# Homework Help: Why take integral of V^2 when deriving Vrms?

1. Jul 23, 2015

### KingDaniel

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

In this case, u = V

3. The attempt at a solution

2. Jul 23, 2015

### SammyS

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

3. Jul 24, 2015

### 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?

4. Jul 24, 2015

### SammyS

Staff Emeritus
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.

5. Jul 24, 2015

### 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?

6. Jul 24, 2015

### SammyS

Staff Emeritus
Yes.

7. Jul 24, 2015

### 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?

Last edited: Jul 24, 2015
8. Jul 24, 2015

### albuser

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.

9. Jul 24, 2015

### SammyS

Staff Emeritus
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 VRMS and IRMS 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
Paverage = VRMS ⋅ IRMS .​

10. Jul 24, 2015

### 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

Last edited: Jul 24, 2015
11. Jul 24, 2015

### 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 = Favg 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 (=\$) = Pavg 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).

12. Jul 24, 2015

### albuser

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.

Last edited: Jul 24, 2015
13. Jul 24, 2015

### SammyS

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

14. Jul 24, 2015

### 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

Last edited: Jul 24, 2015
15. Jul 24, 2015

### lightgrav

more like, the rms should be close to the "alternative", the average absolute value.
standard North American grid has Vpeak = 162V ... so Vpeak2 = 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?

16. Jul 24, 2015

### SammyS

Staff Emeritus
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 .

17. Jul 25, 2015

### 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???

18. Jul 25, 2015

### SammyS

Staff Emeritus
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\$​

19. Jul 26, 2015

### 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..

20. Jul 26, 2015

### vela

Staff Emeritus
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.