# Calculation of the magnetic circuit, Weber-Ampere characteristics

**Given**

W

_{1}=400;

W

_{2}=500;

W

_{3}=1000;

Steel: E11;

l=24 cm;

S=8 cm

^{2};

F

_{1}=2000 A;

F

_{2}=1000 A;

F

_{3}=0 A;

δ

_{3}=1 mm;

δ

_{1}=δ

_{2}=0 mm;

l

_{1}=1,5l;

l

_{2}=l;

l

_{3}=2l;

S

_{1}=S

_{3}=S;

S

_{2}=1,5S.

**Find**

Magnetic flux in branches: Φ

_{1}, Φ

_{2}, Φ

_{3}— ?

## Solution

We set the value of the magnetic inductance 0,5 T in the second branch. Then:

According to the magnetizing curve:

H=H

Field intensity in the air gap:

Magnetic voltages on the resistances gap:

According to the second Kirchhoff's law for magnetic circuits we write down the system of equations and solve it:

We make the same actions for other values of the magnetic inductance (the calculation is computerized in Excel)

According to the specified table of values we make Weber-Ampere characteristics of the circuit:

The diagram shows that fluxes are equal:

According to the magnetizing curve:

H=H

_{2}=f(Φ_{2})=1,7 A/cm;Field intensity in the air gap:

Magnetic voltages on the resistances gap:

According to the second Kirchhoff's law for magnetic circuits we write down the system of equations and solve it:

We make the same actions for other values of the magnetic inductance (the calculation is computerized in Excel)

According to the specified table of values we make Weber-Ampere characteristics of the circuit:

The diagram shows that fluxes are equal:

**Ответ:**