Question

Given a magnetic circuit in Fig. , with an air-gap and two windings wound on the limbs. Air gap hi Ni turns N2 turns At -Magnetic core permeability J mean core length le cross-sectional area Ac

Answer 1

u0 = 1.25e-6;

ur = 100:500:10000;

lc = 10;

g = 0.3;

Ac = 200;

Aa = 5;

N1 = 1000;

N2 = 500;

L = (u0*ur*N1^2*Ac/(lc-g))+(u0*N1^2*Aa/g);

M = (u0*ur*N1*N2*Ac/(lc-g))+(u0*N1*N2*Aa/g);

plot(ur,L,ur,M);title('Self Inductance and Mutual Inductance Vs Permeability');

xlabel('permeability');ylabel('Magnitude');

The plot is as follows