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Problem regarding periodic current functions

  1. Jun 12, 2017 #1
    1. The problem statement, all variables and given/known data
    Three periodic currents have the same ##f=100 Hz##. The amplitude of the second current is ##4 A##. and is equal to half of the amplitude of the third current. Effective value of the third current is 5 times that of the first current. At time ##t_1=2ms## third current increases and changes signs. The first current reached it's amplitude half a period after the second current and the second current is "ahead" in phase to third current by ##\frac{2 \pi}{3}##. Write the functions for the current intensities of the currents.

    2. Relevant equations
    3. The attempt at a solution

    My course of alternating currents has just started and this is my first problem in the book.
    What i know:
    ##f = 100 Hz##
    ##Im_1 = 4A##, the amplitude denoted by ##Im##
    ##Im_3 = 8A##
    ##\frac{Im_3}{\sqrt 2} = 5*\frac{Im_1}{\sqrt 2} \Rightarrow Im_1 = \frac{8}{5} A##
    ##t_1 = 2ms##, moment when ##i_3## changes sign and start to increase.
    Looking for: ##i_1(t)=?, i_2(t)=?, i_3(t)=?##

    The work:
    The phase of the current ##i_3## is ##wt+ψ_3##. If at time ##t_1## the current changes sign than at that time the phase must be ##\frac{3\pi}{2}## since the canonical function of the current deals with a cosine function and the cosine increases at that point.
    ##wt_1+Ψ_3=\frac{3\pi}{2}##
    ##2\pi *100*2+Ψ_3=\frac{3\pi}{2}##
    ##ψ_3 = \frac{11\pi}{10}##
    This is where i disagree with the book. It says it should be ##ψ_3 = -\frac{9}{10} \pi##
    That answer i could obtain by subtracting ##2\pi## from my result but i understand the reason for such a thing. What is wrong here?
     
  2. jcsd
  3. Jun 12, 2017 #2

    ehild

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    Gold Member

    It must be a misprint: the time is 2 ms. ##2\pi *100*0.002+Ψ_3=\frac{3\pi}{2}##
    Nothing is wrong. cos(θ+2π)=cos(θ). Sometimes the phase constant is required between -pi and pi. Look in your notes.
     
  4. Jun 13, 2017 #3
    It doesn't state it specifically but i guess that's how they want it.
    I continued solving the problem and i found the phase of the second current with the fact that:
    ##ψ_2-ψ_3 = \frac{2}{3}\pi \Rightarrow ψ_2 = -\frac{7}{30}\pi##
    The phase of the first current was a bit of a problem. I knew only that the first current reaches it's amplitude half the period after the second and i didn't know how to use that. I helped myself with the book solution and was wondering if you could help me understand it. It goes:
    ##wt_2 + ψ_2 = 0##, i think this means that at time ##t_2## the second current reaches it's amplitude as the cosine is 0.
    ##wt_1+ψ_1 = 0##
    ##t_1 = t_2 + T/2##, the time ##t_1## is half a period later
    ##wt_2 + \frac{wT}{2} = 0##, i dont fully understand this part. Shouldn't this be ##wt_1##? And if so how can it be 0 when it's missing ##ψ_1## term?
    It continued
    ##ψ_1 - ψ_2 + \frac{wT}{2}= 0##, so ##wt_2 = ψ_1 - ψ_2##? Why is this so?
    ##ψ_1 - ψ_2 = \pi \Rightarrow ψ_1 = \frac{23}{30} \pi##
    These are the correct results but i don't understand the reasoning behind the first current's calculations.
     
  5. Jun 13, 2017 #4

    ehild

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    Homework Helper
    Gold Member

    There can be misprints and errors in any book. You are right. ##t_1 = t_2 + T/2##, substitute it into the equation
    ##wt_1+ψ_1 = 0## --> ##wt_2+wT/2+ψ_1= 0##.
    You know that ##wt_2 + ψ_2 = 0##, that is, ##wt_2 = - ψ_2 ##, so
    ##ψ_1- ψ_2+wT/2=0##, and wT/2=pi.
     
    Last edited: Jun 13, 2017
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