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Test: Unbalanced Three Phase Circuits - Electrical Engineering (EE) MCQ


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10 Questions MCQ Test GATE Electrical Engineering (EE) Mock Test Series 2025 - Test: Unbalanced Three Phase Circuits

Test: Unbalanced Three Phase Circuits for Electrical Engineering (EE) 2024 is part of GATE Electrical Engineering (EE) Mock Test Series 2025 preparation. The Test: Unbalanced Three Phase Circuits questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Unbalanced Three Phase Circuits MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Unbalanced Three Phase Circuits below.
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Test: Unbalanced Three Phase Circuits - Question 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.

Detailed Solution for Test: Unbalanced Three Phase Circuits - Question 1

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.

Test: Unbalanced Three Phase Circuits - Question 2

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 IY.

Detailed Solution for Test: Unbalanced Three Phase Circuits - Question 2

The voltage VYB is VYB = 400 ∠ -120⁰V. The impedance Z2 is Z2 = 40 ∠ 60⁰Ω

⇒ IY = (400 ∠ -120o)/(40 ∠ 60o)=(-10 + j0)A.

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Test: Unbalanced Three Phase Circuits - Question 3

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.

Detailed Solution for Test: Unbalanced Three Phase Circuits - Question 3

The line current I1 is the difference of IR and IB. So the line current I1 is I1 = IR – IB = (51.96 + j10) A.

Test: Unbalanced Three Phase Circuits - Question 4

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.

Detailed Solution for Test: Unbalanced Three Phase Circuits - Question 4

The line current I3 is the difference of IB and IY. So the line current I3 is I3 = IB – IY = (-24.646 - j20) A.

Test: Unbalanced Three Phase Circuits - Question 5

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.

Detailed Solution for Test: Unbalanced Three Phase Circuits - Question 5

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.

Test: Unbalanced Three Phase Circuits - Question 6

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.

Detailed Solution for Test: Unbalanced Three Phase Circuits - Question 6

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.

Test: Unbalanced Three Phase Circuits - Question 7

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.

Detailed Solution for Test: Unbalanced Three Phase Circuits - Question 7

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.

Test: Unbalanced Three Phase Circuits - Question 8

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.

Detailed Solution for Test: Unbalanced Three Phase Circuits - Question 8

The line current I2 is the difference of IY and IR. So the line current I2 is I2 = IY – IR = (-27.32 + j10) A.

Test: Unbalanced Three Phase Circuits - Question 9

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 IB.

Detailed Solution for Test: Unbalanced Three Phase Circuits - Question 9

The voltage VBR is VBR = 400 ∠ -240⁰V. The impedance Z3 is Z3 = 10 ∠ -90⁰Ω
⇒ IB = (400 ∠ 240o)/(10 ∠ -90o) = (-34.64-j20)A.

Test: Unbalanced Three Phase Circuits - Question 10

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 IR.

Detailed Solution for Test: Unbalanced Three Phase Circuits - Question 10

Taking VRY = V∠0⁰ as a reference phasor, and assuming RYB phase sequence, we have VRY = 400∠0⁰V Z1 = 20∠30⁰Ω = (17.32 + j10)Ω IR = (400∠0o)/(20∠30o) = (17.32 - j10) A.

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