This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses 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

View Answer

Explanation: 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 Z_{1} = 20∠30⁰Ω, Z_{2} = 40∠60⁰Ω, Z_{3} = 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

View Answer

Explanation: Taking V

_{RY}= V∠0⁰ as a reference phasor, and assuming RYB phase sequence, we have V

_{RY}= 400∠0⁰V Z

_{1}= 20∠30⁰Ω = (17.32+j10)Ω I

_{R}= (400∠0

^{o})/(20∠30

^{o})=(17.32-j10) A.

3. Find the phase current I_{Y} from the question 2.

a) (10-j0) A

b) (10+j0) A

c) (-10+j0) A

d) (-10-j0) A

View Answer

Explanation: The voltage V

_{YB}is V

_{YB}= 400∠-120⁰V. The impedance Z

_{2}is Z

_{2}= 40∠60⁰Ω => I

_{Y}=(400∠-120

^{o})/(40∠60

^{o})=(-10+j0)A.

4. Find the phase current I_{B} from the question 2.

a) (34.64-j20) A

b) (34.64+j20) A

c) (-34.64+j20) A

d) (-34.64-j20) A

View Answer

Explanation: The voltage V

_{BR}is V

_{BR}= 400∠-240⁰V. The impedance Z

_{3}is Z

_{3}= 10∠-90⁰Ω => I

_{B}=(400∠240

^{o})/(10∠-90

^{o})=(-34.64-j20)A.

5. From the question 2, find the line current I_{1}.

a) (-51.96-j10) A

b) (-51.96+j10) A

c) (51.96+j10) A

d) (51.96+j10) A

View Answer

Explanation: The line current I

_{1}is the difference of I

_{R}and I

_{B}. So the line current I

_{1}is I

_{1}= I

_{R}– I

_{B}= (51.96+j10) A.

6. From the question 2, find the line current I_{2}.

a) (-27.32+j10) A

b) (27.32+j10) A

c) (-27.32-j10) A

d) (27.32-j10) A

View Answer

Explanation: The line current I

_{2}is the difference of I

_{Y}and I

_{R}. So the line current I

_{2}is I

_{2}= I

_{Y}– I

_{R}= (-27.32+j10) A.

7. From the question 2, find the line current I_{3}.

a) (24.646+j20) A

b) (-24.646+j20) A

c) (-24.646-j20) A

d) (24.646-j20) A

View Answer

Explanation: The line current I

_{3}is the difference of I

_{B}and I

_{Y}. So the line current I

_{3}is I

_{3}= I

_{B}– I

_{Y}= (-24.646-j20) A.

8. In the question 2 find the power in the R phase.

a) 6628

b) 6728

c) 6828

d) 6928

View Answer

Explanation: The term power is defined as the product of square of current and the impedance. So the power in the R phase = 20

^{2}x 17.32 = 6928W.

9. In the question 2 find the power in the Y phase.

a) 1000

b) 2000

c) 3000

d) 4000

View Answer

Explanation: The term power is defined as the product of square of current and the impedance. So the power in the Y phase = 10

^{2}x 20 = 2000W.

10. In the question 2 find the power in the B phase.

a) 0

b) 1

c) 3

d) 2

View Answer

Explanation: The term power is defined as the product of square of current and the impedance. So the power in the B phase = 40

^{2}x 0 = 0W.

**Sanfoundry Global Education & Learning Series – Network Theory.**

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