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I've done a bit of research but can't find the explanation for this problem, at least mathematically.

I'm calculating W and Wrms of an audio amplifier by knowing output Vpeak and Load resistance. It's all pretty much theoretical so I'm assuming Rload is not inductive.

Having Vpeak = 80 V and Rload = 16 ohm

I would generally use the formula Wpeak = Vpeak x (Vpeak/Rload) which gives me Wpeak= 400W

I convert Vpeak to Vrms by multiplying 80 V x 0.707 = 56.56 Vrms

I would assume Wrms to be Wrms = Vrms x (Vrms/Rload) which gives me Wrms= 199.93 Wrms

now why if I multiply Wrms x 1.414 (which it's the constant to get the peak value from RMS value) it comes out as 282.7 W which is less than 400 W?

Which one is the right Wpeak?

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  • \$\begingroup\$ Because the voltage is squared, i.e., Vp x Vp, and 0.707... x 0.707... [Sq Rt of 2 times itself] = 2. BTW, this assumes sound is sinusoidal, which it is not (not much info in a continuous pure sine wave). \$\endgroup\$ Commented Jun 26, 2022 at 16:55
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    \$\begingroup\$ \$ \Large P = \frac{V_{peak}^2}{2 R_L} = \frac{80V^2}{2* 16\Omega} = 200W \$ \$\endgroup\$
    – G36
    Commented Jun 26, 2022 at 17:42
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    \$\begingroup\$ Rms watts do not exist, this is marketing BS. When you integrate the instantaneous power \$p(t)\$, you obtain an average power expressed in watts. And if you multiply rms currents and voltages when your amp is loaded by a resistance, then you also end up with an average power whose unit is watt. \$\endgroup\$ Commented Jun 26, 2022 at 18:44
  • \$\begingroup\$ Only voltage or current are ever given as RMS quantities; talking about RMS power is nonsensical. \$\endgroup\$
    – Hearth
    Commented Jun 26, 2022 at 18:56
  • \$\begingroup\$ What you refer to as rms power is actually average power and average power is half of peak power. Vrms * Irms = average power = (Vpk * Ipk)/2 = peak power/2 \$\endgroup\$
    – user173271
    Commented Jun 26, 2022 at 19:11

3 Answers 3

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Watts peak makes sense.

Watts RMS does not (because there is no such thing). You are calculating watts average.

I would generally use the formula \$ W_{peak} = V_{peak} \frac {V_{peak}}{R_{load}} \$ which gives me Wpeak= 400W.

That's fine.

I convert Vpeak to Vrms by multiplying 80v x 0.707 = 56.56Vrms

I would assume Wrms to be Wrms = Vrms x (Vrms/Rload) which gives me Wrms= 199.93Wrms

OK, except that that's average power, not RMS power.

Now why if I multiply Wrms x 1.414 (which it's the constant to get the peak value from RMS value) it comes out as 282.7 W which is less than 400 W?

If you multiply your average power, 200 W, by \$ \sqrt 2 \sqrt 2 \$ you'll get 400 W. The reason is that power is proportional to V2.

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    \$\begingroup\$ "The reason is that power is proportional to V2." That's what I needed! thanks. I now understand Wrms not being useful and a thing, I guess I was misled as working in audio you often see Watt RMS on equipment labels to describe the average power the speaker or amp operate at. Why wouldn't they just call it average power? \$\endgroup\$
    – Alex P
    Commented Jun 26, 2022 at 21:23
  • \$\begingroup\$ It's marketing technobabble to make things sound impressive. \$\endgroup\$
    – Transistor
    Commented Jun 26, 2022 at 21:28
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You can measure power at any instant such as peak or averaged over 1 cycle, never RMS power.

THe sine power appears as a sinusoidal power of 2f ranging from 0 to peak with a flat average of 50% of the peak with the product of V(t)I(t).

THe peak load for +/80V into 16 Ohms is 400 Wp and 200 Wavg .

Assuming 100% efficiency, the peak power generated is -400W and -200W average.

Demo

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The term ‘RMS’ applies to voltage and current. It does not apply to power. So that’s part of the problem.

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