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Trouble with cos

  1. May 13, 2010 #1

    db1

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    hi all,

    its been a while since ive used a scientific calculator, anyway i am having trouble with cos. ive been askes to find the magnitude of the current flowing in the neutral conducter.

    V = 400
    Lv=231

    R1= 23
    R2=11.5
    R3=9.2

    I1=10.04
    I2=20
    I3=25.1

    my real trouble is using a scientific calculatern again, so if someone could just show me how you get your answer on the calculater it would realy be great thanks
     
  2. jcsd
  3. May 13, 2010 #2

    berkeman

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    Staff: Mentor

    (thread moved to homework help section)

    Welcome to the PF. Could you please define the terms you've listed, or post a diagram of the circuit? We can offer some hints for you if you can define the problem a bit more precisely.
     
  4. May 13, 2010 #3

    db1

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    well..
    the diagram shows 3 single phase resistive loads connected to a 400v 3 phase supply.
    Determine the magnitude of the current flowing in the neutral conductor by graphical means or by calculation.

    L1= 0º
    L2= -120º
    L3= -240º
    thanks for the help!
     
    Last edited: May 13, 2010
  5. May 13, 2010 #4

    berkeman

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    Staff: Mentor

    So the net current in the Neutral leg is the sum of the 3 sinusoidal currents. Each current depends on the resistance connected on that leg.

    Write the 3 equations for the 3 currents, based on the 3 voltage sinusoids (including their phase shifts that you list), and add them up.

    http://en.wikipedia.org/wiki/3-phase

    .
     
  6. May 13, 2010 #5

    db1

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    my problem is getting the value of cos and sin in the diagram bellow. i havent used a scientific calculater for a good few years. im fine calculating IN once i have cos and sin, im just not sure how to get the 10, -10 and -12.5 for cos and o, -17.32, and 21.65 for sin.
    thanks for your help on this
    ignor the 0.978 and 174 on the corner, thaqt was something else
     
    Last edited: May 13, 2010
  7. May 13, 2010 #6

    db1

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    anyone any ideas on this?
     
  8. May 13, 2010 #7

    berkeman

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    Staff: Mentor

    You need to break each phase down into its sin and cos components, in order to add them to get a single function of the form:

    f(t) = Acos(wt) + B sin(wt)

    Do you remember how to convert from polar to rectangular coordinates? So the voltage vector that is in phase with sin(wt + 120deg) can be expressed as a sum of zero-phase sin(wt) and cos(wt)?
     
  9. May 13, 2010 #8

    db1

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    hi, its breaking them down to get the sin and cos that im having trouble with, i just cant remember how to do it on a calculator, its been drivin me crazy all night. i know how to calculate IN once i get them to a single function. could you explain how i get it on a calculator please?
     
  10. May 13, 2010 #9

    berkeman

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    Staff: Mentor

    It has very little to do with the calculator. It has to do with adding vectors (or phasors). Look at this page on polar coordinates, especially the part about converting from polar to rectangular coordinates:

    http://en.wikipedia.org/wiki/Polar_coordinate_system

    That is what you have to do. You have a zero phase sinusoid:

    F1(t) = 10 sin(wt)

    And two other sinusoids with phase offsets (normally the phase shifts are taken counterclockwise from the +x axis, which is opposite of the way you have drawn them):

    F2(t) = 20 sin(wt + 120deg)

    F3(t) = 25 sin(wt + 240deg)

    In order to add them, you need to get them into rectangular form, so that you can add the components to get the final composite function. Draw the x and y components for each of F1-F3 (F1 is all in the direction of the x axis, which is why it has zero phase offset...). Use trig and the link I gave you, to figure out the components if F2 and F3, and add the components. If you need the final answer back in polar form:

    f(t) = A sin(wt + phi)

    then you user that same link to help you convert the polar form to rectangular form.
     
  11. May 13, 2010 #10

    berkeman

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    Staff: Mentor

    Last edited by a moderator: Apr 25, 2017
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