Graph, emf and flux linkage troubles.

Ok, here's the question. Got two coils, primary 180 turns, secondary 600 turns. There's a graph showing the emf across the secondary coil against time. It's a cos curve with amplitude 5 and a period of 40 microseconds.
Use the graph to estimate the amplitude of the flux linkage in the secondary coil?

Little help, not really got much in the way of working already.

Reply 1

Totally Tom

V1/v2=n1/n2?

Reply 2

ggonex26

Maybe faradays law. emf=flux linkage/time. Taking max emf and period.

Reply 3

*bobo*

emf=5cos((2pi/T)t)= - d(flux linkage)/dt (where T is 40 microseconds)
=> (flux linkage)= -(5T/2pi)sin((2pi/T)t), and go from there...

Reply 4

steelmole

*bobo*

emf=5cos((2pi/T)t)= - d(flux linkage)/dt (where T is 40 microseconds)
=> (flux linkage)= -(5T/2pi)sin((2pi/T)t), and go from there...

What's t in all that? Could you give a little more explanation please?

Edit: I think I get it enough. At least that method. In the mark scheme there is also something involving a quarter of the time period, giving an answer of 5x10^-5 webers in the end. Any idea on how they get that?

Reply 5

*bobo*

t is time.
flux linkage = - INT (5cos((2pi/T)t) dt = - (5T/(2pi))sin((2pi/T)t)
-1 </= sin((2pi/T)t) =/<1

So max. flux linkage= 5T/(2pi)= (5x40x10^(-6))/(2pi) Wb

Reply 6

steelmole

*bobo*

t is time.
flux linkage = - INT (5cos((2pi/T)t) dt = - (5T/(2pi))sin((2pi/T)t)
-1 </= sin((2pi/T)t) =/<1

So max. flux linkage= 5T/(2pi)= (5x40x10^(-6))/(2pi) Wb

Yeah, I see more now. That's a very confusing way to use slashes btw. Thanks.