Basic Electrical Engineering Questions and Answers – Energy

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This set of Basic Electrical Engineering Multiple Choice Questions & Answers (MCQs) focuses 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)
View Answer

Answer: b
Explanation: 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.
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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
View Answer

Answer: c
Explanation: 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
View Answer

Answer: c
Explanation: 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.
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4. Calculate the energy dissipated by the circuit in 50 seconds.
Find the energy dissipated by the circuit in 50 seconds if V is 100 & R is 10
a) 50kJ
b) 50J
c) 100j
d) 100kJ
View Answer

Answer: a
Explanation: 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
View Answer

Answer: c
Explanation: 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.
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6. Calculate the energy in the 10 ohm resistance in 10 seconds.
Find the Energy in 10 ohm resistance in 10 seconds when resistors are connected in parallel
a) 400J
b) 40kJ
c) 4000J
d) 4kJ
View Answer

Answer: b
Explanation: 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
View Answer

Answer: b
Explanation: A battery is a device in which the chemical elements within the battery react with each other to produce electrical energy.
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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
View Answer

Answer b
Explanation: 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.
Find the energy in the 5 ohm resistor in 20 seconds in series connected circuit
a) 21.5kJ
b) 2.15kJ
c) 2.15J
d) 21.5J
View Answer

Answer: a
Explanation: 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.
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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
View Answer

Answer: b
Explanation: 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.

Sanfoundry Global Education & Learning Series – Basic Electrical Engineering.

To practice all areas of Basic Electrical Engineering, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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