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V=100sin(200pi t +pi/4)

  1. Jan 27, 2016 #1
    1. The problem statement, all variables and given/known data
    I have the last 3 parts of this question to answer and I'm struggling with it and would appreciate any advice please or if what i have done is correct thanks.

    2. Relevant equations


    3. The attempt at a solution
    Q1. The time to reach the greatest voltage rate of change

    A. dv/dt =(2 x 10^4pi) cos (200pi t + pi/4)
    maximum occurs when dv/dt = 0 and d^2v/dt^2 is negative
    d^2v/dt^2 = -(4 x 10^6 pi^2) sin (200pi t + pi/4)
    cos (200pi t + pi/4) = 0 when t = 0 and when t = 0 d^2v/dt^2 is negative

    When t = 0 v = 100 as sin(200pi t +pi/t = 1

    Q2. The value of the greatest voltage rate of change

    dv/dt is maximum when cos (200pi t +pi/4) is either 1 or -1
    which occurs when 200pi t + pi/4 = 0 or pi radians
    t is either = 1/800 or 1/800 seconds i.e. t = -0.00125 or 0.00125 seconds

    Q3. Using integral calculus, calculate the RMS value sketch a graph to visualise the process ?
     
  2. jcsd
  3. Jan 27, 2016 #2

    LCKurtz

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    When ##t=0## you get ##\cos(\frac \pi 4)\ne 0##.
    You meant ##\sin(200\pi t +\frac \pi 4)## and that isn't ##1## when ##t=0##.

    What is the integral you have to work?
     
  4. Jan 27, 2016 #3
    The questions i have got are the same as I've written calculus integration. Thanks
     
  5. Jan 27, 2016 #4

    LCKurtz

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    You haven't addressed the mistakes I pointed out. And you haven't shown us that you have at least looked up the integral you need to calculate, much less shown any effort at solving it.
     
  6. Jan 28, 2016 #5

    ChrisVer

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    what is the question?
    in both Q1 and Q2 you solve the same thing...(try to find the time of maximum)....in Q1 as it's pointed out you did it wrong, but in Q2 you got the right answer (if your voltage is as given).
    Also are you looking for the rate of changes or the voltage itself? if you are looking for the rates of changes then you need to maximize dV/dt and not V...
     
  7. Jan 28, 2016 #6
    Hi ChrisVer

    I have the volatge when t is at 0 to be 71v
    The voltage when t = 5 ms -71v

    The question 10.6
    The time to reach the gretaest voltage rate of change- note the greatest rate of change
    - note that the greatest rate of change is when dv/dt is a maximum, ie d^2v/dt^2 = 0

    10.7 The value of the greatest voltage rate of change

    10.8 using integral calculus calculate the RMS value of the voltage sketch graph to visualise process
     
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