Three Phase AC circuits

Three Phase AC circuits

University

30 Qs

quiz-placeholder

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Three Phase AC circuits

Three Phase AC circuits

Assessment

Quiz

Education

University

Practice Problem

Hard

Created by

Ozwin Dsouza

Used 73+ times

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30 questions

Show all answers

1.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Media Image

Calculate the line voltage and Line current in this balanced Y-Y system:

Eline = 1.38 KV

Iline = 0.5312

Eline = 13.8 KV

Iline = 5.312 A

Eline = 7.967 KV

Iline = 5.312 A

Eline = 7.967 KV

Iline = 0.5312

2.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

Calculate phase voltage and phase current in this balanced Y-Y system:

Ephase(source) = 7.967 kV

Iphase(source) = 5.312 KA

Ephase(source) = 7.967 V

Iphase(source) = 5.312 A

Ephase(source) = 7.967 kV

Iphase(source) = 5.312 A

Ephase(source) = 7.967 V

Iphase(source) = 5.312 A

3.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

Calculate the total power consumed in this balanced Y-Y system:

1.2696 kW

126.96 W

12.696 kW

126.96 kW

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In a three-phase system, the voltages are separated by

450

900

1200

1800

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In a three-phase system, when the loads are perfectly balanced, the neutral current is

zero

one-third of maximum

two-thirds of maximum

at maximum

6.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

In a -connected source driving a -connected load, the

load voltage and line voltage are one-third the source voltage for a given phase

load voltage and line voltage are two-thirds the source voltage for a given phase

load voltage and line voltage cancel for a given phase

load voltage, line voltage, and source phase voltage are all equal for a given phase

7.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

Three ideal inductors are connected in star. The combination is supplied from a balanced three phase ac supply. The power consumed by the circuit will be

infinite

3×VL×IL×cosθ\sqrt{3}\times VL\times IL\times\cos\theta

zero

3×Vph×Iph×cosθ3\times Vph\times Iph\times\cos\theta

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