New Electrical Engineering MCQs – Alternating Voltage and Current MCQs ( Electrical Engineering ) MCQs

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New Electrical Engineering MCQs – Alternating Voltage and Current MCQs ( Electrical Engineering ) MCQs

Latest Electrical Engineering MCQs

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Alternating Voltage and Current MCQs ( Electrical Engineering ) MCQs – Electrical Engineering MCQs

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Average and RMS Values of an Alternating Current

1. Find the average value of current when the current that are equidistant are 4A, 5A and 6A.
a) 5A
b) 6A
c) 15A
d) 10A
Answer: a
Explanation: The average value of current is the sum of all the currents divided by the number of currents. Therefore average current = (5+4+6)/3=5A.


2. What is the current found by finding the current in an equidistant region and dividing by n?
a) RMS current
b) Average current
c) Instantaneous current
d) Total current
Answer: b
Explanation: The average value of the current is the sum of all the currents divided by the number of currents.


3. RMS stands for ________
a) Root Mean Square
b) Root Mean Sum
c) Root Maximum sum
d) Root Minimum Sum
Answer: a
Explanation: RMS stands for Root Mean Square. This value of current is obtained by squaring all the current values, finding the average and then finding the square root.


4. What is the type of current obtained by finding the square of the currents and then finding their average and then fining the square root?
a) RMS current
b) Average current
c) Instantaneous current
d) Total current
Answer: a
Explanation: RMS stands for Root Mean Square. This value of current is obtained by squaring all the current values, finding the average and then finding the square root.


5. __________ current is found by dividing the area enclosed by the half cycle by the length of the base of the half cycle.
a) RMS current
b) Average current
c) Instantaneous current
d) Total current
Answer: b
Explanation: The average value of current is the sum of all the currents divided by the number of currents. Hence it can also be found by dividing the area enclosed by the half cycle by the length of the base of the half cycle.


6. What is the effective value of current?
a) RMS current
b) Average current
c) Instantaneous current
d) Total current
Answer: a
Explanation: RMS current is also known as the effective current. RMS stands for Root Mean Square. This value of current is obtained by squaring all the current values, finding the average and then finding the square root.


7. In a sinusoidal wave, average current is always _______ rms current.
a) Greater than
b) Less than
c) Equal to
d) Not related
Answer: b
Explanation: The average value of current is the sum of all the currents divided by the number of currents whereas RMS current is obtained by squaring all the current values, finding the average and then finding the square root. Hence RMS current is greater than average current.


8. For a rectangular wave, average current is ______ rms current.
a) Greater than
b) Less than
c) Equal to
d) Not related
Answer: c
Explanation: The rms value is always greater than the average except for a rectangular wave, in which the heating effect remains constant so that the average and the rms values are the same.


9. Peak value divided by the rms value gives us?
a) Peak factor
b) Crest factor
c) Both peak and crest factor
d) Neither peak nor crest factor
Answer: c
Explanation: Peak and crest factor both mean the same thing. Hence the peak value divided by the rms value gives us the peak or crest factor.


10. Calculate the crest factor if the peak value of current is 10A and the rms value is 2A.
a) 5
b) 10
c) 5A
d) 10A
Answer: a
Explanation: We know that:
Crest factor = Peak value/RMS value.
Substituting the values from the given question, we get crest factor=5.

Average and RMS Values of Sinusoidal & Non-Sinusoidal Currents and Voltages

1. If maximum value of current is 5√2 A, what will be the value of RMS current?
a) 10 A
b) 5 A
c) 15 A
d) 25 A
Answer: b
Explanation: We know, value of RMS current =value of max current/√2
Substituting the value of max current we get, rms current = 5A.


2. If Im is the maximum value of a sinusoidal voltage, what is the instantaneous value?
a) i=Im/2
b) i=Imsinθ
c) i=Imcosθ
d) i=Imsinθ or i=Imcosθ
Answer: d
Explanation: The instantaneous value of a sinusoidal varying current is i=Imsinθ or i=Imcosθ where Im is the maximum value of current.


3. Average value of current over a half cycle is?
a) 0.67Im
b) 0.33Im
c) 6.7Im
d) 3.3Im
Answer: a
Explanation: Average current = ∫0πidθ/π = ∫0πImsinθ dθ/π = 2Im/π =0.67 Im.


4. What is the correct expression for the rms value of current?
a) Irms=Im/2
b) Irms=Im/√2
c) Irms=Im/4
d) Irms=Im
Answer: b
Explanation: Irms2 = ∫0πdθ i2/2π = Im2/2
Irms=Im/√2.


