This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Power”.
1. Which of the following is not an expression power?
Explanation: 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
Explanation: Power is proportional to both voltage and current.
3. A 250V bulb passes a current of 0.3A. Calculate the power in the lamp.
Explanation: 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?
Explanation: 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.
Explanation: 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.
Explanation: P=I2R =52*2=50W.
Explanation: 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.
Explanation: 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.
Explanation: 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?
b) J/s(joules per second)
d) J/h(joules per hour
Explanation: Power = energy/time
SI unit of power = SI unit of energy/SI unit of time = joule/second.
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