Basic Electrical Engineering Multiple Choice Questions on “Power”.
1. Which of the following is not an expression power?
A. P=VI
B. P=I2R
C. P=V2/R
D. P=I/R
Answer d
Clarification: Power is the product of voltage and current. Writing I in terms of V, we get P=V2/R and writing V in terms of I, we get P=I2r.
2. Which of the following statements are true?
A. Power is proportional to voltage only
B. Power is proportional to current only
C. Power is neither proportional to voltage nor to the current
D. Power is proportional to both the voltage and current
Answer: D
Clarification: Power is proportional to both voltage and current.
3. A 250V bulb passes a current of 0.3A. Calculate the power in the lamp.
A. 75W
B. 50W
C. 25W
D. 90W
Answer: A
Clarification: Here, V = 250v and I = 0.3A. P=VI. Which implies that, P=250*0.3=75W.
4. Kilowatt-hour(kWh) is a unit of?
A. Current
B. Power
C. Energy
D. Resistance
Answer: C
Clarification: Power is the energy per unit time. That is, P=E/t. If the unit of power in kW and the unit of time is an hour, then the unit of energy=unit of power*unit of time=kWh.
5. Calculate the power in the 20 ohm resistance.
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A. 2000kW
B. 2kW
C. 200kW
D. 2W
Answer: B
Clarification: Here V = 200v and Resistance( R) = 20ohm. P=V2/R= 2002/20=2000W=2kW.
6. A current of 5A flows in a resistor of 2 ohms. Calculate the energy dissipated in 300 seconds in the resistor.
A. 15kJ
B. 15000kJ
C. 1500J
D. 15J
Answer: A
Clarification: P=I2R =52*2=50W.
E= Pt=50*300=15000J=15kJ.
7. Calculate the power across each 20 ohm resistance.
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A. 1000W, 1000W
B. 500W, 500W
C. 1000kW, 1000kW
D. 500kW, 500kW
Answer: B
Clarification: This is a series connected circuit hence the current across each resistance is the same. To find current: I=V/R=200/20=5A.
To find power: P=I2R=52*20=500W. Since both the resistors have a resistance of 20 ohms, the power across both is the same.
8. Calculate the power across each 10 ohm resistance.
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A. 1000kW, 1000kW
B. 1kW, 1kW
C. 100W, 100W
D. 100kW, 100kW
Answer: B
Clarification: This is parallel connected circuit, hence the voltage across each of the resistors is the same. P =(V2)/R=(1002)/10 = 1000W=1kW. Since both the resistors receive the same amount of voltage, the power in both is the same.
9. Calculate the work done in a resistor of 20 ohm carrying 5A of current in 3 hours.
A. 1.5J
B. 15J
C. 1.5kWh
D. 15kWh
Answer: C
Clarification: To find power: P=I2R=52*20=500W=0.5kW.
To find Work done: W=Pt=0.5*3=1.5kWh.
10. The SI unit of power is?
A. kW(kilo-watt)
B. J/s(joules per seconD.
C. Ws(watt-seconD.
D. J/h(joules per hour
Answer: B
Clarification: Power = energy/time
SI unit of power = SI unit of energy/SI unit of time = joule/second.