Let's break down the solution for both parts of the problem.
### Given data:
- Transformer rating: 3300 / 250V
- Frequency \( f = 50 \, \text{Hz} \)
- Cross-sectional area of the core \( A = 125 \, \text{cm}^2 = 0.0125 \, \text{m}^2 \) (since \( 1 \, \text{cm}^2 = 0.0001 \, \text{m}^2 \))
- Turns on the low voltage winding: \( N_L = 70 \)
- Low voltage \( V_L = 250 \, \text{V} \)
- High voltage \( V_H = 3300 \, \text{V} \)
### (i) Calculate the maximum flux density \( B_{max} \)
The formula for the induced EMF in a transformer winding is given by:
\[
E = 4.44 \times f \times N \times \Phi_{max}
\]
Where:
- \( E \) is the RMS voltage,
- \( f \) is the frequency,
- \( N \) is the number of turns,
- \( \Phi_{max} \) is the maximum flux in the core.
Rearranging for \( \Phi_{max} \):
\[
\Phi_{max} = \frac{E}{4.44 \times f \times N}
\]
For the low voltage side:
- \( E_L = 250 \, \text{V} \)
- \( N_L = 70 \, \text{turns} \)
- \( f = 50 \, \text{Hz} \)
Substitute these into the formula:
\[
\Phi_{max} = \frac{250}{4.44 \times 50 \times 70} = \frac{250}{15540} = 0.0161 \, \text{Wb}
\]
Now, the flux density \( B_{max} \) is given by:
\[
B_{max} = \frac{\Phi_{max}}{A}
\]
Substituting the values of \( \Phi_{max} \) and \( A \):
\[
B_{max} = \frac{0.0161}{0.0125} = 1.29 \, \text{T} \, \text{(Tesla)}
\]
So, the maximum flux density is \( 1.29 \, \text{T} \).
### (ii) Calculate the number of turns on the high voltage winding \( N_H \)
The voltage ratio of a transformer is directly proportional to the turns ratio:
\[
\frac{V_H}{V_L} = \frac{N_H}{N_L}
\]
Rearranging to find \( N_H \):
\[
N_H = \frac{V_H}{V_L} \times N_L = \frac{3300}{250} \times 70 = 924
\]
So, the number of turns on the high voltage winding is \( 924 \).
### Final answers:
1. The maximum flux density \( B_{max} = 1.29 \, \text{T} \).
2. The number of turns on the high voltage winding \( N_H = 924 \).