To calculate the number of turns required on the primary coil of a transformer, we can use the relationship between the magnetic flux density (B), the magnetic flux (Φ), and the area (A) of the core. The formula to find the magnetic flux density is:
\[ B = \frac{\Phi}{A} \]
where:
- \( B \) is the magnetic flux density (in Tesla),
- \( \Phi \) is the magnetic flux (in Weber),
- \( A \) is the area of the core (in m²).
We first need to convert the area of the core from cm² to m²:
\[ A = 35 , \text{cm}^2 = 35 \times 10^{-4} , \text{m}^2 = 0.0035 , \text{m}^2 \]
Next, from the formula for magnetic flux density, we can express the magnetic flux as:
\[ \Phi = B \times A \]
Substituting the values given:
\[ \Phi = 1.95 , \text{T} \times 0.0035 , \text{m}^2 = 0.006825 , \text{Wb} \]
Next, we relate the magnetic flux to the voltage induced in the coil using Faraday's law of electromagnetic induction:
\[ V = N \frac{d\Phi}{dt} \]
In a transformer, the flux is sinusoidal; hence, we consider the peak (maximum) magnetic flux and the frequency (f).
The peak flux linkage can be expressed in terms of the root mean square (RMS) voltage (V) and frequency (f) as:
\[ V = 4.44 N f \Phi \]
Given that the supply voltage is 1400 V (RMS) and the frequency is 50 Hz, substituting the known values into the equation gives:
\[ 1400 = 4.44 N (50) (0.006825) \]
Solving for \(N\):
\[ 1400 = 4.44 N \times 50 \times 0.006825 \]
Calculating the product:
\[ 4.44 \times 50 \times 0.006825 = 1.50735 \]
Thus we can rewrite the equation as:
\[ 1400 = N \times 1.50735 \]
Now solve for \(N\):
\[ N = \frac{1400}{1.50735} \approx 928.30 \]
Since the number of turns must be a whole number, we round to the nearest whole number:
\[ N \approx 928 , \text{turns} \]
Therefore, the number of turns required on the primary coil is approximately 928 turns.
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