Power = ½ V² / R ?

[alphacat]

I sometimes see in some places that power is defined as ½ V² / R.

Where does the ½ come from?

 

[Jony130]

Where did you see this equation

 

[birdman0_o]

That would be half the power

 

[Bob Scott]

I sometimes see in some places that power is defined as ½ V² / R.

Where does the ½ come from?

Searching my brain for experience, I can offer this:

Prms = Vrms² / R. If you are using the peak value of a sinusoidal voltage instead of the RMS value, then Prms = ½Vp² / R, or Prms = Vp² / 2*R. This way, you don't need to know or use the value of Pi.

I use the equation for calculating the maximum power of an audio amp using only the peak voltage and the load impedance.

For example, ignoring voltage drops through semiconductor junctions for now, how much RMS audio power can you get from 14.2V car battery connected to a simple car radio driving a 4 Ohm speaker?

Peak to peak voltage swing will be 14.2Vp-p. Peak output voltage will be half that , or 7.1V. Maximum undistorted RMS power output will be [½ * 7.1²/4]
or 7.1²/8 .

 

[dknguyen]

THe V in your equation is Vpeak for a sinusoid. The actual power formula is (Vrms^2)/2. The 1/2 is coming from the fact that Vrms = Vpeak/sqrt(2), and factoring the sqrt(2) outside of the squared.

 

[mbarazeen]

check that question, may be they have solved for half rectified voltage?? i mean to calculate the power after a series diode.

 

[Carl-NC]

There is no such thing as RMS power. Otherwise correct, Pave = ½ Vp² / R for sinusoids.

 

[Sceadwian]

What do you mean there is no such thing as RMS power? Why would you make such a blatantly incorrect statement?

 

[Jony130]

What do you mean there is no such thing as RMS power? Why would you make such a blatantly incorrect statement?

Well, But it's true, there is no such thing as RMS power.

We only know "true" real power in P=Watts [W] (this power do the work)
"Reactive' power Q = Volt-Amps-Reactive [VAR] and apparent power S=Volt-Amps [VA]

And by definition real power is integrated with respect to the period of the product of voltage multiplied by current (divided by the interval integral).
So for AC signal the "true" power is average power of V_peak*Ipeak.
Thus, "true" power is the average value of the instantaneous power,
and the terms real power, active power, and average power mean the same
thing.

 

[alphacat]

Thank you so much fellas!

It must be P = ½*(Vpeak)^2 / R.

I saw this equation in the course Waves And Transmission Lines, which indeed deals with peak values mostly.

Lots of thanks!

 

[Sceadwian]

Only for a perfect sine wave. Real world circuits are never that perfect. The first diode introduces non-linearities that break the calculations. You can use RMS calculations just as for voltage and current, and with power they're just as important, moreso with non-linear circuits, such as audio amplifiers. Where peak to peak power is completely useless effectively describing output power, RMS power is the real measure.

So now why are there TWO people that say there is no such thing as RMS power?

 

[Jony130]

Term "RMS" refers only to effective value of a AC voltage and current.
And we should not use the term "RMS" to description the real (true) power.
The true power is simple average value of the instantaneous power.
And for this we don't need roots and the square.
So RMS power is inaccurate term, that have no logical justification.

 

[Nigel Goodwin]

Term "RMS" refers only to effective value of a AC voltage and current.
And we should not use the term "RMS" to description the real (true) power.
The true power is simple average value of the instantaneous power.
And for this we don't need roots and the square.
So RMS power is inaccurate term, that have no logical justification.

It's got every justification, and allows you to actually use figures to explain and design things. Which your purely theoretical terms don't.