Question

Q2: 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 densityb) The number of turns on the high voltage winding​

Answer 1

Answer:

Cross sectional Area, A = 125cm2 = 125 * 10-4 m2

low voltage Windings turns,N2 = 70

Frequency f = 50Hz

  E2 = 4.44 m f N2 volt

250 = 4.44 * m * 50 *70

m = 0.01608 weber

As m = Bm * A,

i) Bm = m/A = 0.01608/(125 * 10 4) 1.2864 Wb/m2

ii) E1 = 4.44 m fN1 volt

N1 = 3300/(4.44 * 0.01608 * 50) 924.628 =925 turns

Or N1 = N2(E1/E2) = 924 turns

Explanation:

Answer 2

Given:

• Cross-sectional Area,(A) = $125cm^{2}$  = $125*10^{-4} m^{2}$
• Low Voltage winding turns, ($N_2$) = 70 turns.
• Frequency, (f) = 50Hz
• $E_1 = 3300 V$
• $E_2 = 250 V$

To Find:

• The value of maximum flux density(B)
• The number of turns on the high voltage winding($N_1$)

Solution:

• To find the flux density we are using the EMF equation.
• The EMF equation is given by, $E_2 =$ 4.44Фf$N_2$ volt.
• Substitute the values given and find the maximum flux density.
• 250 = 4.44Ф*50*70
• 250 = 15540Ф
• Ф = 0.01608 weber.
• As we know, Ф = B*A
• B = Ф/A = $\frac{0.01608}{125*10^{-4} }$ = 1.2864$Wb/m^{2}$
• Now we need to find $N_1$ for high voltage using the equation,
• $E_1 =$ 4.44Фf$N_1$ volt
• $N_1 = \frac{3300}{4.44*0.01608*50} = \frac{3300}{3.56976}$
• $N_1 = 924.43 = 925 turns$

The maximum Flux density is 1.2864$Wb/m^{2}$  

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