What is the RMS of this current?

In summary, the author found the average current over a period of 2 seconds by calculating the root-mean-square of the two signals that contributed to that average.
  • #1
yosimba2000
206
9

Homework Statement


Find the average current. The problem and given solution are in Untitled.png.

Homework Equations


RMS = sqrt((1/T)(integral of x^2 dt from 0 to T))

The Attempt at a Solution


From time 0 to 1, the equation describing current is 10t^2.

From time 1 to 2, the equation describing current is 0.

So RMS = sqrt((1/T)integral(10t^2 dt from 0 to 1)) + sqrt((1/T)integral(0 dt from 1 to 2)), where I use T = 1 for both T's because the 10t^2 occurs over 1 second, and the 0 amps also occurs over 1 second.

But I get the wrong answer...

Where did the book answer get 1/2 for the coefficient?
 

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  • #2
yosimba2000 said:
So RMS = sqrt((1/T)integral(10t^2 dt from 0 to 1)) + sqrt((1/T)integral(0 dt from 1 to 2)), where I use T = 1 for both T's because the 10t^2 occurs over 1 second, and the 0 amps also occurs over 1 second.
The first integral gives a result, say X. The second integral gives the result 0.

Adding these results gives X+0 = X, but you have integrated (totally) from 0 to 2, so to find the RMS-value ( root-mean-square ), you must divide by T, where T= 2-0 = 2.
Hence RMS = 1/2 * X.

The mean value is found by dividing by the time for a period, which here is 2s.
 
Last edited:
  • #3
Hesch said:
The first integral gives a result, say X. The second integral gives the result 0.

Adding these results gives X+0 = X, but you have integrated (totally) from 0 to 2, so to find the RMS-value ( root-mean-square ), you must divide by T, where T= 2-0 = 2.
Hence RMS = 1/2 * X.

So if I understand this correctly, I first found the RMS of function 10t^2 from 0-1 seconds, then the RMS of function 0 amps from 1-2.

And then finally I average both RMS values? So it's kind of like averaging the averages (RMS's)?
 
  • #4
To calculate the average over the whole period you should be dividing by the full period. It makes no difference whether you first sum the results of both integrations then divide by 2T, or whether you divide individual integrals by 2T before summing them, but the dividend needs to be 2T.

Suppose the signal2 was 8 for 0<t<1 and 0 for 1<t<2, you wouldn't claim its average over the whole 2 secs to still be 8, but that's the answer your maths would be giving.

You have a second mistake. The sqrt must be performed as the last operation, on the mean of the squares.
 
  • #5
#3: I would formulate it: . . . . .

Well, let NascentOxygen explain.

His english is better than my english.
 
  • #6
Ok, I was confused because I saw that the period was 2 seconds, and I'm used to seeing a period used for one continuous function, not multiple.

As in, if you were treat the curve from 0-2 seconds as 10t^2, you would see it clearly cannot be since 10t^2 won't make that horizontal line. So the curve must be made of two separate functions.

So I thought the book was saying the period of 2 seconds is described by the function 10t^2 alone, hence me questioning the 1/2 when clearly 10t^2 is only over a period of 1 second, and the 0 amps over the next 1 second.

*edit*

Wait I can't average the two RMS's?
 
  • #7
yosimba2000 said:
So if I understand this correctly, I first found the RMS of function 10t^2 from 0-1 seconds, then the RMS of function 0 amps from 1-2.
That is what you did, and that's wrong.

And then finally I average both RMS values? So it's kind of like averaging the averages (RMS's)?
That is what you did, and that also is wrong.

I explained it in words earlier, but here is how you can write it using maths:

RMS = sqrt ( mean of ( sum of squares )))
 
  • #8
Hmm ok it still doesn't make sense to me.

Why do I divide by 2 for period when the 10t^2 is only over 1 second? Likewise with the 0 amps?

I guess I could follow the formula, but I'd like to know the intuition behind it.

Like if I were doing this on the number of rats or something, the period would be switched with the number of samples I have.

Here, isn't my sample 1 second? If I use 2, that implies the 10t^2 function is applied over 2 seconds... but it's only 1 second
 
  • #9
We are interested in knowing the contribution made to the whole period by each signal.

10t^2 existing for 1 sec contributes how much to our 2 sec average? It is best visualised by considering areas under the graph: how high will the rectangle be once you level out the components to show the average?
 
  • #10
Ah ok gotcha,

I started thinking of it in terms of animal samples and it made more sense.

Thanks everyone!
 

Related to What is the RMS of this current?

What is the RMS of this current?

The RMS (Root Mean Square) of a current is the effective or equivalent value of the current that produces the same amount of heat as a DC (direct current) of the same magnitude.

How is the RMS of a current calculated?

The RMS of a current is calculated by taking the square root of the average of the squared values of the current over a given time period.

Why is the RMS value important?

The RMS value is important because it is used to measure the heating effect of an alternating current and is used in calculations for power and energy in AC circuits.

What is the difference between RMS and peak current?

The peak current is the maximum value of the current, while the RMS current is the effective value that takes into account the changing magnitude of the current over time. The RMS value is typically lower than the peak value.

Can the RMS value of a current be negative?

No, the RMS value of a current cannot be negative as it is a measure of the magnitude of the current and does not take into account its direction. However, the instantaneous values of the current can be negative in an alternating current.

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