Question

Suppose \(I_A, I_B\) and \(I_C\) are a set of unbalanced current phasors in a three-phase system. The phase-B zero-sequence current is \(I_{B0} = 0.1\angle 0^\circ\) p.u. If the phase-A current \(I_A = 1.1\angle 0^\circ\) p.u and phase-C current \(I_C = (1\angle 120^\circ + 0.1)\) p.u, then \(I_B\) in p.u is

• A. \(1\angle 240^\circ - 0.1\angle 0^\circ\)
• B. \(1.1\angle 240^\circ - 0.1\angle 0^\circ\)
• C. \(1.1\angle -120^\circ + 0.1\angle 0^\circ\)
• D. \(1\angle -120^\circ + 0.1\angle 0^\circ\)

Hint:

Zero-sequence currents are equal in all three phases—use the sum of currents divided by 3 to find relationships.

Answer 1

The Correct Option is D

Zero-sequence current for any phase is defined as: \[ I_{0} = \frac{I_A + I_B + I_C}{3}. \] Given for phase-B: \[ I_{B0} = \frac{I_A + I_B + I_C}{3} = 0.1\angle 0^\circ. \]

Step 1: Substitute known values.
\[ I_A = 1.1\angle 0^\circ, I_C = 1\angle 120^\circ + 0.1. \]

Step 2: Compute the zero-sequence sum.
\[ I_A + I_B + I_C = 0.3\angle 0^\circ. \]

Step 3: Solve for \(I_B\).
\[ I_B = 0.3 - I_A - I_C. \] After simplification, the result becomes: \[ I_B = 1\angle -120^\circ + 0.1\angle 0^\circ. \] Final Result:
\[ I_B = 1\angle -120^\circ + 0.1\angle 0^\circ. \]