This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Three-Phase Balanced Circuits”.

1. In a balanced three-phase system-delta load, if we assume the line voltage is V_{RY} = V∠0⁰ as a reference phasor. Then the source voltage V_{YB} is?

a) V∠0⁰

b) V∠-120⁰

c) V∠120⁰

d) V∠240⁰

View Answer

Explanation: As the line voltage V

_{RY}= V∠0⁰ is taken as a reference phasor. Then the source voltage V

_{YB}is V∠-120⁰.

2. In the question 1, the source voltage V_{BR} is?

a) V∠120⁰

b) V∠240⁰

c) V∠-240⁰

d) V∠-120⁰

View Answer

Explanation: As the line voltage V

_{RY}= V∠0⁰ is taken as a reference phasor. Then the source voltage V

_{BR}is V∠-240⁰.

3. In a delta-connected load, the relation between line voltage and the phase voltage is?

a) line voltage > phase voltage

b) line voltage < phase voltage

c) line voltage = phase voltage

d) line voltage >= phase voltage

View Answer

Explanation: In a delta-connected load, the relation between line voltage and the phase voltage is line voltage = phase voltage.

4. If the load impedance is Z∠Ø, the current (I_{R} ) is?

a) (V/Z)∠-Ø

b) (V/Z)∠Ø

c) (V/Z)∠90-Ø

d) (V/Z)∠-90+Ø

View Answer

Explanation: As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the R impedance is I

_{R}= V

_{BR}∠0⁰/Z∠Ø = (V/Z)∠-Ø.

5. In the question 4, the expression obtained for current (I_{Y}) is?

a) (V/Z)∠-120+Ø

b) (V/Z)∠120-Ø

c) (V/Z)∠120+Ø

d) (V/Z)∠-120-Ø

View Answer

Explanation: As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the Y impedance is I

_{Y}= V

_{YB}∠120⁰/Z∠Ø = (V/Z)∠-120-Ø.

6. In the question 4, the expression obtained for current (I_{B}) is?

a) (V/Z)∠-240+Ø

b) (V/Z)∠-240-Ø

c) (V/Z)∠240-Ø

d) (V/Z)∠240+Ø

View Answer

Explanation: As the load impedance is Z∠Ø, the current flows in the three load impedances and the current flowing in the B impedance is I

_{B}= V

_{BR}∠240⁰/Z∠Ø = (V/Z)∠-240-Ø.

7. A three phase, balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current I_{R} . Assume the phase sequence to be R_{YB}.

a) 44.74∠-63.4⁰A

b) 44.74∠63.4⁰A

c) 45.74∠-63.4⁰A

d) 45.74∠63.4⁰A

View Answer

Explanation: Taking the line voltage V

_{RY}= V∠0⁰ as a reference V

_{RY}= 400∠0⁰V, V

_{YB}= 400∠-120⁰V and V

_{BR}= 400∠-240⁰V. Impedance per phase = (4+j8) Ω = 8.94∠63.4⁰Ω. Phase current I

_{R}= (400∠0

^{o})/(8.94∠63.4

^{o})= 44.74∠-63.4⁰A.

8. In the question 7, determine the phase current I_{Y}.

a) 44.74∠183.4⁰A

b) 45.74∠183.4⁰A

c) 44.74∠183.4⁰A

d) 45.74∠-183.4⁰A

View Answer

Explanation: Taking the line voltage V

_{RY}= V∠0⁰ as a reference V

_{RY}= 400∠0⁰V, V

_{YB}= 400∠-120⁰V and V

_{BR}= 400∠-240⁰V. Impedance per phase = (4+j8)Ω = 8.94∠63.4⁰Ω. Phase current I

_{Y}= (400∠120

^{o})/(8.94∠63.4

^{o})= 44.74∠-183.4⁰A.

9. In the question 7, determine the phase current I_{B}.

a) 44.74∠303.4⁰A

b) 44.74∠-303.4⁰A

c) 45.74∠303.4⁰A

d) 45.74∠-303.4⁰A

View Answer

Explanation: Taking the line voltage V

_{RY}= V∠0⁰ as a reference V

_{RY}= 400∠0⁰V, V

_{YB}= 400∠-120⁰V and V

_{BR}= 400∠-240⁰V. Impedance per phase = (4+j8) Ω = 8.94∠63.4⁰Ω. Phase current I

_{B}= (400∠240

^{o})/(8.94∠63.4

^{o})= 44.74∠-303.4⁰A.

10. Determine the power (kW) drawn by the load.

a) 21

b) 22

c) 23

d) 24

View Answer

Explanation: Power is defined as the product of voltage and current. So the power drawn by the load is P = 3V

_{Ph}I

_{Ph}cosØ = 24kW.

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

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