Question

A laminated soft iron ring of relative permeability 1000 has a mean circumference of 800 mm and
a cross-sectional area 500 mm2. A radial air-gap of 1 mm width is cut in the ring which is wound
with 1000 turns. Calculate the current required to produce an air-gap flux of 0.5 mWb

Answer 1

Answer:

A radial air gap of 1mm width is cut in the ring which is. would with 1000 turns. Calculate the current required to produce an airgap flux of 0.5 mwb if leakage factor is 1.2 and stacking factor is 0.9. Neglect ... A mild steel ring of 30 cm mean circumference has a cross-sectional area of 6cm2 and has a winding of 500 turns on it.

Explanation:

Answer 2

The given question is not complete. Refer to the question given below:

A laminated soft iron ring of relative permeability of 1000 has a mean circumference of 800 mm and a cross-sectional area of 500 mm2. A radial air gap of 1 mm width is cut in the ring wound with 1000 turns. Calculate the current required to produce an air-gap flux of 0.5 mWb if the leakage factor is 1.2 and the stacking factor is 0.9. Neglect fringing.

Concept:

Current can be determined by the formula- I = Total AT required / Relative Permeability where I is the current

Given:

Relative permeability = 1000

Mean circumference = 800mm

Cross-sectional area = 500mm²

Radial air-gap = 1mm

Turns = 1000

Air-gap flux = 0.5  mWb

Find:

We need to determine the current required to produce an air-gap flux.

Solution:

Total AT required = ФgSg + Ф$_{i}$S$_{i}$ = Фglg/μ₀Ag + Ф$_{i}$l$_{i}$/μ₀μrA$_{i}$B

It has given to us that, air-gap flux, Фs = 0.5mWb = 0.5 ×10⁻³ Wb, lg = 1mm = 1 × 10⁻³m, Ag = 500 mm² = 500 × 10⁻⁶ m²

Flux in the ring is given by the equation, Ф$_{i}$ = 1.2 × 0.5 × 10⁻³ Wb

We can calculate Net-Cross sectional area as Area = A$_{i}$ × Stacking factor = 500 × 10⁻⁶ × 0.9 m²

Therefore, total AT required = 0.5 × 10⁻³ × 1 × 10⁻³/4π × 10⁻⁷ × 500 × 10⁻⁶ + 1.2 × 0.5 × 10⁻³ × 800 × 10⁻³/4π × 10⁻⁷ × 1000 × (0.9 × 500 × 10⁻⁶ )

= 1644

Therefore, I = Total AT required / Relative Permeability

I = 1644/1000

I = 1.64 A

Thus, the current required to produce an air-gap flux of 0.5 mWb is 1.64A.

#SPJ2