Question

A 3300/250V, 50Hz, single phase transformer is built on an iron core having an
effective cross sectional area of 125cm2 and 70 turns low voltage winding. Calculate:

a) The value of maximum flux density

b) The number of turns on the high voltage winding​

Answer 1

Given:

frequency N1 = 50Hz

cross sectional area a = 125 $cm^2$

low winding turns N2 = 70

To find:

1) flux density Bm =?

2) high winding turns N1 =?

Step-by-step explanation:

Electric flux density is the electric flux per unit area, which is a measure of the strength of the normal component of the electric field averaged over the area of integration​ .

we know that

$\displaystyle E_2 = 4.44 \phi_m N_2 \ volt$

3300 = 4.44 × φm ×  70 ×  250

⇒ φm = 0.01608 Wb

1)

As φm = Bm × A

⇒   $\displaystyle \ B_m = \frac{\phi_m}{A}$

            = $\displaystyle \frac{ 0.01608}{(125 \times 10^{-2})^2}$

Bm = 1.2864 Wb/$\mathbf{m^2}$

The value of maximum flux density 1.2864 Wb/$m^2$.

2)

As we know that

$\displaystyle E_1 = 4.44 \phi_m N_1 \ volt$

3300 = 4.44 × N1 × 0.01608 × 250

⇒ N1 = 925

The number of turns on the high voltage winding​ is 925.

Answer 2

Maximum flux density = 1.2864 WB/m2

Number of turns in higher voltage winding = 925