# What is the rms voltage of this waveform?

Discussion in 'Math' started by Hymie, Apr 13, 2018.

1. ### WBahn Moderator

Mar 31, 2012
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If you haven't been shown this stuff before, then there's no reason to kick yourself. It's a simple chain of events once you are aware of it's existence, but if you aren't aware of that it's a lot less obvious.

cmartinez likes this.
2. ### Hymie Thread Starter Member

Mar 30, 2018
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I was being facetious in my above response.

3. ### WBahn Moderator

Mar 31, 2012
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Okay. So which of those equations is an unreasonable leap from the prior one?

The important point is that they aren't a bunch of unrelated equations pulled out of thin air that magically come together to yield some obscure result. I have simply shown in explicit detail the application of the definition of the effective voltage of a waveform.

4. ### Hymie Thread Starter Member

Mar 30, 2018
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The whole point of my question was to highlight the fact that it is not immediately obvious (to me, and perhaps others), that the rms voltage of a half wave rectified waveform is the rms value of the sine wave/√2 (and not ½ of the rms sine wave voltage).

I personally doubt that fact would be apparent to students first studying ac waveforms – without recourse to complex calculations/equations shown within this thread.

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5. ### crutschow Expert

Mar 14, 2008
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Perhaps.
But using the simple observation that cutting out half the waveform obviously gives half the power, and that the RMS voltage is defined as representing the voltage's heating power, which is proportional to the square of the voltage, then it's very simple to calculate the RMS voltage value of the rectified sinewave as 1/√2 times the unrectified RMS voltage.
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6. ### Hymie Thread Starter Member

Mar 30, 2018
462
95
I agree that the simplest proof of the rms voltage (of a half wave rectified sine wave) is to analyse the power dissipation (as you posted earlier), which avoids the need for complex calculations/equations - but it needs a little thought to apply this.

7. ### WBahn Moderator

Mar 31, 2012
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If they are comfortable with just a couple of concepts, then they should be able to answer a question like this without recourse to anything complex at all.

First concept: Power dissipated in a resistor with a DC voltage applied across it is proportional to the square of the voltage.

P = k·(Vdc)²

Second concept: The RMS value of any voltage waveform is equal to the DC voltage value that would dissipate the same power in a resistive load.

P = k·(Vrms)²

Now, given the waveform you provided in this problem, it should be easy to reason out that it will deliver half the power compared to the sinusoidal waveform.

Therefore, since

P_sine = k (Vsine_rms)²

P_half = ½ P_sine = k (V_half)²

it should be readily discernible that

V_half = √( ½ (Vsine_rms)² )

V_half = Vsine_rms / √2

Now complex, memorized and regurgitated equations involved. Just an understanding of the relationship between power and voltage and of what rms voltage means. Which is why I suggested that you start by stating your understanding of these concepts and which you declined to do.

8. ### crutschow Expert

Mar 14, 2008
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That does stop a lot of people, I suppose.