Homework4

# Homework4 - Homework#4 10.23.4 Problem 1 Randall Robert...

This preview shows pages 1–3. Sign up to view the full content.

Homework #4 Randall Robert 10.23.4 9771 Problem 1 An inductor has a magnetic core with relative permeability of μ=8000, cross-section A=25cm² and the length of the magnetic path l=50cm. The inductor is to operate at 240V RMS . Calculate the number of the coil turns N on the condition that the magnetic flux density is not higher than B=1.2[Wb/m²]. Calculate the inductance L of the inductor and its RMS current. φ SAT = 1.2[Wb/m²]*0.0025m² = 0.003Wb R g = 0 R c = l/( μ μ o A) =0.5/(8000* μo*0.0025) =19894[A/Wb] V = (ω/√(2))*N*A*B MAX 240=(377/√(2))N*0.0025*1.2 N=300 I SAT = (φ SAT /N)(R c +R g ) (.003[Wb]/300)(19894[Wb]+0) =0.19894A Z L = 240/ I SAT =j289.5KΩ

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 2 Repeat the problem 1, assuming that the core has an air gap of the width Δ=0.5mm. φ SAT = 1.2[Wb/m²]*0.0025m² φ SAT = 0.003Wb R g = Δ/( μ o A) = 5 E -4/ (μ o 0.0025) = 15955[Wb] R c = l/( μ μ o A) =0.5/(8000* μ o *0.0025) =19894[A/Wb] V = (ω/√(2))*N*A*B MAX 240=(377/√(2))N*0.0025*1.2 N=300 I SAT
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 3

Homework4 - Homework#4 10.23.4 Problem 1 Randall Robert...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online