# Basic Electrical Engineering Questions and Answers – Phasor Diagrams Drawn with RMS Values Instead of Maximum Values

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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

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

2. The rms value is _________ times he maximum value
a) 1.414
b) 0.5
c) 2
d) 0.707

Explanation: We know that the rms value is 1/√2 times the maximum value, hence the rms value 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

Explanation: We know that the rms value is 1/√2 times the maximum value, hence the rms value is 0.707 times the maximum value.
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4. If the phasors are drawn to represent the maximum values instead of the rms values, what would happen to the phase angle between quantities?
a) Increases
b) Decreases
c) Remains constant
d) Becomes zero

Explanation: When phasors are drawn representing the maximum values instead of the rms 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

Explanation: Ammeters and voltmeters are calibrated to read the rms value, hence the phasors are drawn representing the rms values.
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6. If two current phasors, having magnitude 12A and 5A intersect at an angle of 90 degrees, calculate the resultant current.
a) 13 A
b) 10 A
c) 6 A
d) 5 A

Explanation: Using the parallelogram law of addition, I2 = I12 + I22 + 2I1I2 cosθ
I=13 A.

7. If two current phasors, having magnitude 5A and 10A intersect at an angle of 60 degrees, calculate the resultant current.
a) 12.23 A
b) 12.54 A
c) 13.23 A
d) 14.24 A

Explanation: Resultant current can be found using I2 = I12 + I22 + 2I1I2 cosθ
Substituting the values, we get I=13.23 A.

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(36.5°) V
b) 87.2 sin( 0.5°) V
c) 87.2 sin(26.5°) V
d) 87.2 cos(36.5°) V

Explanation: Horizontal component of v1 = 40V
Vertical component of v1=0V
Horizontal component of v2=60cos600
Vertical component of v2=60sin600
Resultant horizontal component=60cos600 + 40 = 70V
Resultant vertical component = 30√3 V
Resultant = 87.2V
tan(ϕ) = 30√3 / 70 => ϕ=36.50
Therefore sum = 87.2 sin( 36.5°) 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(71.5°) V
b) 53 sin( 79..5°) V
c) 53 sin(26.5°) V
d) 53 cos(36.5°) V

Explanation: Horizontal component of v1 = 40V
Vertical component of v1=0V
Horizontal component of v2=-60cos600
Vertical component of v2=-60sin600
Resultant horizontal component=40-30 = 10V
Resultant vertical component = -30√3 V
Resultant v = 53 V
tan(ϕ) = 30√3 / 10 => ϕ=79.50
Therefore sum = 53 sin (79.5°) 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

Explanation: The resultant current can be found by using the parallelogram law of addition I2 = I12 + I22 + 2I1I2 cosθ.

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