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Alternating currents question

  1. Mar 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Two alternating currents i1 and i2 flow into a circuit node,the output current i is given by i1 + i2.

    Calculate the amplitude of i and the first time it occurs.

    i1=5sin(50t +Pi/3)

    i2=6cos50t

    2. Relevant equations


    3. The attempt at a solution
    5 sin ( 50t +60˚)
    = 5 ( sin 50t cos60˚ + sin60˚cos50t )
    = 2.5 sin 50t + 4.33 cos 50t

    2.5 sin 50t + 4.33 cos 50t + 6 cos 50t
    i1+i2 = 2.5 sin 50t +10.33 cos 50t

    R² = 2.5²+10.33²
    R = √113
    = 10.6

    arctan 10.33 / 2.5 = 33.8˚

    so i = 10.6 and occurs when 50t = 33.8˚

    Does this look right?
     
  2. jcsd
  3. Mar 27, 2016 #2

    cnh1995

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    Homework Helper

    This should be more than 45°. Peak current looks correct to me.
     
  4. Mar 27, 2016 #3
    Hi Any ideas what I should do to correct the value?
     
  5. Mar 27, 2016 #4

    cnh1995

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    tan-1 (10.33/2.5)= tan-1 4.13 which is clearly more than 45°(tan 45=1). Calculate the actual value using a calculator.
     
  6. Mar 27, 2016 #5
    76.38°?
     
  7. Mar 27, 2016 #6

    gneill

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    Staff: Mentor

    Your current function isn't a phasor (no imaginary component), so your angle computation is not correct.

    Call your current function f(θ). How would you normally go about finding a minimum or maximum for a function f(θ) with respect to θ?
     
  8. Mar 27, 2016 #7
    Ok so do I need to be looking into something like the following?

    "The result of adding 2 phasors can be summarised below."
    a∠α + b∠β = c∠g
    c = √{ [a cos(a) + b cos(b) ]^2 + [a sin(a) + b sin(b) ]^2}
    g = arctan{ [a sin(a) + b sin(b) ]/ [a cos(a) + b cos(b)] }

    With this in mind would this solution work?

    a angle a + b angle b = c angle g
    c = sqrt{ [a cos(a) + b cos(b) ]2 + [a sin(a) + b sin(b) ]2}
    g = atan{ [a sin(a) + b sin(b) ]/ [a cos(a) + b cos(b)] }

    cos 50t = sin (50t + pi/2)

    a = 5
    b = 6
    angle a = pi/3
    angle b = pi/2

    Thanks
     
  9. Mar 27, 2016 #8

    gneill

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    Staff: Mentor

    No, your current is given as a function in the time domain. It's not a phasor.

    You did fine reducing it to the expression:

    I(t) = 2.5 sin 50t +10.33 cos 50t

    How might you go about finding a maximum for that function?
     
  10. Mar 27, 2016 #9
    From my notes it states that:
    "It is often useful to be able to quote the maximum value of a sinusoidal waveform and the phase angle at which this maximum occurs. This involves only a simple deduction provided the waveform is expressed in one of the now familiar forms:"
    Rsin(x+α)
    or
    Rcos(x+α)

    The first maximum on the sinθ curve occurs at θ=90°, therefore x+α=90°
    The first maximum on the cosθ curve occurs at θ=0° therefore x+α=0°

    Am I going down the right lines?
     
  11. Mar 27, 2016 #10

    gneill

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    If you want to go that route then recognize that you have an expression of the form:

    a⋅sin(θ) + b⋅cos(θ)

    where θ = 50 t ; a = 5/2 ; b = 10.33

    You need to find a way to write this as a function of the form you stated, either Rsin(θ+α) or Rcos(θ+α). That will require using an appropriate trig identity and a bit of a trick with the constants a and b to turn them into sines and cosines of another angle (the phase angle). Do you know that trick?

    What I was hinting at earlier was that you could differentiate the current function, set it to zero and solve for the angle. This will yield the angle of the first maximum (or the first minimum, you should check with either the second derivative test or just plug the angle into the original function to see if the result is positive or negative).
     
  12. Mar 27, 2016 #11
    "What I was hinting at earlier was that you could differentiate the current function, set it to zero and solve for the angle"

    Any further clues you could give me to start doing this?

    Thanks
     
  13. Mar 27, 2016 #12

    gneill

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    Staff: Mentor

    Do you know how to differentiate a function?
     
  14. Mar 27, 2016 #13
    I have limited experience differentiating but Ive had a go.

    Do you mean by the current function:

    I(t) = 2.5 sin 50t +10.33 cos 50t

    i.e

    d/dt(2.5 sin (50t)+10.33cos (50t)

    Answer

    =125cos(50t)-1033 sin(50t)\2

    I do not know what you mean by set it to zero and solve for the angle

    Thanks
     
  15. Mar 27, 2016 #14

    gneill

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    In post #10 I wrote your current function as:

    I(t) = a⋅sin(θ) + b⋅cos(θ)

    where θ = 50 t ; a = 5/2 ; b = 10.33

    Differentiate a⋅sin(θ) + b⋅cos(θ). It's simple because the "50t" is removed from trig arguments and replaced with θ. This θ is the angle you're looking for, when the current reaches a maximum.

    To find a maximum or minimum you find the value of θ that makes the derivative zero.
     
  16. Mar 27, 2016 #15
    Is my calculation to find
    Thanks I will have a crack at this tomorrow. Is my calculation for the amplitude correct?
     
  17. Mar 27, 2016 #16

    gneill

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    Staff: Mentor

    Yes, your amplitude is fine. Your derivative is okay, too. But it's probably easier to leave it in symbolic form with a, b, and θ, solve for the angle in terms of those constants, then plug in the numbers at the end. Thus:

    ##\frac{d}{dθ}(a~ sin(θ) + b~ cos(θ)) = a~ cos(θ) - b~ sin(θ) = 0##

    Solve for θ.
     
  18. Mar 27, 2016 #17

    Charles Link

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    Homework Helper

    If I may add a constructive comment, there is a trick every engineer should really know when they encounter the form ## y=A \cos(\omega t)+B \sin(\omega t) ## It looked like you were heading toward that route in your OP but never got there. And the trick is to factor out sqrt(A^2+B^2) and let A/sqrt(A^2+B^2)=## \cos(\phi) ## and B/sqrt(A^2+B^2)=## \sin(\phi) ## Then y=sqrt(A^2+B^2) ## \cos(\omega t-\phi) ## I first saw this trick in a mechanics text in my sophomore year in college and it has come in handy countless times. (Note that ## \tan(\phi)=B/A ##)
     
  19. Mar 27, 2016 #18

    gneill

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    That is the trick that I was hinting at in post #10.
     
  20. Mar 27, 2016 #19

    NascentOxygen

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    Your very first move when faced with this problem should have been to sketch the curves so you can see exactly what you are dealing with.

    Get to know wolframalpha and it will turn drudgery into fun! :smile:
    http://m.wolframalpha.com/input/?i=plot+5sin(50t+Pi/3)+6cos(50t)+for+0<t<0.01&incCompTime=true

    You can see their sum starts (i.e., when t=0) at an angle that's a little less than the peak, so a little less than 90°.
     
  21. Mar 28, 2016 #20
    Could the problem be looking for an actual time value rather than an angle? i.e in the graph the amplitude 10.6 occurs around 0.005. On top of this using the calculation 2.5sin(50) + 10.33cos(50) I got 8.55. Arctan8.55 = 83.32° which is a little less than the 90° you stated or I am swinging and missing again?
     
    Last edited by a moderator: Mar 28, 2016
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