This set of Network Theory Multiple Choice Questions & Answers (MCQs) focuses on “Reactive Power”.
1. The reactive power equation (Pr) is?
a) Ieff2 (ωL)sin2(ωt+θ)
b) Ieff2 (ωL)cos2(ωt+θ)
c) Ieff2 (ωL)sin(ωt+θ)
d) Ieff2 (ωL)cos(ωt+θ)
View Answer
Explanation: If we consider a circuit consisting of a pure inductor, the power in the inductor is reactive power and the reactive power equation (Pr) is Pr =Ieff2 (ωL)sin2(ωt+θ).
2. Reactive power is expressed in?
a) Watts (W)
b) Volt Amperes Reactive (VAR)
c) Volt Ampere (VA)
d) No units
View Answer
Explanation: Reactive power is expressed in Volt Amperes Reactive (VAR) and power is expressed in watts and apparent power is expressed in Volt Ampere (VA).
3. The expression of reactive power (Pr) is?
a) VeffImsinθ
b) VmImsinθ
c) VeffIeffsinθ
d) VmIeffsinθ
View Answer
Explanation: The expression of reactive power (Pr) is VeffIeffsinθ Volt Amperes Reactive (VAR). Reactive power = VeffIeffsinθ Volt Amperes Reactive (VAR).
4. The power factor is the ratio of ________ power to the ______ power.
a) average, apparent
b) apparent, reactive
c) reactive, average
d) apparent, average
View Answer
Explanation: The power factor is the ratio of average power to the apparent power. Power factor = (average power)/(apparent power).
5. The expression of true power (Ptrue) is?
a) Pasinθ
b) Pacosθ
c) Patanθ
d) Pasecθ
View Answer
Explanation: True power is the product of the apparent power and cosθ. The expression of true power (Ptrue) is Pacosθ. True power = Pacosθ.
6. The average power (Pavg) is expressed as?
a) Pasecθ
b) Patanθ
c) Pacosθ
d) Pasinθ
View Answer
Explanation: The average power is the product of the apparent power and cosθ. The average power (Pavg) is expressed as Pacosθ. Average power = Pacosθ.
7. The equation of reactive power is?
a) Pacosθ
b) Pasecθ
c) Pasinθ
d) Patanθ
View Answer
Explanation: The reactive power is the product of the apparent power and sinθ. The equation of reactive power is Pasinθ. Reactive power = Pasinθ.
8. A sinusoidal voltage v = 50sinωt is applied to a series RL circuit. The current in the circuit is given by I = 25sin (ωt-53⁰). Determine the apparent power (VA).
a) 620
b) 625
c) 630
d) 635
View Answer
Explanation: The expression of apparent power (VA) is Papp= VeffIeff = (Vm/√2)×(Im/√2). On substituting the values Vm = 50, Im = 25, we get apparent power = (50×25)/2 = 625VA.
9. A sinusoidal voltage v = 50sinωt is applied to a series RL circuit. The current in the circuit is given by I = 25sin (ωt-53⁰). Find the power factor.
a) 0.4
b) 0.5
c) 0.6
d) 0.7
View Answer
Explanation: In sinusoidal sources the power factor is the cosine of the phase angle between the voltage and the current. The expression of power factor = cosθ = cos53⁰ = 0.6.
10. A sinusoidal voltage v = 50sinωt is applied to a series RL circuit. The current in the circuit is given by I = 25sin (ωt-53⁰). Determine the average power.
a) 365
b) 370
c) 375
d) 380
View Answer
Explanation: The average power, Pavg = VeffIeffcosθ. We know the values of Veff, Ieff are Veff = 625 and Ieff – 0.6. So the average power = 625 x 0.6 = 375W.
Sanfoundry Global Education & Learning Series – Network Theory.
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