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Power = current * voltage sometimes? 
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#1
Aug2705, 08:51 PM

PF Gold
P: 7,120

I thought power is current * voltage in all cases... but someone told me its only true "sometimes". Now the people on the forum that I heard this from are usually wrong all the time so i wanted to check it out.



#2
Aug2705, 10:42 PM

P: 282

I have heard the same thing, in certain circumstances (most notably with AC circuits) P = V(I) supposedly gives the wrong answer. As for why, that was never explained to me clearly and I was never given a direct example to see. This was mentioned to me as a side note when introducing the formula P = I^2(R) which apparently always gives the correct answer.



#3
Aug2705, 10:43 PM

PF Gold
P: 7,120

But arent they all derivations of eachother? I know theres something called RMS but i dont thinkt hats what there talking about.



#4
Aug2705, 10:55 PM

P: 282

Power = current * voltage sometimes?
I have always had a problem with this too. If you substitute R=V/I into the equation that "always works" you get:
P = I^2(V/I) Which simplifies to...P=I(V)! There are only two possibilities I can think of: one, that I was told the wrong thing and that you can use V(I) or two, that there is some sort of problem with measuring "voltage" in ac circuits and then applying it to this equation (maybe because the sign for Voltage changes direction in AC?). If you determine the resistance on say, a resistor connected to a DC circuit, and then remove that resistor and place it on an AC circuit I can see how using P=I^2(R) would be much easier to measure (you wouldn't have to deal with the problem of measuring voltage in this instance), but at the same time I am not sure what they mean when they say using V(I) will give you the wrong answer. 


#5
Aug2805, 02:02 AM

Sci Advisor
HW Helper
P: 3,144

I think the problem you are asking about has to do with the fact that Ohm's law is an approximation that can fail when the fields are too strong. For example, in an isotropic medium, the electrical current density is usually written in terms of the electric field as [itex]\vec j = \sigma \vec E[/itex] (for an isotropic medium) where [itex]\sigma[/itex] is the electrical conductivity. Notice that it is a linear relationship and is valid provided the electric field isn't too intense.
Generally, however, the conductivity can depend on the strength of the electric field such as in a conductor or plasma. It is a property of Coulomb collisions between electrons and ions (the source of resistivity!) that the faster electrons move the less effective the scattering becomes. The oscillatory velocity of electrons produced, for example, by an electromagnetic wave depends on the amplitude of that wave and, therefore, the conductivity also depends on the amplitude of the applied electric field. Similar effects occur during dielectric breakdown that occurs during an electrical discharge. 


#6
Aug2805, 05:43 AM

Mentor
P: 7,320

P=IE is directly related to Ohms law, and therefore only holds for purely resistive DC circiuts. In AC circiuts you must consider the reactive impedance. The impedance is dependent upon the frequency as well as the circiut resistance, inductance and capacatance. P=IE will also have a reactive component in AC circiuts, making it different from that of a pure resistive DC circiut.



#7
Aug2805, 07:17 AM

P: n/a

The instant power in an electric circuit is [tex]p(t) = v(t).i(t)[/tex].
This is allways valid. What happens with reactive circuits and sinusoidal voltages and currents is that the peak value of the power is different of the product of the peak values of voltage and current. [tex]P_{max} = V_{max} . I_{max} . cos(\phi)[/tex] Where [tex]\phi[/tex] is the phase angle between voltage and current. 


#8
Aug2805, 08:52 AM

Sci Advisor
PF Gold
P: 1,479

First, check: http://www.physicsforums.com/showthread.php?t=86053 Secondly, it depends of what kind of system are you analyzing. For instance, P isn't VI in triphase systems or in noncompletely resistive AC systems. 


#9
Aug2905, 11:28 AM

P: 1,295

P = IV [tex]\frac{energy}{time} = \frac{charge}{time} \frac{energy}{charge}[/tex] This is always true, anything else in an engineer's confusion. 


#10
Aug2905, 12:09 PM

Mentor
P: 41,477

The instantaneous values of V & I always satisfy P = IV. But if I and V represent the rms values in an AC circuit, then to get the average power you need to consider the phase difference between them: [itex]P_{ave} = I V \cos \phi[/itex], where [itex]\phi[/itex] is the phase angle by which the voltage leads the current.
I just realized that SGT already said this same thing! D'oh! 


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