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Power in terms of work and energy 
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#1
Nov2112, 06:42 AM

P: 132

Is power a quantity defined by:
[tex]\frac{dU}{dt}[/tex] and [tex]\frac{dW}{dt}[/tex] Is it just defined to be it, or can it be derived in terms of the other (I mean, dU/dt in terms of dW/dt and vice versa)? I now there's physical motivation to it, but sometimes I just can't help trying to ponder how these equations came about, because textbooks sometimes won't explain explicitly the context about how they are being used. 


#2
Nov2112, 07:41 AM

P: 1,969

What are you asking?
Is it if the power is defined in terms of work or/and energy rate? Or if the rate of doing work is equal to the rate of change in energy (U)? 


#3
Nov2112, 07:57 AM

Mentor
P: 11,744

In general, it's often useful to discuss the rate at which some quantity is produced, consumed, or transferred.
Water flowing in a pipe: liters per second. Electric charge flowing in a wire, i.e. electric current: coulombs per second = amperes. Work done by a motor: joules per second = watts. 


#4
Nov2212, 07:23 AM

P: 132

Power in terms of work and energy
@nasu: I guess it's both, I want to know if dU/dt is equivalent to dW/dt in general,and what is the actual general definition of power?



#5
Nov2212, 07:34 AM

P: 1,969

See first law of thermodynamics, for example. In isothermal expansion of an ideal gas, dU=0 but dW is not zero. But it is not clear what exactly you have in mind when you write "U". As work results in some energy transfer, the rate of doing work is measured by power as well. 


#6
Nov2212, 07:40 AM

P: 132

Opps, yeah, I meant energy when I wrote 'U'. Anyways, I wonder if I can suggest that we look at it in the context of electromagnetism. Let's say I have a wire that has an increasing current in it, how exactly is the energy changing? Is it because of the changing potentials?



#7
Nov2212, 08:03 AM

P: 1,969

I don't really understand what your problem is. What are you trying to derive?
Power is defined as rate of energy transfer, change etc. Any time when you have some sort of energy changing, you can call the rate of change "power". It does not have to be the total energy or any specific sort of energy. If you want to focus on potential energy only, you can associate a power with it. 


#8
Nov2312, 09:08 AM

P: 132

I'm sorry if I sounded very ambiguous. But I think I got it now.
I'm so hung up with the equation [tex]W= \Delta U [/tex]. See, the derivation for the energy associated with capacitors goes somewhere like this: Q=CV and let the lower case v and q be the any finite charge and voltage during the charging process of a capacitor. Keeping in mind that dW/dq = v: [tex] dW=\frac{q dq}{C} [/tex] The rest of the derivation goes with integrating to the final work done, does an appropriate change of integration variables and limits and says that this is the amount of energy associated with the charged capacitor (with the energy = 0 when the capacitor is uncharged). What confused me is that, I tried to substitute W=(delta)U in one of the first steps of the derivation, but that does not give me the same answer; now, I think that the fatal mistake here is that I thought the work required to charge the capacitor is done by a conservative force (where I tried setting the initial potential energy at zero), but I think this is not the case (hence I could not use the equation). I think this is already hinted when the textbook said that the work required to charge the capacitor is equal to the work done by the electric field when it is discharging. Now, I may have only made the discussion messier, but I'd really appreciate anyone's insight on this. 


#9
Nov2312, 09:23 AM

P: 1,969

The work is equal with the negative change in PE only for conservative forces.
when the capacitor discharges some energy is converted into heat (resistive hating). Actually, if I remember correctly, half of the energy is lost, no matter what is the value of the resistor. This is of course, neglecting radiation. I think that this problem was discussed somewhere on the forum, in the past. At least once. 


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