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Power - Meaning, Unit, Formula, FAQs

Power - Meaning, Unit, Formula, FAQs

Edited By Vishal kumar | Updated on Sep 25, 2024 04:18 PM IST

Power in physics refers to the amount of work done or energy transferred per unit of time. It is expressed in watts (W) and a watt is the equivalent of one joule per second. Power measures, how fast the energy is used or converted to another form. In electrical systems, for e.g. power = voltage x current (in a charge-neutral case, what this really tells you is how much electrical energy is being consumed, or being pumped, per unit of time. Power is important in running the various types of equipment and machines we use in everyday life. Generally, the higher the power rating, the more work a device can do in less time.

Power - Meaning, Unit, Formula, FAQs
Power - Meaning, Unit, Formula, FAQs

In this article, we are going to read about power and different types of power and also see some solved examples, which belong to the chapter work, energy, and Power, which is one of the important chapters in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), more than fifteen questions have been asked on this concept. And for NEET almost six questions were asked from this concept.

Background wave

Let's read this entire article to gain an in-depth understanding of the concept of power.

Define Power

Power is defined as the rate at which work is done or energy is transferred.

ML2T3

  • M: Stands for mass
  • L: Stands for length
  • T: Stands for time

Units - Watt or Joule/sec (in SI), Erg/sec (in CGS)

Average power

Pavg=ΔWΔt=0tPdt0tdt

Pavg  : Average power
ΔW : Change in work done
Δt : Change in time
P: Power at a given time

The average power is computed over a time interval by taking the total work done divided by the total time taken.

Instantaneous Power:

P=dWdt=Fv

- P: Instantaneous power
- dW/dt : Rate of change of work
- F : Force vector
- v : Velocity vector

Instantaneous power is the dot product of the force applied and the velocity of the object. This represents the power being delivered at any specific moment in time.

Power and Kinetic Energy:

P=dKdt

- P: Power
- dK/dt : Rate of change of kinetic energy

Power can also be defined as the rate of change of kinetic energy, connecting work-energy principles to power.

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Solved Example Based on Power

Example 1: An engine of a car of mass m = 1000 Kg changes its velocity from 5 m/s to 25 m/s in 5 minutes. The power (in KW) of the engine is

1) 5

2) 2

3) 1

4) 4

Solution:

Calculate the change in kinetic energy (ΔK) :

ΔK=121000((25)2(5)2)ΔK=121000(62525)ΔK=121000600=300,000 J
Now, calculate the power:

P=ΔKΔt=300,000300P=1000 W=1 kW
Final Answer: Hence, the power of the engine is 1 kW.
Correct Option: (3).

Example 2: A constant power-delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time 't' is proportional to :

1) t2/3
2) t
3) t3/2
4) t1/2

Solution:

Power delivered by the machine is constant:

P=FV
Substitute F=ma and a=dVdt :

P=mVdVdt
Given P=C (constant):

VdVdt=Cm
Integrate:

V22=CmtV2tVt1/2
Since V=dxdt :

dxdtt1/2

Integrate again:

xt3/2
Final Answer:
Correct Option: (3).

Example 3: Sand is being dropped from a stationary dropper at a rate of 0.5 kg−1 on a conveyor belt moving with a velocity of 5 ms−1. The power needed to keep the belt moving with the same velocity will be :

1) 1.25 W
2) 2.5 W
3) 6.25 W
4) 12.5 W

Solution:

The power needed to maintain the belt's motion is:

P=dmdtv2
Substitute the given values:

P=0.5(5)2P=0.525=12.5 W
Final Answer:
The power required is 12.5 W . Correct Option: (4).

Example 4: Sand is being dropped from a stationary dropper at a rate of 0.5 kg1 on a conveyor belt moving with a velocity of 5 ms1. The power needed to keep the belt moving with the same velocity will be:

1) 1.25 W
2) 2.5 W
3) 6.25 W
4) 12.5 W

Solution:

The power needed to keep the conveyor belt moving with the same velocity is given by:

P=dmdtv2


Substituting the values:
- dmdt=0.5 kg/s
- v=5 m/s

P=(0.5)(5)2P=0.525=12.5 W


Final Answer:
The power required is 12.5 W .
Correct Option: (4).

Frequently Asked Questions (FAQs)

1. In terms of electricity, what is the difference between Power and energy?

The energy supplied by source in maintaining the flow of electric current is called electrical energy meanwhile the time rate at which electric energy is consumed by an electrical device is called electric power.

2. In terms of physics, what is power?

Power is a unit of measurement for the rate of energy transmission per unit of time. A scalar quantity is Power. In electrical engineering, Power refers to the rate at which electrical energy flows into or out of a given component.

3. What exactly is Power? What is the SI unit for it?

It is defined as the rate at which work is completed. The watt is a unit of measurement.

4. Define the term "average power."

Average power is the ratio of total effort or energy consumed to total time when a machine or person undertakes different quantities of work or utilizes energy at different periods of time.

5. What are Electric Generators?

Electric generators transform mechanical energy (the Power of motion) from an external source into electrical energy.

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Questions related to

Get answers from students and experts

If you have an eye power of -2, whether you can apply for RRB exams depends on the medical standards required for the specific post you’re aiming for.


For example:


A-3 Medical Standard: Your power should not exceed 2 diopters. So, if your power is exactly -2, you may qualify as long as your vision is corrected to 6/9 in both eyes with glasses.


B-1 Medical Standard: Here, the limit is 4 diopters, so you would qualify as well if your vision meets the required correction standards.



Other vision-related criteria like color vision, binocular vision, and field of vision also matter. It's best to carefully check the medical standards for the job you're applying for.

(-7/10)^3 = (-7)^3 / (10)^3 = -343 / 1000 = -0.343

Explanation:


For any negative fraction, if the exponent raised is an odd number, then the result obtained will be negative. Here we have,

Cube the numerator: (-7)^3 = -343

Cube the denominator: (10)^3 = 1000

Divide the results: -343 / 1000 = -0.343

Thus, (-7/10)^3 is equal to -0.343.

Hello,

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Hope it helps !

To solve remainder problem you can use pattern recognition method,eulers therom,modular exponentiation, online calculator,etc.just simply simplify large powers into small ones,use pattern and cycle to reduce calculation and keep practicing to improve your skills.from the mod calculation the answer of this equation is 65.if you need calculation please specify in next question.

Hello,


To calculate the total power in the AM wave and power in each sideband, we can use the following formulas:


1. Total Power (Pt) = Carrier Power (Pc) x [1 + (μ^2)/2]


where μ is the modulation index (80% in this case)


1. Power in each sideband (Ps) = (μ^2/4) x Pc


Given:

Pc = 500 W (carrier power)

μ = 0.8 (modulation index)


Calculations:


1. Total Power (Pt) = 500 x [1 + (0.8^2)/2]

= 500 x [1 + 0.64/2]

= 500 x 1.32

= 660 W


2. Power in each sideband (Ps) = (0.8^2/4) x 500

= (0.64/4) x 500

= 0.16 x 500

= 80 W


So, the total power in the AM wave is 660 W, and the power in each sideband is 80 W.

Hope this helps,

Thank you

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