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At 03:14 PM 4/2/02, Paul Giusti wrote:length/mu*A
sorry but who is harnwell and bleaney and bleaney?
These are the authors of some texts used in American schools.
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
mm^2= 5*10^-3/ 1200 * mu0 * 16*10^-3
=207.2 At/Wb
You determined the flux length was 5 mm and the cross section area was 16
and mu0 was 1.257E-6 N/A^2 Your handling of scientific notation wasi
unusual.
You wrote 5*10^-3 for 50 mm instead of 5*10^-2.
But later, when expressing the numbers in decimal fractions, the numbers
looked better.
There you expressed the same length as 0.05
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,
andhave writen beside it " series/paralell circuit, total S= centre+1/2 side
reluctances(assuming symetrical)." i did not understand what this meant
amhave 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
whatnot lazy, looking for someone to do it for me. i am just confused as to
book.to do.
today however, i had an idea after seeing a different question in a
diagram,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
howeverwhich 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,
totalmy answer for this depends on the answer for reluctance being correct.
paul
The central difficulty you expressed was in visualizing reluctance,
analogous to working out circuit resistance but of a magnetic path. If
you had a resistor R1, whose ends were connected by two parallel side
resistors R2 each, you would not have much trouble in visualizing the
circuit resistance through the center limb as R1 + R2/2limb
But I should apologize to you. You sent me a diagram so that I could place
it in a web accessible position.
Unfortunately, this was a diagram embedded in a Microsoft Word 97 product,
which guaranteed I could not extract it to the web (unwilling as I am to
keep up the race to next years' products of the Evil Kingdom) It was by
all accounts a conventional shell transformer (i.e. one with a central
and two side limbs) and the only dimension not mentioned elsewhere was thereluctances
side limb cross section, which you gave above.
To summarize - refreshing on number representation, so you can quickly
check that
1.23E-3 = 1.23*10^-3 = 0.00123 might be helful, and the idea of
being treated like series and parallel resistances would not be hard to
keep in mind.
Do you want to persevere with the remaining parts of the problem? I am
finding your
attack helpful.
Brian W
Brian Whatcott
Altus OK Eureka!