Question

Exercise 12 The magnetic circuit has A-12 cm2; -50m, 20000; N, - 500 turns, N-1000 tums First part The reluctance of the magnetic circuit is: R- The magnetic flux due to the currently is 4- NA R The flux linkages of the two coils are: NA A NA The self-inductance of collis 42,14 The mutual inductance is Second part Similar to first part Figure 25: Circuit for problem 12 (a) The first coil is supplied with a currenti - 10 A, whereas the second coil is lett un-cerised (open circuited) Calculate the self-inductance of coil 1 and the mutual inductance La, between the two coils. [1SH; 30H) (b) The first coil is de wergised, and the second coll is connected to a source from which it drws a current - 8 A. Calculate the self-inductance La of coil 2 and the mutual inductance La between the two coils. [60H:30H)

Part 2

Exercise 12 The magnetic circuit has A-12 cm2; -50m, 20000; N, - 500 turns, N-1000 tums First part The reluctance of the magnetic circuit is: R- The magnetic flux due to the currently is 4- NA R The flux linkages of the two coils are: NA A NA The self-inductance of collis 42,14 The mutual inductance is Second part Similar to first part Figure 25: Circuit for problem 12 (a) The first coil is supplied with a currenti - 10 A, whereas the second coil is lett un-cerised (open circuited) Calculate the self-inductance of coil 1 and the mutual inductance La, between the two coils. [1SH; 30H) (b) The first coil is de wergised, and the second coll is connected to a source from which it drws a current - 8 A. Calculate the self-inductance La of coil 2 and the mutual inductance La between the two coils. [60H:30H)

Please do both examples I will rate

Answer 1

1. Given: Area of the one turn of the coil(A_r) is = 12cm² = 12*10^-4m² ( 1cm = 10^-2m)

Thickness of the coil, t_r = 50 cm = 50*10^-2m

Relative permeability of the material of the core,$\mu$_r = 20000

We know that the permeability of vacuum/air is
HO
=
AII* 10
.

Then, the reluctance of magnetic material is ,

t, R, Ho *Hy * A

  

R. 50 * 10-2 411 * 10-7* 20000 * 12 * 10-4

  

50 * 105 R, 9611

   $R_{r} = \frac{50*10^{5}}{96*3.14 }$

$R_{r} = \frac{50*10^{5}}{301.44}$

   = 0.1658705*10⁵
   = 16587.05 H^-1

a) Magnetic flux due to the current i₁(10A),    ^{$\phi _{1}=\frac{N_{1}*i_{1}}{R_{r}}$}

  ^{$\phi _{1}=\frac{500*10}{16587.05}$}

= 0.30144HA

Flux linkage of coil 1,$\lambda _{11}=N_{1}\phi _{1}$

$\lambda _{11}=0.30144*500$

   = 150.72HA

Self-inductance of coil 1 is:$\L _{11}=\lambda _{11}/i_{1}$

$\L _{11}=150.72/10$

= 15.072H = 15H(approx.)   

Flux linkage of coil 2,$\lambda _{21}=N_{2}\phi _{1}$

$\lambda _{11}=0.30144*1000$

   = 301.44HA

Mutual-inductance of coil 2 is:$\L _{11}=\lambda _{21}/i_{1}$

$\L _{11}=301.44/10$

= 30.144H = 30H(approx.)

b)

Magnetic flux due to the current i₂(8A),    ^{$\phi _{2}=\frac{N_{2}*i_{2}}{R_{r}}$}

  ^{$\phi _{2}=\frac{1000*8}{16587.05}$}

= 0.482304HA

Flux linkage of coil 2,$\lambda _{22}=N_{2}\phi _{2}$

$\lambda _{22}=0.482304*1000$

   = 482.304HA

Self-inductance of coil 2 is:$\L _{22}=\lambda _{22}/i_{2}$

$\L _{22}= 482.304/8$

= 60.288H = 60H(approx.)

Flux linkage of coil 1,$\lambda _{12}=N_{1}\phi _{2}$

$\lambda _{12}=0.482304*500$

   = 241.152HA

Mutual-inductance of coil 1 is:$\L _{12}=\lambda _{12}/i_{2}$

$\L _{12}=241.152/8$

= 30.144H = 30H(approx.)

Please note: the subscript used in symbols for thickness, area and reluctance was not clearly visible, so r is used as subscript. Both the examples were same.