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sorry but who is harnwell and bleaney and bleaney?
i have tried to do some
of it but as you can see, this is why i got stuck.:
firstly, i tried to calculate the total reluctance using S=flux length/mu*A
= 5*10^-3/ 1200 * mu0 * 16*10^-3
=207.2 At/Wb
i then calculated the self inductance using L=N^2/S
=400^2/207.2
= 772.2 H
this is where i got confused and started e-mailing physics lecturers. i
noticed in my notes, for a similar shaped circuit as the one i have got, i
have writen beside it " series/paralell circuit, total S= centre+1/2 side
reluctances(assuming symetrical)." i did not understand what this meant and
have looked throught books but cannot seem to find an explanation.
if that means what i think it does then the total reluctance would be
(S=flux length/mu*A) + (2*S=flux length/mu*A)
=(2.5^-3/1200*mu0*16*10^-3) + (2*( 5*10^-3/ 1200*mu0*10*10^-3))
i am not sure however and genuinely cannot find anything in any books. i am
not lazy, looking for someone to do it for me. i am just confused as to what
to do.
today however, i had an idea after seeing a different question in a book.
what i did was :
S1= core length/Ur*U0*A = 0.025/(1200 *U0 * 0.16) =103.6 at/wb
S2= 2*(flux path lenght / Ur*U0*A) = 2* (0.05/ 1200 *U0 *0.1) =2* 331.57
at/wb
= 663.15 at/wb
therefore total reluctance = 663.15+766.75 at/wb
for S2, i multiplied this by 2 because i treated them as two seperate
shapes. you may not be able to see what i mean as you do not have a diagram,
which is with the question, and i am not allowed to send attachments.
then i worked out the self inductance by using the formula L= N^2/S, however
my answer for this depends on the answer for reluctance being correct.
paul