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

Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

A bulb of $2.5  V$, draws a current of $0.5  A$. If the bulb is switched on for 2 minutes, calculate the energy released by the bulb.

  1. $1.25 J$
  2. $260 J$
  3. $5 J$
  4. $150 J$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

P.D $= 2.5   V$
Current $= 0.5   A$
$t = 2$ minute $= 2 \times 60 = 120  s$
$ E = V \times I \times t = 2.5 \times 0.5 \times 120 = 150   J$.

Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

A table lamp of power 60 W consumed 9 (commerical) units of electricity in the month of April. For how many hours per day, on an average, was the lamp in use?

  1. 1 h

  2. 3 h

  3. 5 h

  4. 6 h
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

1 unit is $3.6\times 10^6J$

let us suppose it has been used for y hours a day
energy consumed in y hours =$60\times y\times 60\times 60$ , y in hours, so converting it into seconds

now it has been used for whole month so,
$60\times y\times 60\times 60\times  30=9\times 3.6\times 10^6$

$y=5 hours$

Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

A table lamp of power 60 W consumed 9 (commercial) units of electricity in the month of April. For how many hours per day, on an average, was the lamp in use?

  1. $1\ h$
  2. $3\ h$
  3. $5\ h$
  4. $6\ h$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$9\ units = 9kWh$
For the month of April (30 days), we get an average of $ \cfrac{9000}{30} =300 Wh$ energy per day.
For an appliance of $60\ W$, this clearly corresponds to $ t = \cfrac{E}{P} = \cfrac{300}{60} = 5\ h $

Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

An air conditioner is rated $240\ V, 1.5\ kW$. The air conditioner is switched on $8\ hrs$ each day. What is electrical energy consumed in $30\ days$?

  1. $2.88\ kWh$
  2. $360\ kWh$
  3. $120\ kWh$
  4. $240\ kWh$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy consumed is given by:

$E= Pt$
    $= 1.5 KW \times  30 \times 8 hrs= 360 KWh $

Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

An electric bulb is connected to a $20$ V battery of negligible internal resistance. The resistance offered by the bulb is $5$ $\Omega$. The electrical energy consumed by the bulb in $3$ hours is _____ kWh.

  1. $0.12$
  2. $0.24$
  3. $0.06$
  4. $0.03$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power consumed by bulb is $P=\dfrac{V^2}{R}$


Energy$=E=Pt=\dfrac{V^2}{R}t=\dfrac{20^2}{5}W\times 3h$

$\implies E=240Wh=0.24kWh$

Answer-(B)

Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

What should be the bill for the month of April for a heater of resistance 60.5 $\Omega$ connected 220 V mains, The cost of energy is Rs 2 per kWh and the heater is used for 3 hours daily?

  1. $Rs\ 144$
  2. $Rs\ 222$
  3. $Rs\ 662$
  4. $Rs\ 238$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Power of heater $=P=\dfrac{V^2}{R}$


Energy used in one month$=E=Pt=\dfrac{V^2}{R}\times 3\times 30$

$\implies E=\dfrac{220\times 220}{60.5}\times 90Wh=\dfrac{484\times 9}{60.5}kWh=72kWh$

Hence cost$=72\times 2=Rs144$

Answer-(A)

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

An electric iron uses a power of $1320 W$ when set to higher temperature. If set to lower temperature one third of higher temperature current is used. If iron is connected to a potential of $220 V$, then power used to lower temperature is _____ .

  1. $220 W$
  2. $440 W$
  3. $660 W$
  4. $880 W$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Power, $P = VI$ (power at high temperature) 


$1320 = 220 I$ (current at high temperature)

Current at high temperature, $I _t = 6A $

$I _{LT} = \dfrac{6}{3} = 2A$ (current at low temperature) 

$P _{LT} = 220 \times2 \ W$

$ = 440 \ W$ (power at low temperature)

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

A student in town in India, where the price per unit ($1 unit = 1 kW-hr$) of electricity is $Rs.5.00$, purchases a $1 kVA$ UPS (uninterrupted power supply) battery. A day before the exam, 10 friends arrive to the student's home with their laptops and all connect their laptops to the UPS. Assume that each laptop has a constant power requirement of $90 W$. Consider the following statements :


I. All the 10 laptops can be powered by the UPS if connected directly.
II. All the 10 laptops can be powered if connected using an extension box with a $3 A$ fuse.
III. If all the 10 friends use the laptop for 5 hours, then the cost of the consumed electricity is about $Rs.22.50.$

Select the correct option with the true statements.

