Question about AC, frequency, and watts.

no.

no.

sure, anytime

The average power dissipated is I^2R where I is the rms current and R is the real part of the load (assuming the load is complex ie with inductance). So if both rms currents are the same (which you said they were) and R is the same then the power is the same. Average power is also given by VIcos(phi) where V and I are both rms voltage and current and cos(phi) is the power factor ie phi is the angle bewteen V and I. This gives the same answer as previous.

However, if two ac voltages of the same rms value but different frequencies are applied to a load R+jwL

where R is resistance L is inductance and w is 2pif rads/s then the current for each case will be different.

rms current I=3DV/sqrt(R^2+(wL)^2) and hence I will be smaller if w is larger and hence the power will be smaller. So a larger freq dissipates smaller power for a given load. The power for both cases will be

P =3DR . V^2/(R^2+(wL)^2)

Hardy

It doesn't if you measuring what's called apparent power. But, with AC lines people usually think in terms of RMS power. It which case the frequency doesn't have anything to do with power throughput.

OK Greenxenon I strongly suggest you get a physics text book on E&M and read it. Do all the problems. And then come back and ask some questions. Volts times Amps is Watts. I don't care what the frequency. However if you want to calculate the number of photons in the 50 Hz signal you will find that there are more than in the 60 Hz signal. Same is true for any frequency or wavelength

George Herold

The current drawn is frequency dependent if the load is inductive (and resistive). So therefore the power is also frequency dependent. For a purely resistive load however you are right.

Hardy

Idiot

Solve the equations, moron.

Which is why real scientists and engineers spend their time working on perpetual motion machines, free energy, channeled schematic diagrams, UFO tracking, debunking relativity, time travel, ghost busting, aether theory, faster than light communications, warp drives, etc. ...

Idiot.

(Not because your answer is wrong, but because you responded to GreenXenon at all! )

Idiot

Do not respond to GreenXenon, moron.

Actually, it's 'average', not 'RMS'. RMS power is a useless quantity. RMS is applicable to voltage & current, but not power.

Paul Cardinale

Not to disagree with the point you're trying to make, but as an aside, RMS power can be used to describe temperature rise situations such as motor duty cycle.

dave y.

You are correct if your voltage is ripple free DC. However DC doesn't have frequency, so your answer doesn't apply. AC power really can not be explained as simply Volts X Amps (P=VI) due to the inductive, capacitive components that can be present in an AC circuit. You can use P=V(sq)/R or P=VIFp.