Question #161612

An iron ring has a mean diameter of 15 cm, a cross-section of 20 cm ^ 2 and a radial air gap of 0.5 mm cut on it. The ring is uniformly wound with 1500 turns of insulated wire and a magnetizing current of 1A produces a flax of 1 mWb in the air gap. Calculate: (i) the reluctance of the magnetic circuit and fii the relative normoabilit of t [5]

Expert's answer

Given quantities:

"d = 0.15m \\space \\space L = \\pi d = 0.4712m"

"A = 20*10^{-4}m^2"

"Lg = 0.5mm"

"N = 1500"

"I = 1A"

Solution:

a) the reluctance of the magnetic circuit

R = (0.4712/u * 20 * 10^{-4}) + (0.0005/u_{o} *20 *10^{-4}) ........ expression 1

Flux = MMF/R = 1500/R

R = 1500/(1*10^{-3})

= 1.5*10^{-6} A/wb expression 2

b) the relative permeability

Equating expression 1 and 2

1.5 *10^{-6} = (0.4712/u * 20 * 10^{-4}) + (0.0005/u_{o} *20 *10^{-4})

By simplifying above

ur = 143.9

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