Exam Help

Discussion in 'Physics' started by Hello, Dec 27, 2008.

  1. Hello

    Thread Starter Active Member

    Dec 18, 2008
    82
    0
    Can you please have look at the attached file.
    I am having difficulty with Q6 part (iv), not sure if i have the correct answer!

    I derived the expression from part (iii) and found; E1/N1 = E2/N2 = 4.44fϕ.
    I used the expression in part (iv) to get E2 = 1300 volts and
    Area = 6.00 x 10^-3 m^2.

    I used another method to check my answer,
    e.m.f = N. dϕ/dt = N.B.A.ω (where, ω = radian frequency)
    Found Area = 4.24 x 10^-3 m^2 !!!

    Any help is greatly appreciated.
     
    Last edited: Dec 27, 2008
  2. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    The area equals:

    A=V/(2*pi*f*N*B)

    V=peak value of voltage
    f=supply frequency
    N=turns number
    B=peak value of magnetic flux

    I find A=4.24 x 10^-3 m^2

    I think you made a mistake with your first equation.
     
  3. Hello

    Thread Starter Active Member

    Dec 18, 2008
    82
    0
    Can you please have a look at this attachment, pages 11 and 12.
    Thats the method i used, but when i compare it with the alternative method (the one you used) they give different answers!
     
  4. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    In the way your notes show you have to use the RMS value of the supply voltage whereas in the way I told you you have to use the peak value of the supply voltage. You don't know exactly if the problem gives you the RMS or peak value but usually is the RMS value.
     
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