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What is the RMS of this current?

  1. Jan 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the average current. The problem and given solution are in Untitled.png.

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

    3. 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?
     

    Attached Files:

  2. jcsd
  3. Jan 23, 2016 #2

    Hesch

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    Gold Member

    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: Jan 23, 2016
  4. Jan 23, 2016 #3
    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)?
     
  5. Jan 23, 2016 #4

    NascentOxygen

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    Staff: Mentor

    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.
     
  6. Jan 23, 2016 #5

    Hesch

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    Gold Member

    #3: I would formulate it: . . . . .

    Well, let NascentOxygen explain.

    His english is better than my english.
     
  7. Jan 23, 2016 #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?
     
  8. Jan 23, 2016 #7

    NascentOxygen

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    Staff: Mentor

    That is what you did, and that's wrong.

    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 )))
     
  9. Jan 23, 2016 #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
     
  10. Jan 23, 2016 #9

    NascentOxygen

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    Staff: Mentor

    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?
     
  11. Jan 23, 2016 #10
    Ah ok gotcha,

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

    Thanks everyone!
     
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