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

How much units of energy is consumed in operating ten 50 watt bulbs for 10 hours per day in a month (30 days)

  1. 100 kwh

  2. 150 kwh

  3. 160 kwh

  4. 500 kwh
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given :    $P = 0.05$ kW               $t = 10$ hours per day 
Number of bulbs     $n = 10$
Energy consumed per day       $E= nPt = 10 \times 0.05\times 10 = 5$ kwh  per day
Total energy consumed in 30 days    $E _t = 5\times 30 = 150$ kWh
Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

How much energy in kilowatt hour is consumed in operating ten 50 watt bulbs for 10 hours per day in a month (30 days)

  1. 1500

  2. 5000

  3. 15

  4. 150
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Energy consumed $=10\times 50\times 10\times 30\times 3600 J$
$[1 Kwh=3600\times 1000J]$
$=\frac {10\times 50\times 10\times 30\times 3600}{3600\times 1000}kWh=150$


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

A geyser of $2.5 kw$ is used for $8$ hours daily. Calculate the monthly consumption ( 30 days) of electrical energy units. Also calculate the cost of electricity units consumed in a month if rate per unit is $3.50$

  1. $Rs. 2100.00$
  2. $Rs. 155.00$
  3. $Rs. 150.00$
  4. $Rs. 30.00$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Daily usage of electricity   $E _1 = 2.5 \ kw\times 8 = 20 \ kwh = 20 \ units$

Energy used in one month   $E = 30E _1 = 30\times 20 = 600\ units$
Cost    $ = Rs. 600\times 3.5 = Rs 2100.00$

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

An electric heater operating at $220\ V$ boils $5\ l$ of water in $5\ \text{minutes}$. If it is used on a $110\ V$ line, it will boil the same amount of water in:

  1. $10\ \text{minutes}$
  2. $20\ \text{minutes}$
  3. $5\ \text{minutes}$
  4. $1\ \text{minute}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Heat required to boil the water will be same in the two cases.
Heat produced by heater is:
$Q = \cfrac{V^2}{R} t$
$V^2 = \cfrac{QR}{t}$
$V^2 \propto \dfrac{1}{t}$

$\cfrac{t _2}{t _1} = \cfrac{V _1^2}{V _2^2}$
$t _2 = \dfrac{220^2 \times 5}{110^2}$
$t _2 = 20\ min$
Multiple choice physics effects of electric current thermal effect of electric current heating effect of electric current electric current and its effects

The energy expended in $1\  kW$ electric heater in $30\ \text{seconds}$ will be:

  1. $\displaystyle 3\times 10^4\ J$
  2. $\displaystyle 3\times 10^4\ erg$
  3. $\displaystyle 3\times 10^4\ eV$
  4. $0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\text{Energy = Power} \times \text{time}$
$\text{Energy} = 1000 \times 30 = 30000\ J = 3 \times {10}^{4}\ J$

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

Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply. The energy supplied in kWh to the three heaters in 5 hours is :

  1. $3.75 kWh$
  2. $4 kWh$
  3. $0.6 kWh$
  4. $5.74 kWh$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Power is given as: $P=VI$


Substituting, $I=\dfrac{V}{R}$ in the above formula, we get, $P=\dfrac { { V }^{ 2 } }{ R } $

Given that the voltage is $250\ V$ and the power is $60\ W$, the resistance of the bulb is calculated as follows.

$R=\dfrac { { V }^{ 2 } }{ P } =\dfrac { { 100 }^{ 2 } }{ 250 } =40\Omega$

Hence, the resistance of each resistor is 40 ohms.

The energy consumed by the appliance in kWh is given by the formula: $P=\dfrac { { V }^{ 2 } }{ R } \times t=\dfrac { { 100 }^{ 2 } }{ 40 } \times t= 1.25\ kWh$.

When three heater are connected in parallel, then the total energy consumed is given as $1.25\ kWh\times 3=3.75\ kWh.$.

Hence, the total power consumed is 3.75 kWh.

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

A geyser is rated 1500 W, 250 V. This geyser is connected to 250 V mains. The cost of energy consumed at Rs. 4.20 per kWh for 5 hours will be :

  1. $Rs. 250$
  2. $Rs. 300$
  3. $Rs. 310$
  4. $Rs. 315$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The electric power in watts associated with a complete electric circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. 


The power is given by the product of applied voltage and the electric current.

 That is, $P=VI. The\ power\ of\ the\ geyser\ is\ given\ as\ 1500 W$

The energy consumed by the geyser in kWh is given by the formula $Q=P\times t\quad =\quad 1500W\times 50hours\quad =\quad 75\quad kWh$. 

Hence, the energy consumed by the geyser in 5 hours is 75 kWh. 

The cost of energy consumed per kWh is given as Rs. 4.20.

That is, $4.20\times 75\quad kWh=\quad Rs.315$

Hence, the total cost is given as Rs. 315.

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

A lamp of $100\ W$ and a heater of $1\ kW$ are in simultaneous use for $10\ hrs$. The units of electricity consumed according to the meter in the house is:

  1. $10$
  2. $9$
  3. $11$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For lamp, electrical energy consumed is $E _L = P _Lt =100\times 10 = 1\ kWh$

For heater, electrical energy consumed is $E _H = P _Ht =1000\times 10 = 10\ kWh$
Hence, total electrical energy consumed is: $E = E _L+E _H=11\ kWh$

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

A motor of $50\ W$ runs for $20\ hrs$. How many 'units' ($kWh$) of electrical energy are consumed?

  1. $5\ kWh$
  2. $2\ kWh$
  3. $1\ kWh$
  4. $2.5\ kWh$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given :       $P = 0.050\ kW$ 

                   $t = 20\ h$

Energy consumed,      
$E = Pt =0.050\times 20 = 1\ kWh$

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

Which of these will consume most units of energy in a day?

  1. A $60\ W$ bulb used for $6\ hrs$.
  2. A $30\ W$ bulb used for $12\ hrs$.
  3. A $20\ W$ bulb used for $24\ hrs$.
  4. A $100\ W$ bulb used for $2\ hrs$.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Energy consumed per day is: $E = Pt$

(A) :     $E _A = 0.06\times 6 = 0.36\ kWh$
(B) :     $E _B = 0.03\times 12 = 0.36\ kWh$ 

(C) :     $E _C = 0.02\times 24 = 0.48\ kWh$ 
(D) :     $E _D = 0.10\times 2 = 0.2\ kWh$ 

Thus bulb of option C will consume more energy.

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

Calculate the total power of 2 fans, if each of them draws a current of $5   A$ at a p.d of $200   V$.

  1. $2000 \omega$
  2. $1000 V$
  3. $250 \omega$
  4. $2 \omega$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$I = 5A,   p.d = 200   V$
Power $= V \times I = 200 \times 5 = 1000   \omega$
For 2 fans $= 2 \times 1000 = 2000   \omega$.