Why take integral of V^2 when deriving Vrms?

[KingDaniel]

Homework Statement

Please explain why we take the "integral of V^2" when deriving the root mean square of Voltage in alternating current?
I do not understand why we integrate at all, instead of taking V as it is and squaring it.

Also, when we DO integrate, why do we take the integral of "V^2", i.e: int(V^2), instead of squaring the integral of V, i.e: [int(V)]^2 ?

Lastly, adding on to the integral issue, even in other formulas that we were taught in high school, like Work=Fx, in university we now have an "integral of F dx" = Work. I don't understand the function of the integral in such cases. I'm only comfortable with it being the area under a graph, and when dealing with velocity and acceleration...but other than those, I don't quite understand the function of the integral.

Homework Equations

In this case, u = V

3. The Attempt at a Solution

 

[SammyS]

Homework Statement

Please explain why we take the "integral of V^2" when deriving the root mean square of Voltage in alternating current?
I do not understand why we integrate at all, instead of taking V as it is and squaring it.

Also, when we DO integrate, why do we take the integral of "V^2", i.e: int(V^2), instead of squaring the integral of V, i.e: [int(V)]^2 ?

Lastly, adding on to the integral issue, even in other formulas that we were taught in high school, like Work=Fx, in university we now have an "integral of F dx" = Work. I don't understand the function of the integral in such cases. I'm only comfortable with it being the area under a graph, and when dealing with velocity and acceleration...but other than those, I don't quite understand the function of the integral.

Homework Equations

In this case, u = V

3. The Attempt at a Solution

What is the definition of $\ V_\text{RMS} \ ?$

 

[KingDaniel]

@SammyS , Vrms is the square root of the mean of V^2. So in this case it looks like V= int(V) dt ...so why did they just square the V instead of squaring the whole integral?

 

[SammyS]

@SammyS , Vrms is the square root of the mean of V^2. So in this case it looks like V= int(V) dt ...so why did they just square the V instead of squaring the whole integral?

It's not the square of the mean, it's the mean of the square.

You sum the square (that's what the integral does), then divide to get the mean of the square, then take the square root.

 

[KingDaniel]

@SammyS , that's what I said, the square root of the mean of V^2...(V^2 = square of V). Okay, so when we sum V^2, does that mean we're taking the sum of all the V^2 at different values of t? And that's why we divide by the period to get an average?

 

[SammyS]

@SammyS , that's what I said, the square root of the mean of V^2...(V^2 = square of V). Okay, so when we sum V^2, does that mean we're taking the sum of all the V^2 at different values of t? And that's why we divide by the period to get an average?

Yes.

 

[KingDaniel]

@SammyS , is this a correct way of thinking about this:
The reason we find Vrms is because the average value of V is zero, yet we know for sure that there is in fact a Voltage and Current, so it can't be zero? And that's why why use Vrms instead, and not the average of V, because Vrms makes the most sense?
If so, shouldn't the Vrms value be only SLIGHTLY greater than zero, and not MUCH greater?

 

[albuser]

@SammyS , is this a correct way of thinking about this:
The reason we find Vrms is because the average value of V is zero, yet we know for sure that there is in fact a Voltage and Current, so it can't be zero? And that's why why use Vrms instead, and not the average of V, because Vrms makes the most sense?
If so, shouldn't the Vrms value be only SLIGHTLY greater than zero, and not MUCH greater?

It's true that the average of V would be zero because the integral of v(t) from 0 to T is zero, but when you square the function and take the average of it it's no longer zero. Draw out v(t) and v(t)^2 and you'll see what I mean.

 

[SammyS]

@SammyS , is this a correct way of thinking about this:
The reason we find Vrms is because the average value of V is zero, yet we know for sure that there is in fact a Voltage and Current, so it can't be zero? And that's why why use Vrms instead, and not the average of V, because Vrms makes the most sense?
If so, shouldn't the Vrms value be only SLIGHTLY greater than zero, and not MUCH greater?

That's the basic idea.

Other forms of finding a non-zero meaningful measure could be used.

You can simply take the mean of the absolute value of the voltage (or current). For that matter, you can use the amplitude (peak value for positive peaks).

We're generally dealing with sinusoidal voltage and current so there is an advantage to the RMS idea.

If voltage and current are in phase, using V_RMS and I_RMS are very handy for finding average power dissipated.