5. Average value of current over a full cycle is?
a) 0.67Im
b) 0
c) 6.7Im
d) 3.3Im
Answer: b
Explanation: Average of sine or cosine over a period is zero so, average value of current over full cycle is zero.


6. What is the correct expression for the form factor?
a) Irms * Iav
b) Irms / Iav
c) Irms + Iav
d) Irms – Iav
Answer: b
Explanation: The correct expression for form factor is Irms/Iav where Irms is the rms value of the current and Iav is the average current.


7. For a direct current, the rms current is ________ the mean current.
a) Greater than
b) Less than
c) Equal to
d) Not related to
Answer: c
Explanation: For a direct current, the mean current value is the same as that of the rms current.


8. For a direct current, the rms voltage is ________ the mean voltage.
a) Greater than
b) Less than
c) Equal to
d) Not related to
Answer: c
Explanation: For a direct current, the mean voltage value is the same as that of the rms voltage.


9. What is the value of the form factor for sinusoidal current?
a) π/2
b) π/4
c) 2π
d) π/√2
Answer: a
Explanation: For sinusoidal current, Irms=Im/√2
Iav=√2 Im/π
So, form factor = Irms/Iav = π/2.

 

Alternating Voltage And Current MCQs




10. If the maximum value of the current is 5√2 A, what will be the value of the average current?
a) 10/π A
b) 5/π A
c) 15/π A
d) 25/π A
Answer: a
Explanation: We know, the value of the average current = value of max current *√2 /π
Substituting the value of max current we get, rms current = 10/π A.

Representation of an Alternating Quantity by a Phasor

1. For addition and subtraction of phasors, we use the _________ form.
a) Rectangular
b) Polar
c) Either rectangular or polar
d) Neither rectangular nor polar
Answer: a
Explanation: For addition and subtraction of phasors, we use the rectangular form because in the rectangular form we can only add the real part and the complex part separately to get the total value.


2. For multiplication and division of phasors, we use ____________ form.
a) Rectangular
b) Polar
c) Either rectangular or polar
d) Neither rectangular nor polar
Answer: b
Explanation: For multiplication and division of phasors, we use the polar form because in the polar form we just multiply or divide the values and add or subtract the angles.


3. If a voltage of 2+5j and another voltage of 3+ 6j flows through two different resistors, connected in series, in a circuit, find the total voltage in the circuit.
a) 2+5j V
b) 3+6j V
c) 5+11j V
d) 5+10j V
Answer: c
Explanation: The total voltage in the circuit is the sum of the two voltages where we add the real parts and imaginary parts separately.
Therefore, Vtotal= 5+11j V.


4. Find the total current in the circuit if two currents of 4+5j flow in the circuit.
a) 4+5j A
b) 4A
c) 5A
d) 8+10j A
Answer: d
Explanation: The total current in the circuit is the sum of the two currents where we add the real parts and imaginary parts separately.
Therefore, Itotal= 8+10j A.


5. What is the correct expression of ω?
a) ω=2π
b) ω=2πf
c) ω=πf
d) ω=2f2
Answer: b
Explanation: The correct expression for ω is ω=2πf where f is the frequency of the alternating voltage or current.


6. Find the value of ω if the frequency is 5Hz?
a) 3.14 rad/s
b) 31.4 rad/s
c) 34 rad/s
d) 341 rad/s
Answer: b
Explanation: The expression for ω is ω=2*π*f.
Substituting the value of f from the question, we get ω=31.4 rad/s.


7. When one sine wave passes through the zero following the other, it is _________
a) Leading
b) Lagging
c) Neither leading nor lagging
d) Either leading or lagging
Answer: b
Explanation: The sine wave is said to lag because it passes though zero following the other, hence it crosses zero after the first wave, therefore it is said to lag.


8. 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
Answer: a
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.


9. The time axis of an AC phasor represents?
a) Time
b) Phase angle
c) Voltage
d) Current
Answer: b
Explanation: The time axis while measuring an AC sinusoidal voltage or current represents the phase angle when converting it to a phasor.


10. The length of the phasor represents?
a) Magnitude of the quantity
b) Direction of the quantity
c) Neither magnitude nor direction
d) Either magnitude or direction
Answer: a
Explanation: The length of the phasor arrow represents the magnitude of the quantity, whereas the angle between the phasor and the reference represents the phase angle.

Phasor Diagrams Drawn with RMS 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
Answer: a
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
Answer: d
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
Answer: a
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.


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


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

New Electrical Engineering MCQs – Alternating Voltage and Current MCQs ( Electrical Engineering ) MCQs

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