250+ TOP MCQs on Energy and Answers

Basic Electrical Engineering Multiple Choice Questions on “Energy”.

1. Which among the following is a unit for electrical energy?
A. V(volt)
B. kWh(kilowatt-hour)
C. Ohm
D. C(coloumB.
Answer: B
Clarification: Kilowatt is a unit of power and hour is a unit of time. Energy is the product of power and time, hence the unit for Energy is kWh.

2. A bulb has a power of 200W. What is the energy dissipated by it in 5 minutes?
A. 60J
B. 1000J
C. 60kJ
D. 1kJ
Answer: C
Clarification: Here, Power = 200w and time = 5min. E=Pt => E= 200*5= 1000Wmin=60000Ws= 60000J= 60kJ.

3. Out of the following, which one is not a source of electrical energy?
A. Solar cell
B. Battery
C. Potentiometer
D. Generator
Answer: C
Clarification: Solar cell converts light energy to electrical energy. Battery converts chemical energy to electrical energy. Generator generates electrical energy using electromagnetic induction. A potentiometer is an instrument used for measuring voltage and consumes electrical energy instead of generating it.

4. Calculate the energy dissipated by the circuit in 50 seconds.
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A. 50kJ
B. 50J
C. 100j
D. 100kJ
Answer: A
Clarification: Here V = 100 and R = 10. Power in the circuit= V²/R = 100²/10 = 1000W.
Energy= Pt= 1000*50 = 50000J = 50kJ.

5. Which among the following is an expression for energy?
A. V²It
B. V²Rt
C. V²t/R
D. V²t²/R
Answer: C
Clarification: Expression for power = VI, substituting I from ohm’s law we can write, P=V²/R. Energy is the product of power and time, hence E=Pt = V²t/R.

6. Calculate the energy in the 10 ohm resistance in 10 seconds.
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A. 400J
B. 40kJ
C. 4000J
D. 4kJ
Answer: B
Clarification: Since the resistors are connected in parallel, the voltage across both the resistors are the same, hence we can use the expression P=V²/R. P=200²/10= 4000W. E=Pt = 4000*10=40000Ws = 40000J = 40kJ.

7. A battery converts___________
A. Electrical energy to chemical energy
B. Chemical energy to electrical energy
C. Mechanical energy to electrical energy
D. Chemical energy to mechanical energy
Answer: B
Clarification: A battery is a device in which the chemical elements within the battery react with each other to produce electrical energy.

8. A current of 2A flows in a wire offering a resistance of 10ohm. Calculate the energy dissipated by the wire in 0.5 hours.
A. 72Wh
B. 72kJ
C. 7200J
D. 72kJh
Answer b
Clarification: Here I (current) = 2A and Resistance(R) = 10ohm. Power = I²R = 2²*10=40. Energy = Pt = 40*0.5*60*60 = 72000J=72kJ.

9. Calculate the energy in the 5 ohm resistor in 20 seconds.
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A. 21.5kJ
B. 2.15kJ
C. 2.15J
D. 21.5kJ
Answer: A
Clarification: The current in the circuit is equal to the current in the 5 ohm resistor since it a series connected circuit, hence I=220/(5+10)=14.67A. P=I²R = 14.67²*5=1075.8W. E=Pt = 1075.8*20 = 21516J=21.5kJ.

10. Practically, if 10kJ of energy is supplied to a device, how much energy will the device give back?
A. Equal to10kJ
B. Less than 10kJ
C. More than 10kJ
D. Zero
Answer: B
Clarification: Practically, if 10kJ of energy is supplied to a system, it returns less than the supplied energy because, some of the energy is lost as heat energy, sound energy etc.