Question

A single-phase transformer has 300 turns on its primary side and 750 turns on its secondary side, the maximum flux density in the core is 1 wb/m². Calculate: (i) the net cross-sectional area of the core and (ii) the emf induced in the secondary side.

Answer 1

To calculate both the net cross-sectional area of the core and the EMF induced in the secondary winding of the transformer, we need to use the fundamental relationships in transformer theory and Faraday’s law of electromagnetic induction.

### (i) Net Cross-Sectional Area of the Core

We can determine the net cross-sectional area of the core by using the formula related to the EMF induced in the primary winding and the core's magnetic flux.

Faraday’s Law states that the induced EMF (\( E \)) in a coil is given by:
\[
E = 4.44 \, f \, N \, \Phi_{\text{max}}
\]
Where:
- \( E \) = Induced EMF in volts (V),
- \( f \) = Frequency of the AC supply (in Hz),
- \( N \) = Number of turns in the winding,
- \( \Phi_{\text{max}} \) = Maximum flux in the core (in Weber, Wb).

The maximum flux \( \Phi_{\text{max}} \) is also related to the flux density \( B_{\text{max}} \) and the core area \( A \) by:
\[
\Phi_{\text{max}} = B_{\text{max}} \times A
\]
Where:
- \( B_{\text{max}} \) = Maximum flux density (in Weber per square meter, Wb/m²),
- \( A \) = Cross-sectional area of the core (in m²).

Rearranging for \( A \):
\[
A = \frac{\Phi_{\text{max}}}{B_{\text{max}}}
\]

We need to calculate \( \Phi_{\text{max}} \). This depends on the frequency and the primary EMF \( E_1 \). We can use the relationship:
\[
E_1 = 4.44 \, f \, N_1 \, \Phi_{\text{max}}
\]
Rearranging for \( \Phi_{\text{max}} \):
\[
\Phi_{\text{max}} = \frac{E_1}{4.44 \, f \, N_1}
\]

To proceed further, we need the following additional information:
- The applied voltage on the primary side \( E_1 \),
- The supply frequency \( f \).

Without these, we can’t directly calculate \( A \), but we know the relationships and can compute it when the missing data is provided.

---

### (ii) EMF Induced in the Secondary Side

To calculate the EMF induced in the secondary side (\( E_2 \)), we use the **turns ratio** and the relationship between the primary and secondary windings.

The transformer turns ratio is given by:
\[
\frac{N_2}{N_1} = \frac{E_2}{E_1}
\]
Where:
- \( N_2 \) = Number of turns on the secondary side,
- \( N_1 \) = Number of turns on the primary side,
- \( E_2 \) = EMF induced in the secondary winding,
- \( E_1 \) = EMF induced in the primary winding.

Given:
- \( N_1 = 300 \) turns (primary),
- \( N_2 = 750 \) turns (secondary).

Rearranging to find \( E_2 \):
\[
E_2 = E_1 \times \frac{N_2}{N_1}
\]
\[
E_2 = E_1 \times \frac{750}{300} = E_1 \times 2.5
\]

Thus, the EMF induced in the secondary winding is 2.5 times the EMF in the primary winding.

#### Final Step:
- If the primary voltage \( E_1 \) is known, you can calculate the exact value of \( E_2 \) using this formula.

---

In summary:
1. **The net cross-sectional area of the core** depends on the primary voltage, frequency, and turns, but can be calculated once the primary voltage \( E_1 \) and frequency \( f \) are known.
2. **The EMF induced in the secondary side** is 2.5 times the primary voltage \( E_1 \), based on the turns ratio \( \frac{N_2}{N_1} \).

Answer 2

To solve this problem, we will use the following formulas and steps.

### Given Data
- Number of turns on the primary side (\(N_p\)): 300
- Number of turns on the secondary side (\(N_s\)): 750
- Maximum flux density in the core (\(B_{max}\)): 1 Wb/m²

### (I) Net Cross-Sectional Area of the Core

The flux density \(B\) is given by:
\[ B = \frac{\Phi}{A} \]

Where:
- \(\Phi\) is the magnetic flux (in Weber)
- \(A\) is the cross-sectional area of the core (in m²)

Rearranging the formula to solve for \(A\):
\[ A = \frac{\Phi}{B} \]

First, we need to find the magnetic flux \(\Phi\). To find \(\Phi\), we use the formula for the emf induced in the primary winding and secondary winding of the transformer:

\[ E_p = 4.44 \cdot f \cdot N_p \cdot \Phi \]
\[ E_s = 4.44 \cdot f \cdot N_s \cdot \Phi \]

The voltage ratio between primary and secondary sides of the transformer is given by:
\[ \frac{E_s}{E_p} = \frac{N_s}{N_p} \]

Since the primary and secondary voltages are proportional to the number of turns, and we assume a typical frequency \(f\), we can equate:
\[ \frac{E_s}{E_p} = \frac{750}{300} = 2.5 \]

Now, the maximum flux density \(B\) and flux \(\Phi\) relationship is:
\[ B_{max} = \frac{\Phi_{max}}{A} \]

From the formula:
\[ \Phi_{max} = B_{max} \cdot A \]

Therefore:
\[ A = \frac{\Phi_{max}}{B_{max}} \]

Let's calculate \(\Phi_{max}\). If we assume \(f\) (frequency) is 50 Hz, which is common, then:
\[ \Phi_{max} = \frac{E_p}{4.44 \cdot f \cdot N_p} \]

For simplicity, let’s use a practical example or assume:
\[ E_p = 1 \text{ V (as we are interested in the cross-sectional area and not the exact value of emf)} \]

So:
\[ \Phi_{max} = \frac{1}{4.44 \cdot 50 \cdot 300} \]

Plugging this value into:
\[ A = \frac{\Phi_{max}}{B_{max}} \]

\[ A = \frac{\frac{1}{4.44 \cdot 50 \cdot 300}}{1} \]
\[ A \approx \frac{1}{66600} \text{ m}^2 \]
\[ A \approx 1.5 \times 10^{-5} \text{ m}^2 \]

### (II) The EMF Induced in the Secondary Side

The emf induced in the secondary side is given by:
\[ E_s = 4.44 \cdot f \cdot N_s \cdot \Phi \]

We have \(\Phi\) from:
\[ \Phi = B_{max} \cdot A \]

Substituting \(A\):
\[ \Phi = 1 \cdot 1.5 \times 10^{-5} \text{ Wb} \]

So:
\[ E_s = 4.44 \cdot 50 \cdot 750 \cdot 1.5 \times 10^{-5} \]
\[ E_s \approx 4.44 \cdot 50 \cdot 750 \cdot 1.5 \times 10^{-5} \]
\[ E_s \approx 4.99 \text{ V} \]

### Summary
1. **Net Cross-Sectional Area of the Core**: Approximately \(1.5 \times 10^{-5} \text{ m}^2\)
2. **EMF Induced in the Secondary Side**: Approximately 4.99 V