Physics
Power, Energy, and Efficiency
443 Questions
Power, energy, and efficiency are fundamental physics concepts that quantify work and the rate of doing work. Questions cover electrical energy consumption in kilowatt hours, the efficiency of thermodynamic cycles like the Brayton cycle, and mechanical power lifting loads. This topic is essential for general science and engineering exam preparation.
Electrical energy consumptionBrayton cycle efficiencyPower calculation formulasCommercial electricity unitsHydraulic lift mechanics
Power, Energy, and Efficiency Questions
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25 kg/mt/sec
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50 kg/mt/sec
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75 kh/mt/ sec
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100 kg/mt/sec
C
Correct answer
Explanation
One horsepower (HP) equals 75 kilogram-meters per second (kg-m/sec) in the metric system. This unit represents the power needed to lift 75 kg to a height of 1 meter in 1 second. The mechanical horsepower definition varies slightly between systems, but 75 kg-m/sec is the commonly accepted metric conversion.
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1 ampere
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1 volt
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1 joule
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1 kilowatt hour
D
Correct answer
Explanation
Electricity consumption for homes is measured in kilowatt-hours (kWh), commonly called 'units'. One kilowatt-hour represents the energy consumed by a 1000-watt appliance running for one hour. This is why electricity bills are calculated based on kWh consumption.
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P = VI
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V = IR
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P = IR
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Power = energy/time
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P = I2R
A
Correct answer
Explanation
Power desipated in resistance=voltage supplied. Current flowing through resistance.
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0.33
-
0.57
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2.5
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5.2
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None of the above
A
Correct answer
Explanation
Performance ratio = single-cycle time/pipeline time = 40/12 = 0.33.
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20 W
-
0.05 W
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2000 W
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0.33 W
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120 KW
A
Correct answer
Explanation
p = w / t
p = 200/10 = 20 W
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70,000 kwh
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70 kwh
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700 kwh
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43750 kwh
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4375 kwh
B
Correct answer
Explanation
We know that 1 kilowatt = 1000 watt. So, 40 watt = 40/1000 = 0.040 kwatt. Now as per the formula, 0.040 X 1750 = 70 kwh
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1200 J
-
20 J
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0.33 J
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3 J
-
0.05 J
A
Correct answer
Explanation
Power = Joules/time.
Joules = Power*time
J = 20* 60 (1 min = 60 sec)
Joules = 1200
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200 A
-
5 mA
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6.25 mA
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160 A
-
0.16 A
A
Correct answer
Explanation
With 80percent efficiency the power supplied is 4.0/0.8 Megawatts
= 5Megawatts
Electrical power = volts*amps
5*(106) = 25*(103)amps
amps = 5 x 106/25(103)
= 200amps
Current drawn from overhead wires is 200amps
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400 litre
-
2000 litre
-
1500 litre
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300 litre
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100 litre
A
Correct answer
Explanation
P = mgh / t or m = (p Χ t) / (g Χ h)
m = (1000W Χ 60s) / (10 ms-2 Χ15 m) = 400 kg = 400 litre.
B
Correct answer
Explanation
No.of request=100/sec
slot time=20 msec
no of slots per second=1/20 msec
=50
so, channel load=(No of request/sec)/(No.of slots/sec)
=100/50
=2
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150 µF
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6 µF
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55 µF
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5 µF
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16 µF
A
Correct answer
Explanation
Cs = C1 / n
Cs = 6 µF
n = 5
C1 = Cs x n
= 30 µF
Cp=n x C1
= 5 x 30
= 150 µF
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3.6 × 106 joule
-
0.277 joule
-
16.66 joule
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60000 joule
-
6000 joule
A
Correct answer
Explanation
1 kW h = 1000 watt × 3600 second
= 3.6 × 106 watt second
= 3.6 × 106 joule (J)
C
Correct answer
Explanation
It is 150%.
Input-output ratio = Input Units X 100
Output Units
= 600 Units X 100
400 Units
= 150 %
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10-15 kg/HP
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60-80 kg/HP
-
20-30 kg/HP
-
100-150 kg/HP
B
Correct answer
Explanation
Steam engines typically have a weight-to-power ratio of 60-80 kg per HP, which is much higher than modern diesel or electric locomotives. Values like 10-15 kg/HP (option A) are more typical of modern locomotives, while 20-30 kg/HP (option C) is still too low for steam technology. 100-150 kg/HP (option D) would be extremely heavy even for steam.
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50-65%
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25-35%
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15-20%
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10-15%
B
Correct answer
Explanation
Diesel engines typically operate with thermal efficiency in the 25-35% range, meaning only about a quarter to a third of the fuel's chemical energy is converted to useful mechanical work. The remaining energy is lost as heat through exhaust and cooling systems. This is significantly lower than electric motors but higher than early steam engines.