Three-Phase Circuits MCQs

Practice Three-Phase Circuits MCQs with four options, answers and explanations for focused Electronics & Telecommunication Engineering revision.

Question 1

Network TheoryThree-Phase Circuits

In a balanced three-phase star-connected load, the relation between line voltage VL and phase voltage Vph is?

AVL=Vph
BVL=√3 Vph
CVL=3Vph
DVL=Vph/√3

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Correct Answer: B. VL=√3 Vph

Explanation: For a balanced star connection, line voltage magnitude is √3 times phase voltage magnitude.

Reference: Standard network theory: three-phase circuits

Question 2

Network TheoryThree-Phase Circuits

In a balanced three-phase delta-connected load, the relation between line current IL and phase current Iph is?

AIL=Iph
BIL=√3 Iph
CIL=3Iph
DIL=Iph/√3

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Correct Answer: B. IL=√3 Iph

Explanation: For a balanced delta connection, line current magnitude is √3 times phase current magnitude.

Reference: Standard network theory: three-phase circuits

Question 3

Network TheoryThree-Phase Circuits

The total real power in a balanced three-phase load is?

AVL IL cos(phi)
B√3 VL IL cos(phi)
C3 VL IL only
DVL IL sin(phi)

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Correct Answer: B. √3 VL IL cos(phi)

Explanation: Balanced three-phase real power is P=√3 VL IL cos(phi).

Reference: Standard network theory: three-phase circuits

Question 4

Network TheoryThree-Phase Circuits

A balanced three-phase load operates at VL = 400 V, IL = 10 A and unity power factor. Its total real power is closest to?

A2.31 kW
B4.00 kW
C6.93 kW
D12.0 kW

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Correct Answer: C. 6.93 kW

Explanation: P=√3×400×10×1 ≈ 6928 W = 6.93 kW.

Reference: Standard network theory: three-phase circuits

Question 5

Network TheoryThree-Phase Circuits

In a balanced three-phase star-connected load, the relationship between line voltage VL and phase voltage Vph is?

AVL=Vph
BVL=sqrt(3)Vph
CVL=3Vph
DVL=Vph/sqrt(3)

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Correct Answer: B. VL=sqrt(3)Vph

Explanation: For a balanced star connection, line voltage magnitude is sqrt(3) times phase voltage magnitude.

Reference: Standard network theory: star connection

Question 6

Network TheoryThree-Phase Circuits

A balanced three-phase load draws line current 10 A from a 400 V line at power factor 0.8. Its total real power is closest to?

A3.2 kW
B4.0 kW
C5.54 kW
D6.93 kW

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Correct Answer: C. 5.54 kW

Explanation: P=sqrt(3) VL IL cos(phi)=1.732×400×10×0.8≈5542 W=5.54 kW.

Reference: Standard network theory: three-phase power numerical

Question 7

Network TheoryThree-Phase Circuits

A balanced three-phase delta-connected load has a phase current of 8 A. The line current magnitude is approximately?

A4.62 A
B8 A
C13.86 A
D24 A

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Correct Answer: C. 13.86 A

Explanation: For a balanced delta connection, IL=sqrt(3)Iph≈1.732×8=13.86 A.

Reference: Standard network theory: delta line current numerical

Question 8

Network TheoryThree-Phase Circuits

For a balanced three-phase delta-connected load, the phase voltage is related to line voltage by?

AVph=VL
BVph=VL/sqrt(3)
CVph=sqrt(3)VL
DVph=3VL

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Correct Answer: A. Vph=VL

Explanation: In a balanced delta connection, each phase is connected directly across two lines, so phase voltage equals line voltage.

Reference: Standard network theory: delta connection

Question 9

Network TheoryThree-Phase Circuits

A balanced three-phase star load has line current 12 A. The phase current is?

A4 A
B6.93 A
C12 A
D20.78 A

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Correct Answer: C. 12 A

Explanation: In a star connection, line current equals phase current, so Iph=IL=12 A.

Reference: Standard network theory: star line and phase current

Question 10

Network TheoryThree-Phase Circuits

A balanced three-phase system supplies 10 kW at 400 V line voltage and unity power factor. The line current is closest to?

A8.33 A
B14.43 A
C25 A
D43.3 A

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Correct Answer: B. 14.43 A

Explanation: P=sqrt(3)VLILcos(phi). At unity power factor, IL=10000/(1.732×400)≈14.43 A.

Reference: Standard network theory: three-phase current numerical

Question 11

Network TheoryThree-Phase Circuits

A balanced three-phase load takes 6 kW at a power factor of 0.6 lagging. Its reactive power is?

A3.6 kVAR
B4.8 kVAR
C6 kVAR
D8 kVAR

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Correct Answer: D. 8 kVAR

Explanation: For cos(phi)=0.6, sin(phi)=0.8 and tan(phi)=4/3. Hence Q=P tan(phi)=6×4/3=8 kVAR.

Reference: Standard network theory: three-phase reactive power

Question 12

Network TheoryThree-Phase Circuits

In the two-wattmeter method for a balanced three-phase load, total real power is equal to?

AW1-W2
BW1+W2
Csqrt(3)(W1+W2)
D3(W1-W2)

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Correct Answer: B. W1+W2

Explanation: For a balanced three-phase load measured by the two-wattmeter method, total real power is the algebraic sum W1+W2.

Reference: Standard network theory: two-wattmeter method

Question 13

Network TheoryThree-Phase Circuits

In the two-wattmeter method, if the balanced load has unity power factor, the two wattmeters ideally read?

AEqual positive values
BEqual and opposite values
COne zero and one positive value
DBoth zero

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Correct Answer: A. Equal positive values

Explanation: At unity power factor phi=0, the readings are proportional to cos(30°-phi) and cos(30°+phi), so they are equal and positive.

Reference: Standard network theory: two-wattmeter unity power factor

Question 14

Network TheoryThree-Phase Circuits

In the two-wattmeter method for a balanced load, one wattmeter reads zero when the power factor is?

A1
B0.866
C0.5
D0

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Correct Answer: C. 0.5

Explanation: One reading becomes zero at phi=60°, for which cos(phi)=0.5.

Reference: Standard network theory: two-wattmeter power factor

Question 15

Network TheoryThree-Phase Circuits

A balanced three-phase load is measured by two wattmeters reading 4 kW and 2 kW. The total real power is?

A2 kW
B4 kW
C6 kW
D8 kW

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Correct Answer: C. 6 kW

Explanation: Total three-phase real power in the two-wattmeter method is W1+W2=4+2=6 kW.

Reference: Standard network theory: two-wattmeter numerical