250+ TOP MCQs on Three-Phase Unbalanced Circuits and Answers

Network Theory Multiple Choice Questions on “Three-Phase Unbalanced Circuits”.

1. If the system is a three-wire system, the currents flowing towards the load in the three lines must add to ___ at any given instant.
A. 1
B. 2
C. 3
D. zero
Answer: D
Clarification: If the system is a three-wire system, the currents flowing towards the load in the three lines must add to zero at any given instant.

2. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Determine the phase current I_R.
A. (17.32-j10) A
B. (-17.32-j10) A
C. (17.32+j10) A
D. (-17.32+j10) A
Answer: A
Clarification: Taking V_RY = V∠0⁰ as a reference phasor, and assuming RYB phase sequence, we have V_RY = 400∠0⁰V Z₁ = 20∠30⁰Ω = (17.32+j10)Ω I_R = (400∠0^o)/(20∠30^o) = (17.32-j10) A.

3. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the phase current I_Y.
A. (10-j0) A
B. (10+j0) A
C. (-10+j0) A
D. (-10-j0) A
Answer: C
Clarification: The voltage V_YB is V_YB = 400∠-120⁰V. The impedance Z₂ is Z₂ = 40∠60⁰Ω => I_Y = (400∠-120^o)/(40∠60^o)=(-10+j0)A.

4. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the phase current I_B.
A. (34.64-j20) A
B. (34.64+j20) A
C. (-34.64+j20) A
D. (-34.64-j20) A
Answer: D
Clarification: The voltage V_BR is V_BR = 400∠-240⁰V. The impedance Z₃ is Z₃ = 10∠-90⁰Ω => I_B = (400∠240^o)/(10∠-90^o)=(-34.64-j20)A.

5. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I₁.
A. (-51.96-j10) A
B. (-51.96+j10) A
C. (51.96+j10) A
D. (51.96+j10) A
Answer: C
Clarification: The line current I₁ is the difference of I_R and I_B. So the line current I₁ is I₁ = I_R – I_B = (51.96+j10) A.

6. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I₂.
A. (-27.32+j10) A
B. (27.32+j10) A
C. (-27.32-j10) A
D. (27.32-j10) A
Answer: A
Clarification: The line current I₂ is the difference of I_Y and I_R. So the line current I₂ is I₂ = I_Y – I_R = (-27.32+j10) A.

7. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I₃.
A. (24.646+j20) A
B. (-24.646+j20) A
C. (-24.646-j20) A
D. (24.646-j20) A
Answer: C
Clarification: The line current I₃ is the difference of I_B and I_Y. So the line current I₃ is I₃ = I_B – I_Y = (-24.646-j20) A.

8. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the R phase.
A. 6628
B. 6728
C. 6828
D. 6928
Answer: D
Clarification: The term power is defined as the product of square of current and the impedance. So the power in the R phase = 20² x 17.32 = 6928W.

9. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the Y phase.
A. 1000
B. 2000
C. 3000
D. 4000
Answer: B
Clarification: The term power is defined as the product of square of current and the impedance. So the power in the Y phase = 10² x 20 = 2000W.

10. The three impedances Z₁ = 20∠30⁰Ω, Z₂ = 40∠60⁰Ω, Z₃ = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the B phase.
A. 0
B. 1
C. 3
D. 2
Answer: A
Clarification: The term power is defined as the product of square of current and the impedance. So the power in the B phase = 40² x 0 = 0W.