250+ TOP MCQs on Reactive Power and Answers

Network Theory Multiple Choice Questions on “Reactive Power”.

1. The reactive power equation (P_r) is?
A. I_eff² (ωL)sin2(ωt+θ)
B. I_eff² (ωL)cos2(ωt+θ)
C. I_eff² (ωL)sin(ωt+θ)
D. I_eff² (ωL)cos(ωt+θ)
Answer: A
Clarification: If we consider a circuit consisting of a pure inductor, the power in the inductor is reactive power and the reactive power equation (P_r) is P_r =I_eff² (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
Answer: B
Clarification: 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. V_effI_msinθ
B. V_mI_msinθ
C. V_effI_effsinθ
D. V_mI_effsinθ
Answer: C
Clarification: The expression of reactive power (P_r) is V_effI_effsinθ Volt Amperes Reactive (VAR). Reactive power = V_effI_effsinθ 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
Answer: A
Clarification: 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 (P_true) is?
A. P_asinθ
B. P_acosθ
C. P_atanθ
D. P_asecθ
Answer: B
Clarification: True power is the product of the apparent power and cosθ. The expression of true power (P_true) is P_acosθ. True power = P_acosθ.

6. The average power (P_avg) is expressed as?
A. P_asecθ
B. P_atanθ
C. P_acosθ
D. P_asinθ
Answer: C
Clarification: The average power is the product of the apparent power and cosθ. The average power (P_avg) is expressed as P_acosθ. Average power = P_acosθ.

7. The equation of reactive power is?
A. P_acosθ
B. P_asecθ
C. P_asinθ
D. P_atanθ
Answer: C
Clarification: The reactive power is the product of the apparent power and sinθ. The equation of reactive power is P_asinθ. Reactive power = P_asinθ.

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
Answer: C
Clarification: The expression of apparent power (VA. is P_app= V_effI_eff = (V_m/√2)×(I_m/√2). On substituting the values V_m = 50, I_m = 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
Answer: C
Clarification: 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
Answer: C
Clarification: The average power, P_avg = V_effI_effcosθ. We know the values of V_eff, I_eff are V_eff = 625 and I_eff – 0.6. So the average power = 625 x 0.6 = 375W.