In this case the instantaneous power dissipated will be non-negative at all times, being v(t)⋅i(t). The average power dissipated will simply be

P_average = V_RMS ⋅ I_RMS .​

 

[KingDaniel]

@albuser , I understand that. What I mean is don't we want a value that's close to zero? Because after taking the mean of v(t)^2 and taking the square root of that, if Vpeak is really high, we would get a Vrms value that's quite greater than zero. But isn't the reason that we use Vrms in the first place and not Vaverage because they are "similar"...since we squared V, took its mean then took the square root of that(to compensate for squaring it in the first place).

e.g: a random appliance has alternating current flowing through it and since it's alternating, we want to find the "average" voltage and current. Now, if we find the regular "average" we get zero. BUT we know for sure that there is a voltage and current flowing through the appliance. So we find a method to find the "average" voltage and current, and that's Vrms and Irms. So, my question is, shouldn't the Vrms and Irms be close to the regular average (zero)?
@SammyS , @lightgrav

 

[lightgrav]

Work should NEVER be taught as Fx ... an "x" value is a place, not a distance d=Δx.
So if the Force is not quite uniform at different places (graph F vs x to see), you need to use Work = F_avg Δx.
The Area under the curve F(x), is how you find the location-average of the Force.
(since speed is probably not constant, it is NOT the same as the time-averaged Force)

In a Resistive circuit, I = V/R , so the Power P(t) = V² /R ... you care most about the Energy (=$) = P_avg Δt

average V is exactly 0, so contains no information at all.
the square root is taken to get the units back into simply Volt (instead of Volt² , whatever that means).

 

[albuser]

So, my question is, shouldn't the Vrms and Irms be close to the regular average (zero)?

Like Sammy said, when the voltage and current are in phase the average power will typically be much greater than zero, but the average of V and I are both zero, so we must come up with a way to find the power dissipated without using just the averages of V and I. Imagine two sine functions each with the same period representing V and I. If both are positive (or negative) at the same time their product will always be non-negative, thus the average of their product will be greater than zero.

 

[SammyS]

@albuser ...

So, my question is, shouldn't the Vrms and Irms be close to the regular average (zero)?
@SammyS , @lightgrav

The square of the voltage is never a negative number, so how can averaging it result in something close to zero?

 

[KingDaniel]

@SammyS, I didn't say it squaring would give a negative number. It could be close to zero and be positive, i.e: slightly above the t-axis on a voltage-time graph.
I'm basically asking if there's a more accurate method than using Vrms, as its value is far from the regular "average" of zero, ie: it's far above the t-axis

 

[lightgrav]

more like, the rms should be close to the "alternative", the average absolute value.
standard North American grid has V_peak = 162V ... so V_peak² = 26 440 V² .
only for the small fraction of a cycle that V ≤ 1V does the squaring procedure reduce the numerical value (of course, it changes the units, too)
If you average the 26 440 V² with the 0 [V²], you get 13 220 V² ... which implies an RMS of 115 V.

This compares "close" that to the average absolute value, right?

 

[SammyS]

@SammyS, I didn't say it squaring would give a negative number. It could be close to zero and be positive, i.e: slightly above the t-axis on a voltage-time graph.
I'm basically asking if there's a more accurate method than using Vrms, as its value is far from the regular "average" of zero, ie: it's far above the t-axis

I didn't say you said that it would give a negative.

The only way to get a small number when summing a lot of values, many of which are relatively large, is for there to be a sufficient number of negative numbers to cancel those positive numbers.

In fact, $\displaystyle \ \left(V_0\sin(\omega t) \right)^2=\frac{V_0^2}{2}\left( 1-\cos(2\omega t) \right)\ .$

The mean value for that (over integer multiples of the period) is simply (V₀²)/2 .

 

[KingDaniel]

Okay, thanks. So, basically, we use the rms values because it just so happens that the energy of the waveform is the same as the energy of a waveform of a DC current with Voltage equal to Vrms?

 

[SammyS]

Okay, thanks. So, basically, we use the rms values because it just so happens that the energy of the waveform is the same as the energy of a waveform of a DC current with Voltage equal to Vrms?

That's the idea. Specifically:

For DC, the power is given by $\displaystyle \ P=V\cdot I=\frac{V^2}{R}=I^2R\ .$

For a sinusoidal wave form, the average power is given by similar relations if you use the RMS values.

$\displaystyle \ P=V_\text{rms}\cdot I_\text{rms}=\frac{(V_\text{rms})^2}{R}=(I_\text{rms})^2R\$​

 

[CWatters]

The integral of a curve gives you the area under a curve.

If you plot three curves.. V, the absolute value of V, and Vrms you can see visually that Vrms is the area under the curve. The shaded areas are the same..

 

[vela]

@SammyS , Vrms is the square root of the mean of V^2. So in this case it looks like V= int(V) dt ...so why did they just square the V instead of squaring the whole integral?

It sounds like you're saying
$$\sqrt{\int V^2\,dt} = \int \sqrt{V^2}\,dt = \int V\,dt.$$ This would be akin to saying $\sqrt{a^2+b^2} = a+b$, which, I hope, you know is incorrect.