Class notes
Class notes EEB231
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Simplified notes to master dc generators,electromagnetic inductance,electric and magnetic circuits,coefficient of coupling all explained in the best possible way
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Analogy of Magnetic – Electric circuits• (a) In Electric circuits: Voltage forces current to go thru’ the resistor
• (b) In Magnetic circuits: mmf forces flux to go thru’ the reluctance
Series reluctance
• R total= R 1+R 2+R 3+….+
F = NI =ϕR total =Hl+Hairgaplairgap,
where, from: B= µH, H=NI/l
H can be determined in both cases( H = B/µ; Hairgap=B/𝜇0 )
,• Parallel magnetic circuit:
• (i)Parallel reluctance
1 1 1
• = R1 + + +….+
R total 1 R 2 R 3
• (ii) magnetic flux, Φ = R 𝐹
total
• (iii) Total mmf (At) required for the circuit= (mmf of central core)+ (mmf of one o
the outer legs)
• where: mmf (At) = 𝑁𝐼=H𝑙; H= B/𝜇= B/𝜇𝑟 𝜇0 ; B = ϕ/A
F = NI =ϕR total =Hl+Hairgaplairgap,
where, from: B= µH,
H can be determined in both cases( H = B/µ; Hairgap=B/𝜇0 )
, Example
• A ring has a diameter of 21 cm and a cross-sectional area
cm2. The ring is made up of semicircular sections of cas
and cast steel, with each joint having a reluctance equal
airgap of 0.2 mm. Find the ampere-turns required to prod
flux of 8x10-4 Wb. The relative permeabilities of cast stee
cast iron are 800 and 166 respectively. Neglect fringing
leakage effects.
• Solution:
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