Power Dissipation in a Resistor: Calculate Average

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To calculate the average power dissipated in a resistor with a current of i=8sin(ωt) and a voltage of v=200sin(ωt), the correct approach involves using the root-mean-square (RMS) values of voltage and current. The average power can be found using the formula Paverage = Vrms * Irms, where Vrms and Irms are the RMS values of voltage and current, respectively. The RMS value for a sinusoidal function is calculated as the peak value divided by √2. Alternatively, the average value of the power function P(t) = 1600sin²(ωt) can be determined by integrating it over one complete cycle and dividing by the period. The expected answer for average power dissipation is 800W.
dortec
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A resistor draws a current i=8sinwt at a voltage of v=200sinwt. Calculate the average power dissipated in the resistor.

What i did is p=ui = 1600sin^2 (wt) and i got stuck:P i don't think it's the right equation.. the answer should be 800W .. and that's nothing like it.
Could someone help me ?
 
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Hi dortec! :smile:

(have an omega: ω and try using the X2 tag just above the Reply box :wink:)

Hint: for a resistor, the power factor, cosφ, is 1, and so Paverage = VrmsIrms :wink:
 
Can you rephrase what you just said please. i actually didnt understand. wat's rms? and what do u mean by have an omega.. can u re-explain please:S
 
Hi dortec! :smile:

rms means "root-mean-square" … do you know what that is?

(if you don't, you can get the same result by finding the average value of sin2ωt, which you get by integrating it from 0 to 2π)

(and where you typed w, I assumed you would have preferred ω. :wink:)
 
nope i don't know what root-mean-square:S and my bad for the omega thing:P.. can u please show me how to solve it?
 
ok, never mind if you haven't done root-mean-square …

just find the average value of your original P(t) = 1600sin2ωt, which you get by integrating it from 0 to 2π/ω, and dividing by … ? :smile:

(alternatively, just write 1600sin2ωt in terms of cos2ωt and/or sin2ωt, and then the average is obvious :wink:)
 
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