Power - Meaning, Unit, Formula, FAQs
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.
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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.
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.
Dimension- ML2T−3
Units - Watt or Joule/sec (in SI), Erg/sec (in CGS)
Average power
Pav=ΔwΔt=∫0tp⋅dt∫0tdt
Instantaneous power
P=dwdt=P=F→⋅v→
Where, F→→ force v→→ velocity
Power is expressed as the rate of change of kinetic Energy
P=dkdt
Where,
dk→ change in kinetic energy dt→ interval of time
i.e. power is equal to the scalar product of force with velocity
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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:
Power expressed as the rate of change of kinetic Energy -
Power = Work done / time = change in K.E / time
Power = dk/dt
P=12m[(25)2−(5)2]5×60=12×1000×6005×60=1000 WP=1KW
Hence, the answer is the 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 machine is, P
Force acting on the machine, F
\text { As we know, } P=\vec{F} \cdot \vec{V}
As force and velocity have the same direction the angle between force and velocity is
So P=FV
or P=maV=mdVdtV
Given that the machine is delivering constant power.
So P=C (constant)
P=C
FV=C
MdVdtV=C
Integrating both sides, we get
V22∝tV∝t1/2dxdt∝t1/2 Again integrate both sides x∝t3/2
Hence, the answer is the 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 keep the belt moving with the same velocity will be
P=(dmdt)v2=(0.5)×25P=12.5 W
Hence, the answer is the option (4).
Example 4: A particle of mass M is moving in a circle of fixed radius R in such a way that its centripetal acceleration at time t is given by n2 R t2 where n is a constant. The power delivered to the particle by the force acting on it is:
1) Mn2R2t
2) MnR2t
3) MnR2t2
4) 12Mn2R2t2
Solution:
Centripetal acceleration, ac=n2rt2 where, ac=v2r
⇒v2r=n2rt2⇒v=nrt…
tangential acceleration, at=dvdt=nr…
Tangential force acting on the particle, F=Mat=Mnr
Power delivered, P=F→⋅v→=Fvcosθ
∴P=Fv=(Mnr)×nrt(∵θ=0∘)⟹P=Mn2r2t
Hence, the correct option is option 1.
Example 5: 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 keep the belt moving with the same velocity will be
P=(dmdt)v2=(0.5)×25P=12.5 W
Hence, the answer is the option (4).
Conclusion
Power in physics is the rate at which work is done or energy is transferred, measured in watts (W), which is equivalent to one joule per second. It can be calculated as the product of voltage and current in electrical systems and as the rate of change of kinetic energy in mechanical systems. Power is crucial for efficiently operating various equipment and machines. Understanding power is essential for exams like JEE Main and NEET, where multiple questions are often asked on this concept.
Frequently Asked Questions (FAQs)
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.
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.
It is defined as the rate at which work is completed. The watt is a unit of measurement.
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.
Electric generators transform mechanical energy (the Power of motion) from an external source into electrical energy.
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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
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Job Assessment:
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In summary, while there are challenges, it's possible to secure a ground staff position if your visual impairments can be managed effectively and do not compromise safety or job performance. Consult with potential employers about the specific requirements and consider roles that align with your abilities.
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