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Pengwuino

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- Thread starter Pengwuino
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Pengwuino

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- #2

Renge Ishyo

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- #3

Pengwuino

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- #4

Renge Ishyo

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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

Tide

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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

Integral

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- #7

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

Clausius2

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Pengwuino said:

It depends of what kind of power do you want to measure:

First, check: https://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 tri-phase systems or in non-completely resistive AC systems.

- #9

Crosson

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P=IE is directly related to Ohms law, and therefore only holds for purely resistive DC circiuts.

Power is the rate of doing work, energy per time.

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

Doc Al

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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!*

Last edited:

- #11

Clausius2

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Crosson said:Power is the rate of doing work, energy per time.

P = IV

[tex]\frac{energy}{time} = \frac{charge}{time} \frac{energy}{charge}[/tex]

This is always true, anything else in an engineer's confusion.

Maybe did you mean "anything else IS an engineer's confusion"?

If so, read again SGT, mine and Doc Al post. If you keep on thinking I'm wrong, then anything else IS a god knows who the hell's confusion.

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