Test: AC Power Analysis


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10 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Test: AC Power Analysis

Test: AC Power Analysis for Electrical Engineering (EE) 2022 is part of GATE Electrical Engineering (EE) 2023 Mock Test Series preparation. The Test: AC Power Analysis questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: AC Power Analysis MCQs are made for Electrical Engineering (EE) 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: AC Power Analysis below.
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Test: AC Power Analysis - Question 1

 The value of  for the waveform shown below is

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For the square wave shown in given figure,

Test: AC Power Analysis - Question 2

 For the idealized full wave rectifier system the average and rms value of the voltage is

Test: AC Power Analysis - Question 3

The instantaneous power delivered to the 5 Ω resistor at t = 0 in the given circuit is

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At t = 0, the given circuit is as shown below.


∴ 

Test: AC Power Analysis - Question 4

For a sinusoidal waveform, the ratio of average to rms value is

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For a sinusoidal waveform, 

Test: AC Power Analysis - Question 5

An impedance of (3 + j 4) Ω is connected in parallel with a resistance of 10 Ω. The ratio of power loss in these parallel circuits is

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Test: AC Power Analysis - Question 6

A 0.2 HP induction motor runs at an efficiency of 85%. If the operating power factor is 0.8 lag, the reactive power taken by the motor is

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∴ Reactive power taken by motor is

Test: AC Power Analysis - Question 7

In an a.c. circuit, v = 100 sin (ωt + 30°) V, i = 5 sin(ωt- 30°) A. The apparant and reactive power in the circuit are respectively

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Test: AC Power Analysis - Question 8

The inductive reactance in series with Z in the circuit shown below has a value of 25 Ω If the voltage drop a cross Z is 179 volts, the power dissipated in the circuit is 320 W.

The reactive part of Z is 

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Now, power dissipated,

Test: AC Power Analysis - Question 9

The values of current I1 and I2 in the circuit shown below are given by:

1A = (0.8 + j0.2) A and I2 = -(1 - j0.5) A

The total power lost in the 5 Ω resistor is

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Power lost in the 5Ω resistor is

(As current through 5 Ω resistor = I1 -I2)

Test: AC Power Analysis - Question 10

​The power factor of the circuit shown below is

Detailed Solution for Test: AC Power Analysis - Question 10

Given, ω = 1 rad/sec. So, the circuit can be redrawn as shown below.

Total impedance of the circuit is

∴ p.f. of the given circuit = cos90° = 0

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