Do we get 2x the frequency when doing P=VI for AC Circuits?

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SUMMARY

The discussion centers on the relationship between power (P), voltage (V), and current (I) in AC circuits, specifically exploring the equation P = VI. It is established that while voltage and current share the same frequency (wt), the power frequency is effectively doubled, resulting in P = VIcos(2wt + psi). The conclusion is that power oscillates at twice the frequency of the voltage and current due to the nature of their product, which peaks at both positive and negative voltage peaks.

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yosimba2000
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I know that when you get a current from a voltage (like calculating current through resistor), the current and voltage equations have the same frequency. Does this still hold for power P = VI ?

So assume V in Euler form, V(t) = Vcos(wt+theta) + jVsin(wt+theta) = V*ej(wt+theta) and in Phasor, V<theta = Vcos(wt+theta)

And assume I in Euler form, I(t) = Icos(wt+phi) + jIsin(wt+phi) = I*ej(wt+phi) and in Phasor, I<phi = Icos(wt+phi)

Then P = V(t)*I(t) = V*ej(wt+theta) * I*ej(wt+phi)

To get P = VI*ejwt+jwt+jphi+jtheta, call jphi + jtheta = jpsi
Then P = VI *e2jwt+jpsi
Then P = VI *ej(2wt+psi)
Then P = VIcos(2wt+psi)+jVIsin(wt+psi)

Real component is what we want, so P = VIcos(2wt+psi)

2wt (power frequency) isn't the same as wt (the frequency of current and voltage). Did I get the right result?
 
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Think of the power as peaking on the positive voltage peak and the negative voltage peak, which is twice the frequency.
 
meBigGuy answered it

here's a picture to help

age=http%3A%2F%2Felectrical4u.com%2Felectrical%2Fwp-content%2Fuploads%2F2013%2F03%2Factive-power.png


any time volts and amps have same sign their product is positive
and that happens twice per line cycle

see - you learned that in first year algebra. Much of learning is really just discovering what we already know.
 
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That's the picture I was looking for!
 
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Not to hijack the thread but notice that there is now also a DC component. This is like AM modulation in that 2 signals are multiplied together to get sum and difference frequencies. In this case it so happens that the two input signals are the SAME frequencies.
 
The trig identity for squaring a sinewave (since power is E^2/R) shows the DC and the 2x frequency
sin2(x) = ½[1 – cos(2x)]
 
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meBigGuy said:
½[1 – cos(2x)]
sin^2x.jpg


I envy you who are fluent in math.
 

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