This set of Basic Electrical Engineering Questions and Answers for Aptitude test focuses on “Phasor Diagrams Drawn with r.m.s. Values Instead of Maximum Values”.

1. Ammeters and voltmeters are calibrated to read?

a) RMS value

b) Peak value

c) Average value

d) Instantaneous value

View Answer

Explanation: Ammeters and voltmeters are calibrated to read the rms value because the rms value is the most accurate average value.

2. The rms value is _________ times he maximum value

a) 1.414

b) 0.5

c) 2

d) 0.707

View Answer

Explanation: We know that the rms value is i/root(2) time the maximum value, hence the rms vale is 0.707 times the maximum value.

3. The rms value is 0.707 times the _________ value.

a) Peak

b) Instantaneous

c) Average

d) DC

View Answer

Explanation: We know that the rms value is i/root(2) time the peak value, hence the rms value is 0.707 times the peak value.

4. If the phasors are drawn to represent the rms values, instead of the maximum values, what would happen to the phase angle between quantities?

a) Increases

b) Decreases

c) Remains constant

d) Becomes zero

View Answer

Explanation: When phasors are drawn representing the rms values instead of the maximum value, the shape of the diagram remains unaltered and hence the phase angle remains the same.

5. Usually phasor diagrams are drawn representing?

a) RMS value

b) Peak value

c) Average value

d) Instantaneous value

View Answer

Explanation: Ammeters and voltmeters are calibrated to read the rms value, hence the phasors are drawn representing the rms values.

6. A phasor has frozen at 30 degrees, find the value of the phase angle?

a) 30 degrees

b) 60 degrees

c) 120 degrees

d) 180 degrees

View Answer

Explanation: The value of the phase angle is the value at which the phasor stops or freezes. Here, it freezes at 30 degree, hence the phase angle is 30 degrees.

7. If two current phasors, having magnitude 5A and 10A intersect at an angle of 60 degrees, calculate the resultant current.

a) 5A

b) 10A

c) 25A

d) 20A

View Answer

Explanation: The resultant current can be found by using the parallelogram law of addition. Hence Iresultant= I1I2cos(theta), where theta is the angle between I1 and I2.

8. The instantaneous values of two alternating voltages are given as _________

v1=60sinθ and v2=40sin(θ − π/3). Find the instantaneous sum.

a) 87.2 sin(23.5°) V

b) 87.2 sin( 0.5°) V

c) 87.2 sin(-23.5°) V

d) 87.2 cos(23.5°) V

View Answer

Explanation: Horizontal component os v1= 60V

Vertical component of v1=0V

Horizontal component of v2=40cos60=20V

Vertical component of v2=-20sin60=-34.64V

Resultant horizontal component=80V

Resultant vertical component= -34.64V

Resultant v= 87.2V

tan(phi)=-34.64/80

phi= -23.5

Therefore sum= 87.2 sin( 23.5°) V.

9. The instantaneous values of two alternating voltages are given as:

v1=60sinθ and v2=40sin(θ − π/3). Find the instantaneous difference.

a) 53 sin(30.9°) V

b) 53 sin(40.9°) V

c) 53 cos(30.9°) V

d) 53 cos(40.9°) V

View Answer

Explanation: Horizontal component os v1= 60V

Vertical component of v1=0V

Horizontal component of v2=40cos60=20V

Vertical component of v2=-20sin60=-34.64V

Resultant horizontal component=40V

Resultant vertical component= 34.64V

Resultant v= 53

tan(phi)=34.64/40

phi= 40.9

Therefore sum= 53 sin (40.9°) V.

10. The resultant of two alternating sinusoidal voltages or currents can be found using ___________

a) Triangular law

b) Parallelogram law

c) Either triangular or parallelogram law

d) Neither triangular nor parallelogram law

View Answer

Explanation: The resultant current can be found by using the parallelogram law of addition. Where the magnitude is found by squaring the magnitudes, adding the squares and finding the square root of the sum and the theta is the angle between the two quantities.

**Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.**

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