Question

A 680-turn coil is wound on the central limb of a cast steel frame as shown in Fig.  with all dimensions in cms. A total flux of 1.6 mWb is required in the air-gap. Find the current in the magnetizing coil. Assume uniform flux distribution and no leakage. Data for B-H curve for cast steel is given.

Answer 1

φ = 1.6 mWb through air-gap and central limb 

φ/2 = 0.8 mWb through yokes 

Corresponding flux densities are : 

B_g = B_c = 1.6 mWb/(16 × 10⁻⁴) = 1.0 Wb/m²

By = 0.8 m Wb/(16 × 10⁻⁴ ) = 0.50 Wb/m² 

MMF-Calculations : 

(a) For Air gap : For B_g of 1 Wb/m² , H_g = 1.0/μ_o 

AT_g = Hg × l_g = [1/(4π × 10⁻⁷ )] × (0.1 × 10⁻² ) 

= 796 amp turns 

(b) For Central limb : AT_c = H_c × l_c = 900 × 0.24 = 216 

∴ For B_c = 1.00, H_c from data = 900 AT/m 

The yokes are working at a flux-density of 0.50 Wb/m² . From the given data and the corresponding plot, interpolation can be done for accuracy. 

H_y = 500 + [(0.5 − 0.45)/(0.775 − 0.45)] × 200 

= 530 AT/m 

F_y = 530 × 0.68 = 360 

Total mmf required = 796 + 216 + 360 = 1372 

Hence, coil-current = 1372/680 = 2.018 A