Q.1 An iron ring with 10 cm diameter and 15 cm2 in cross-sectionis bound with 25

Question

Q.1
An iron ring with 10 cm diameter and 15 cm2 in cross-sectionis bound with 250 turns of wire of flux density of 1.5 Wb/m2 and permeability 500. Findthe exciting current, inductance and stored energy. Find corresponding quantity whenthere is a 2 mm air-gap.
An iron ring with 10 cm diameter and 15 cm2 in cross-sectionis bound with 250 turns of wire of flux density of 1.5 Wb/m2 and permeability 500. Findthe exciting current, inductance and stored energy. Find corresponding quantity whenthere is a 2 mm air-gap. An iron ring with 10 cm diameter and 15 cm2 in cross-sectionis bound with 250 turns of wire of flux density of 1.5 Wb/m2 and permeability 500. Findthe exciting current, inductance and stored energy. Find corresponding quantity whenthere is a 2 mm air-gap.

Explanation / Answer

Given d = 10cm = 0.1m r = d/2 = 0.1 / 2 = 0.05m A = 15 cm2 = 0.0015 m2 N = 250 = 500 B = 1.5 Wb /m2 The flux density in the ring is given by B = Ni / 2r Then the exciting current will be i = 2rB / N i = (2)(3.14)(0.05) (1.5) / (500) (250) i = 3.768A The inductance is given by L = N2 A / 2r L = (500) (250)2 (0.0015) / (2)(3.14)(0.05) L =149283H Energy stored is E = Li2 / 2 E = (149283)(3.768)2 / 2 E = 1.059J The inductance is given by L = N2 A / 2r L = (500) (250)2 (0.0015) / (2)(3.14)(0.05) L =149283H Energy stored is E = Li2 / 2 E = (149283)(3.768)2 / 2 E = 1.059J

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