  1. I only

  2. I and II only

  3. I and III only

  4. II and III only
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
power supplied by battery $1\ kVA=1000\ W$

$1$ laptop required $90\ W$.

and all $10$ laptop are connected directly so,

$P _T=90\ W\times 10=900\ W$

$1\ unit=1\ KW\ hr$ cost $=5\ Rs$.

So, If $10$ Laptops are used for $5\ hr$ well consumed

$unit=\dfrac{900\times 5\times 3600}{3600000}=4.5$

Total pay amount in $5\ hr=4.5\ unit\times 5$
                         $=22.5\ rupees$
Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

An electric bulb is rated 250 W, 230 V. The energy consumed in one hour is :

  1. $9 \times 10^5$ J
  2. $15 \times 10^5$ J
  3. $25 \times 10^5$ J
  4. $23 \times 10^5$ J
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Energy ($power\times time$) is measured in Joules and by including time (t) in the power formulae, the energy dissipated by a component or circuit can be calculated.


Energy dissipated Q= Pt.

In the question, the time taken is given as 1 hour = 60 minutes, that is $60\times 60=3600\quad seconds$ and the power is 250 W.

Hence, the energy consumed by the bulb is $250\quad W\times 3600\quad s=9.00,000\quad J.\quad That\quad is,\quad 9\times { 10 }^{ 5 }\quad Joules$. 

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

An air conditioner is rated $240\;V,\;1.5\;kW$. The air conditioner is switched on for $8$ hours each day. How much electrical energy is consumed in $30$ days ?

  1. $360\;kW\;h$
  2. $8.64\;kW\;h$
  3. $120\;kW\,h$
  4. $240\;kW\;h$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

If a certain amount of power is dissipated for a given time, then energy is dissipated. Energy (powertime) is measured in Joules and by including time (t) in the power formulae, the energy dissipated by a component or circuit can be calculated.
Energy dissipated = Pt .
In this case, the power dissipated is 1.5 kW and it runs for 8 hours a day.
So, in a day, the energy dissipated is 1.5 * 8 = 12 Kwh.
For 30 days, the power dissipated is 12 * 30 = 360 Kwh.
Hence, the electrical energy is consumed by the air conditioner in 30 days is 360 Kwh.

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

A washing machine rated $300W$ is operated for one hour/day. If the cost of a unit is Rs $3.00$ then the cost of the energy to operate a washing machine for the month of March is

  1. $Rs 25.60$
  2. $Rs 27.50$
  3. $Rs 27.90$
  4. $Rs 26.90$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In this case, the washing machine that is rated $ 300\ W$ is operated for $1$ hr/day. The electric bill has to be calculated for the month of March. That is, the total number of hours the washing machine is operated in this month is $31$ hours as this month has $31$ days.
The total energy that is used by the washing machine in a day is given by the formula $Q=Pt$.
Here, the power is $300\ W$ and the time used is $1\ hour$.
So, $Q=300W\times 1h=300Wh= 0.3\ kWh$
Therefore, the total number of $kWh$ for a month is $0.3kWh\times 31=9.3\ kWh $
The cost of $1\ kWh$ is given as $Rs. 3$.

Hence, the total cost for a month is $=9.3kWh\times 3=Rs. 27.90$

Multiple choice physics electrical circuits domestic electric circuits and safety precautions domestic and commercial circuit electric power

If the length of filament of a heater is reduced by  10%, the power of the heater will 

  1. increase by about 9%

  2. increase by about 11%

  3. increase by about 19%

  4. increase by about 10%
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$P=\dfrac { { V }^{ 2 } }{ R } \ R=\dfrac { \rho l }{ A } \ \therefore P=\dfrac { { V }^{ 2 }A }{ \rho l } $

$ \therefore P\alpha \dfrac { 1 }{ l } \quad $ [Keep in $\dfrac{v^2A}{\rho }$ constant]
$\therefore$ with reduction of $l$ by $10\%$ power will increase by $10\%$.