You're at least closer. How much total energy would be dumped into a 1 Ω resistor over the course of the waveform? What is the average power being dumped into at 1 Ω resistor? What DC voltage would dump that same average power into that same 1 Ω resistor?
Sir Correct me if i'm wrong power being dumped is p=v^2/r p=(9.7467)^2/1kiloohm 9.7467 Dc voltage would dump the same power I figured out the rms value to find dc equivalent voltage as average of ac is zero so i take the Rms by formula v=sqrt1/Tintegral(v)^2dt
This would be true only if Vrms = 9.7467 V, and it doesn't. You can't just assume that your answer is correct when the goal is to determine that your answer is correct. Don't cram everything together, the equation "v=sqrt1/Tintegral(v)^2dt" is almost unreadable. Assuming you mean v = sqrt( (1/T) integral((v^2)dt) ), then please SHOW how you used this to get 9.7467 V. We have absolutely no way to point out where you are going wrong as long as you refuse to show your work. That should not be a difficult concept to understand.
I agree with WBahn that you haven't shown your work very well, but I can see in the image attached to post #1 that your remaining problem is a simple arithmetic mistake. You are evaluating: when you should be evaluating:
So I wondered if it would work for a 1Ω resistor, would it also work for a 33Ω resistor? And indeed it does, but it would still be simpler to calculate the RMS voltage as ½√350.
Agreed. Suggesting that he use a specific resistor value was just intended to make the concepts extremely concrete. But the concept is that it works for ANY resistor value (including a non-specified symbol resistance R) and you can easily show that there is no need to even use that. It all falls out from the definition of "effective" power (not even "rms" power -- the "rms" is a direct consequence of the definition of "effective" power).