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Find the RMS value of current I

  1. May 22, 2016 #1
    1. The problem statement, all variables and given/known data
    In the network of sinusoidal current, as shown in the figure, the current i2
    is phase delayed for the angle of 3π/4 behind the current i.
    In the moments in which the current i2 is minimal,
    current value of current i1 is sqrt(2) A. This value is two times lower than the maximum
    value of the current i1, and in those moments current i1
    and growing.
    Calculate the effective value of the current I.
    Captureelektrijada.jpg [/PLAIN] [Broken]
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    2. Relevant equations


    3. The attempt at a solution
    osnovi_elektrijada.png
    capture screen

    Is there any way to find the angle Φ2 since I know that in the moments when i2 is minimal value of the current i1 is sqrt(2)?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. May 22, 2016 #2

    rude man

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    Homework Helper
    Gold Member

    Did you ask what "effective current" means? I couldn't even guess.
     
  4. May 23, 2016 #3
    It's RMS value( Imax/ sqrt(2) ) .

    I tried to solve it this way now:
    i = Imax *sin(wt)
    i1 = I1max *sin(wt + ψ1)
    i2 = I2max *sin(wt+ 3*pi/4 )

    at time t = t1 when current i2 is minimal.
    -I2max = I2max * sin(w*t1 + 3*pi/4) -------> wt1 = -5*pi /4
    2*i(t1) = I1max ------> I1max = 2*sqrt(2)

    sqrt(2) = 2sqrt(2) * sin (w*t2 + ψ1) / arcsin
    ψ1 = 17*pi / 12

    which is incorrect the the correct phase delay of the current i1 is ψ1 = 45°
     
  5. May 23, 2016 #4

    rude man

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    What is A? It's not related to anything else in the problem.
    ???
    Totally confusing problem statement.
     
  6. May 23, 2016 #5

    NascentOxygen

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    Staff: Mentor

    By "is minimal" for a sinusoid should we understand it to indicate a zero-crossing or a negative peak?
     
  7. May 23, 2016 #6

    NascentOxygen

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    Staff: Mentor

    A is the abbreviation of "amperes".
     
  8. May 24, 2016 #7

    NascentOxygen

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    Staff: Mentor

    i2 is delayed, so phase should be negative.
     
  9. May 25, 2016 #8
    Sorry for the late replay guys,I had exam this week.
    I think that they mean for minimal as value when it is a negative peak of the current.
    Correct.
    Tried calculations now with negative sign of ψ2 = -3π/ 4 and the result that I got is ψ1 = -π / 12 , which gives ψ1 - ψ2 = 2π/3 as it is correct.Now need to figure out how to find RMS value of the current I.
     
    Last edited: May 25, 2016
  10. May 25, 2016 #9
    Got the right solution now:

    elektrijada_osnovi_sinusoida.png

    Their solution
    osnovi_elektrijada.jpg

    How did they get this expression ? It looks way simpler than my approach.
     
  11. May 25, 2016 #10

    NascentOxygen

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    Staff: Mentor

    Looks like they applied The Sine Rule. The 3 currents form a closed triangle.
     
  12. May 26, 2016 #11
    Yes you are right
    osnovi_elektrijada2.png

    Thank you for your help.
     
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