# Exam Help

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

1. ### Hello Thread Starter Active Member

Dec 18, 2008
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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.

• ###### DEN109.pdf
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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!

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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.