87) mA

(c) -14 sin(ωt + 36.87) mA

(d) -10 cos(ωt + 36.87) mA

I got this question in an interview for job.

Query is from Advanced Problems on Network Theory topic in chapter Network Theory and Complex Circuit Diagram of Network Theory

Answer 1

Correct choice is (d) -10 cos(ωt + 36.87) mA

To explain I would say: Applying KCL, we get, I1 + I2 + I3 = 0

∴ -6 sin(ωt) + 8 cos(ωt) + I3 = 0

∴ I3 = 6 sin (ωt) – 8 cos (ωt)

= 10[sin (ωt).sin (36.86) – cos (ωt) cos (36.86)]

=-10[cos (ωt) cos (36.86) – sin (ωt) sin (36.86)]

= -10 cos (ωt + 36.86)

[As, cos (A+B) = cosA.cosB – sinA.sinB].