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This mock test of AC Power Analysis - 1 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam.
This contains 10 Multiple Choice Questions for Electrical Engineering (EE) AC Power Analysis - 1 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The value of for the waveform shown below is

Solution:

For the square wave shown in given figure,

QUESTION: 2

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

Solution:

QUESTION: 3

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

Solution:

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

∴

QUESTION: 4

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

Solution:

For a sinusoidal waveform,

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

Solution:

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

Solution:

∴ Reactive power taken by motor is

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

Solution:

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

Solution:

Now, power dissipated,

QUESTION: 9

The values of current I_{1} and I_{2} in the circuit shown below are given by:

1A = (0.8 + j0.2) A and I_{2} = -(1 - j0.5) A

The total power lost in the 5 Ω resistor is

Solution:

Power lost in the 5Ω resistor is

(As current through 5 Ω resistor = I_{1} -I_{2})

QUESTION: 10

The power factor of the circuit shown below is

Solution:

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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