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Average value of a function and such.

  1. Mar 2, 2009 #1
    1. The problem statement, all variables and given/known data
    The voltage V in an electrical outlet is given as a function of time, t, by the function V = V0 cos(120πt), where V is in volts and t is in seconds, and V0 is a positive constant representing the maximum voltage.
    (a) What is the average value of the voltage over 3 seconds?
    (b) Engineers do not use the average voltage. They use the root mean square voltage defined by V = √avg of V^2. Find V in terms of V0. (Take the average over 3 seconds.)
    (c) The standard voltage in an American house is 110 volts, meaning that V = 110. What is V0?

    the V0 is really V_o. and that is pi up there.

    any help is appreciated. this is my set up:

    (V_o)/3 [sin(120 pi t)/(120pi)][0 to 3]
     
  2. jcsd
  3. Mar 2, 2009 #2
    (a) The average over 3 seconds can be found by taking the integral of V0 cos(120 PI t) from t=0 to t=3, and dividing your answer by 3. NOTE: you divide by the change in time, not by the change in argument to the cosine function. NOTE: this should come out to be very close to zero.

    (b) Same as (a), but now the function you're integrating from 0 to 3 seconds is:
    (V0)^2 cos^2(120 PI t)
    You may want to use the fact that cos(2x) = cos^2(x) - sin^2(x) = 2cos^2(x) - 1.
    Once you get the integral, then just divide by 3, and that's <V^2>.
    Then take the sqrt and you're done.

    (c) Well,

    110 = <V^2> = f(V0)

    is the equation you need to solve. You found f(V0) in part (b).
     
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