Physics Questions and Answers – Current Electricity – Electrical Energy & Power

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “Current Electricity – Electrical Energy & Power”.

1. The SI unit of electrical energy is ____________
a) kilojoule (KJ)
b) joules (J)
c) watt (W)
d) kilowatt (KW)
View Answer

Answer: b
Explanation: Electric energy is defined as the total electric work done or energy supplied by the source of emf in maintaining the current in an electric circuit for a given time. The SI unit of electrical energy is joule (J). The commercial unit of electric energy is kilowatt-hour (kWh).
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2. 1 kWh = ___________
a) 3.6 × 106 J
b) 3.6 × 105 J
c) 0.36 × 106 J
d) 0.36 × 105 J
View Answer

Answer: a
Explanation: The commercial unit of electric energy is kilowatt-hour (kWh).
1 kWh = 1000 Wh = 3.6 × 106J = one unit of electricity consumed. The electric energy used in factories, industries and houses are measured in kWh.

3. Calculate the number of units of electricity used if a bulb of 100 W is kept on for 5 hours.
a) 1 unit
b) 0.1 unit
c) 5 unit
d) 0.5 unit
View Answer

Answer: d
Explanation: The number of units of electricity consumed is
n = \(\frac {(total \, wattage \, \times \, time \, in \, hour)}{1000}\)
Total wattage = 100 W      Time in hour = 5 hours
Therefore, n = \(\frac {100 \times 5}{1000}\)
= 0.5 units
So, the number of units of electricity consumed is 0.5 units.
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4. Electric energy is dependent on time.
a) True
b) False
View Answer

Answer: a
Explanation: Yes, electric energy is dependent on time.
Electric energy = electric power × time = P × t. Electric energy is basically the energy derived from electric charge. It can be calculated by multiplying the electric power of the body with the time taken for the power emission.

5. Kabir bought 5 new light bulbs of 50 W each in addition to the 7 bulbs he already had in his house which were also 50 W each. Calculate his electricity bill, if he keeps the new bulbs on for 5 hours and the older bulbs on only for 3 hours, and the cost of one unit of electricity is Rs. 60.
a) Rs.136
b) Rs.137
c) Rs.138
d) Rs.139
View Answer

Answer: c
Explanation: New bulbs = 5; Old bulbs = 7; Total wattage of new bulbs = 5 × 50;
Total wattage of old bulbs = 7 × 50; Time the new bulbs are kept on = 5 hours;
Time the old bulbs are kept on = 3 hours
The number of units of electricity consumed by new bulbs (n1) = \(\frac {(total \, wattage \, \times \, time \, in \, hours)}{1000}\)
= \(\frac {50 \times 5 \times 5}{1000}\) = 1.25
The number of units of electricity consumed by old bulbs (n2) = \(\frac {50 \times 7 \times 3}{1000}\) = 1.05 Total bill of electricity = number of units of electricity consumed × amount for one unit of electricity
= (n1 + n2) × 60
= (1.25 + 1.05) × 60
= 138
Therefore, his electricity bill will be Rs. 138.
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6. Identify the correct formula of electric power.
a) Electric power = \(\frac {time \, taken}{electric \, work \, done}\)
b) Electric power = \(\frac {electric \, work \, done}{time \, taken}\)
c) Electric power = electric work done × time taken
d) Electric power = \(\frac {1}{electric \, work \, done}\)
View Answer

Answer: b
Explanation: Electric power is defined as the rate at which work is done by the source of emf in maintaining the current in the electric circuit. So, the formula of electric power is:
Electric power (P) = \(\frac {electric \, work \, done}{time \, taken}\)

7. Which one of the following is the practical unit of power?
a) Watt (W)
b) Kilowatt hour (kWh)
c) Horse power (hp)
d) Kilojoule (kJ)
View Answer

Answer: c
Explanation: The practical unit of power is horse power (hp). Kilo watt is also another practical unit of power. 1 kilowatt = 1000 watt; 1 hp = 746 watt. It is usually used in reference to the output of engines or motors.
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8. The power consumed by a 300 V bulb, having a resistance of 100 ohms, is 3 Watts.
a) True
b) False
View Answer

Answer: b
Explanation: Electric power can also be calculated or represented in other ways, such as: Power = voltage × current; Power = current2 × resistance; Power = \(\frac {voltage^2}{resistance}\). In this case, we can use the equation – Power = \(\frac {voltage^2}{resistance}\). Voltage = 300 V; Resistance = 100 ohms
Power = \(\frac {300 \times 300}{100}\)
= 900 Watts
Therefore, the power consumed by a 300 V bulb, having a resistance of 100 ohms, is 900 Watts.

9. One watt is equal to __________
a) one kilowatt per second
b) one kilo joule per second
c) one joule per second
d) one joule per minute
View Answer

Answer: c
Explanation: Watt is the SI unit of power. Power = \(\frac { work \, done}{time}\). The SI unit of work done is the same as energy, that is, joule and that of time is seconds. Therefore, one watt is equal to one joule per second.
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10. An engine uses 30 A of current. The resistance offered is 15 ohms. Calculate the power consumed by the engine in horse power.
a) 18 hp
b) 19 hp
c) 17 hp
d) 13500 hp
View Answer

Answer: a
Explanation: Current used = 30 A; Resistance = 15 ohms
The required equation is: Power = current2 × resistance
= 30 × 30 × 15
= 13,500 Watts
We know that, 1 Watt = 746 horse power (hp). So, 13,500 Watts = 18.096 hp, which can approximately be equal to 18 hp.
Therefore, the engine consumes 18 hp power.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter