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Apparent Power is MCQ
By BYJU'S Exam Prep
Updated on: September 25th, 2023
The apparent power combines active and reactive power consumed in the circuit. If P is the active or true power consumed by the circuit, Q is the reactive power consumed in the circuit, then apparent power or complex power S is given by; the apparent power is the combination of active and reactive power consumed in the circuit. If P is the active or true power consumed by the circuit, Q is the reactive power consumed in the circuit, then apparent power or complex power S is given by,
S=P+jQ
Here, S=VI
P=VIcos∅=Scos∅
Q=VIsin∅=Ssin∅
S=√(P2+Q2)
Here we will go through a few basic MCQs based on apparent power; these may help us understand the concept of Apparent power.
Table of content
Apparent Power MCQ1
The units of the apparent power are
- Watts
- VAR
- VA
- None
Answer: C. VA
Explanation:
The apparent power can be expressed in VA(Volt-ampere), the active power(P) can be expressed in watts, whereas reactive power can be expressed in VAR (Volt-ampere reactive).
Hence option C is the correct answer.
Apparent Power MCQ2
A source Vs=200cosωt delivers power to the load at 0.8 power factor lagging. The reactive power is 300 VAR. The apparent power supplied by the source is_______.
- 500 VA
- 400 VA
- 250 VA
- 600 VA
Answer: A. 500 VA
Solution:
Given that,
Reactive power (Q) = 300 VAR
Power factor cos ∅ =0.8⇒∅=36.860
Q=Ssin∅ =300
S(0.6)=300
S=500 VA
Hence option A is the correct answer.
Apparent Power MCQ3
Three-loads are connected in parallel across a single-phase, 1200 V, 50 Hz supply.
Load 1: Capacitive load, 10 kW and 40 kVAR
Load 2: Inductive load, 35 kW and 120 kVAR
Load 3: Resistive load of 15 kW
What is the total complex power of the circuit?
- (60+j80) kVA
- (10+j80) kVA
- (10-j160) kVA
- (60+j160) kVA
Answer: A.(60+j80) kVA
Solution:
Given,
For load1 which is a capacitive load, P1=10 kW , Q1=-40 kVAR
S1=P1+jQ1=10-j40 kVA
For load2, P2=35 kW , Q2=120 kVAR
S2=P2+jQ2=35+j120 kVA
For load3, P3=15 kW, Q3=0 kVAR
S3=P3+jQ3=15 kVA
Total complex power in the circuit is,
S=S1+S2+S3=10-j40+35+j120+15+j0=60+j80 kVA
Hence option A is the correct answer.
Apparent Power MCQ4
A 230 V source supplies power to two loads connected in parallel. One load draws 10 kW at 0.8 power factor leading, and the second load draws 10 kVA at 0.8 power factor lagging. The complex power delivered by the source is
- (18+j1.5) kVA
- (18-j1.5) kVA
- (20+j1.5) kVA
- (20-j1.5) kVA
Answer: B (18-j1.5) kVA
Solution:
Given that for load 1,
P1=10 kW,cos∅1=0.8 leading
P1=S1cos∅1=10
S1(0.8)=10
S1=12.5 kVA
S1=√(P12+Q12)⇒Q1=√(S12-P12)=7.5 kVAR
S1=(10-j7.5) kVA (*given leading load)
For load2,
S2=10 kVA,cos ∅2=0.8 lagging
P2=S2cos ∅2=8 kW
Q2=S2sin ∅2=6 kVAR
S2=8+j6 kVA
Total complex power in the circuit is,
S=S1+S2=10-j7.5+8+j6=18-j1.5 kVA
Hence option B is the correct answer.
Apparent Power MCQ5
A voltage source of 100 V is applied to an impedance Z=(3+j4). What are the values of active power, reactive power, and apparent power respectively?
- 1200 W, 1200 VAR and 2200 VA
- 1600 W, 1600 VAR, and 2000 VA
- 1200 W, 1600 VAR, and 2000 VA
- 1600 W, 1200 VAR, and 2200 VA
Answer: C.1200 W, 1600 VAR, and 2000 VA
Solution:
Given that,
Z=(3+j4)
⇒∅=tan-1(4/3)=53.060
V=100 V
I=V/Z=100/√(32+42)=20 A
S=VI=200*10=2000 VA
P=VIcos ∅ =2000(0.6)=1200 W
Q=VIsin ∅ =2000(0.8)=1600 VAR
Hence option C is the correct answer.
