Numerical on Magnetic Circuits Hindi

Q. A Coil Of 500 turns resistance of 20 ohm is wound uniformly on an iron ring of mean circumference 50 cm. and cross- sectional area 4 cm2 . It is connected to 24 V d.c. supply, Relative permeability of the material = 800, Find (i) MMF (ii) Magnetising force (iii) Total flux (iv) Reluctance.

 Ans: Given : No. of turns (N) = 500 Resistance (R) = 20 Ω Mean circumference = (pd) =50 cm = 0.5 m Cross Sectional area (a) = 4 cm2 =4x10-4 m2 Applied voltage (v) = 24 V Relative permeability (µr) = 800 To find : I) MMF IN ii) Magnetising force (H) iii) Total flux (ϕ) iv) Reluctance (S) 

Soln. (i) mmf = IN I= V/R = 24/20 = 1.2 Amp

 mmf = 1.2 x 500 mmf = 600 AT

 (ii) Magnetising force H = mmf/l = 600/0.50 =1200 AT/ m

(iii) Total flux (ϕ) = mmf/s Mean circumference (pd) = l = 0.50 m

 Reluctance (S) = l/(µoµr a) = 0.50/(4p x 〖10〗^(-7) x 800 x 4 x 〖10〗^(-4) ) = 1.24x106 AT/ Wb

 (ϕ) = mmf/s = 1200 /1.24 x 106= 9.67 x 〖10〗^(-4) Webers = 0.967 x10-3 Wb = 0.967 mWb Total

flux (ϕ) = 0.967 mWb (iv) Reluctance (s) = 1.24x106 AT/ Wb

(i)  MMF = 600 AT

(ii) Magnetising force =1200 AT / m

(iii) Total flux = 0.967 mWb

 (iv) Reluctance = 1.24 x106 AT/ Wb