Rms

Is my answer alright

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WBahn

Joined Mar 31, 2012
24,969
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?
 

Thread Starter

Zulkifal

Joined Feb 5, 2016
18
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
 

WBahn

Joined Mar 31, 2012
24,969
Sir Correct me if i'm wrong
power being dumped is p=v^2/r p=(9.7467)^2/1kiloohm
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.

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
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.
 

The Electrician

Joined Oct 9, 2007
2,751
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:
RMS1.png

when you should be evaluating:
RMS2.png
 

RBR1317

Joined Nov 13, 2010
509
How much total energy would be dumped into a 1 Ω resistor over the course of the waveform?
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.

RMS-33-resistor.png
 

WBahn

Joined Mar 31, 2012
24,969
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.

View attachment 100314
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).
 
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