# Network Theory Questions and Answers – Three-Phase Balanced Circuits

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 VRY = V∠0⁰ as a reference phasor. Then the source voltage VYB is?
a) V∠0⁰
b) V∠-120⁰
c) V∠120⁰
d) V∠240⁰
View Answer

Answer: b
Explanation: As the line voltage VRY = V∠0⁰ is taken as a reference phasor. Then the source voltage VYB is V∠-120⁰.

2. In a balanced three-phase system-delta load, if we assume the line voltage is VRY = V∠0⁰ as a reference phasor. Then the source voltage VBR is?
a) V∠120⁰
b) V∠240⁰
c) V∠-240⁰
d) V∠-120⁰
View Answer

Answer: c
Explanation: As the line voltage VRY = V∠0⁰ is taken as a reference phasor. Then the source voltage VBR 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

Answer: c
Explanation: In a delta-connected load, the relation between line voltage and the phase voltage is line voltage = phase voltage.
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4. If the load impedance is Z∠Ø, the current (IR) is?
a) (V/Z)∠-Ø
b) (V/Z)∠Ø
c) (V/Z)∠90-Ø
d) (V/Z)∠-90+Ø
View Answer

Answer: a
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 IR = VBR∠0⁰/Z∠Ø = (V/Z)∠-Ø.

5. If the load impedance is Z∠Ø, the expression obtained for current (IY) is?
a) (V/Z)∠-120+Ø
b) (V/Z)∠120-Ø
c) (V/Z)∠120+Ø
d) (V/Z)∠-120-Ø
View Answer

Answer: d
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 IY = VYB∠120⁰/Z∠Ø = (V/Z)∠-120-Ø.
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6. If the load impedance is Z∠Ø, the expression obtained for current (IB) is?
a) (V/Z)∠-240+Ø
b) (V/Z)∠-240-Ø
c) (V/Z)∠240-Ø
d) (V/Z)∠240+Ø
View Answer

Answer: b
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 IB = VBR∠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 IR. Assume the phase sequence to be RYB.
a) 44.74∠-63.4⁰A
b) 44.74∠63.4⁰A
c) 45.74∠-63.4⁰A
d) 45.74∠63.4⁰A
View Answer

Answer: a
Explanation: Taking the line voltage VRY = V∠0⁰ as a reference VRY = 400∠0⁰V, VYB = 400∠-120⁰V and VBR = 400∠-240⁰V. Impedance per phase = (4+j8) Ω = 8.94∠63.4⁰Ω. Phase current IR = (400∠0o)/(8.94∠63.4o )= 44.74∠-63.4⁰A.
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8. A three-phase balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current IY.
a) 44.74∠183.4⁰A
b) 45.74∠183.4⁰A
c) 44.74∠183.4⁰A
d) 45.74∠-183.4⁰A
View Answer

Answer: c
Explanation: Taking the line voltage VRY = V∠0⁰ as a reference VRY = 400∠0⁰V, VYB = 400∠-120⁰V and VBR = 400∠-240⁰V. Impedance per phase = (4+j8)Ω = 8.94∠63.4⁰Ω. Phase current IY = (400∠120o)/(8.94∠63.4o)= 44.74∠-183.4⁰A.

9. A three-phase balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current IB.
a) 44.74∠303.4⁰A
b) 44.74∠-303.4⁰A
c) 45.74∠303.4⁰A
d) 45.74∠-303.4⁰A
View Answer

Answer: b
Explanation: Taking the line voltage VRY = V∠0⁰ as a reference VRY = 400∠0⁰V, VYB = 400∠-120⁰V and VBR = 400∠-240⁰V. Impedance per phase = (4+j8) Ω = 8.94∠63.4⁰Ω. Phase current IB = (400∠240o)/(8.94∠63.4o) = 44.74∠-303.4⁰A.
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10. Determine the power (kW) drawn by the load.
a) 21
b) 22
c) 23
d) 24
View Answer

Answer: d
Explanation: Power is defined as the product of voltage and current. So the power drawn by the load is P = 3VPhIPhcosØ = 24kW.

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