Question

A 250/125 V, 5 KVA single-phase transformer has primary resistance of 0.2 Ω and reactance of 0.75 Ω. The secondary resistance is 0.05 Ω and reactance of 0.2 Ω. Determine its regulation while supplying full load on 0.8 leading P.F.

Answer 1

To calculate the voltage regulation of a transformer, we use the formula:

\[
\text{Voltage Regulation} = \frac{V_{nl} - V_{fl}}{V_{fl}} \times 100
\]

Where:
- \( V_{nl} \) is the no-load voltage (voltage at open circuit)
- \( V_{fl} \) is the full-load voltage

However, the above formula requires the full-load and no-load voltages, which can be derived from the impedance of the transformer and the power factor. Since you're given the resistances, reactances, and the power factor, we can use an alternative method to calculate the voltage regulation.

### Given:
- Primary Voltage (\( V_1 \)) = 250 V
- Secondary Voltage (\( V_2 \)) = 125 V
- Apparent Power \( S = 5 \, \text{kVA} \)
- Primary Resistance (\( R_1 \)) = 0.2 Ω
- Primary Reactance (\( X_1 \)) = 0.75 Ω
- Secondary Resistance (\( R_2 \)) = 0.05 Ω
- Secondary Reactance (\( X_2 \)) = 0.2 Ω
- Power Factor = 0.8 (leading)

### Step 1: Calculate the equivalent impedance referred to the primary side

The secondary resistances and reactances are referred to the primary side using the turns ratio \( a \), which is given by:

\[
a = \frac{V_1}{V_2} = \frac{250}{125} = 2
\]

Now, refer the secondary resistances and reactances to the primary side:

- Referred secondary resistance: \( R_2' = R_2 \times a^2 = 0.05 \times 2^2 = 0.2 \, \Omega \)
- Referred secondary reactance: \( X_2' = X_2 \times a^2 = 0.2 \times 2^2 = 0.8 \, \Omega \)

### Step 2: Calculate total equivalent impedance on the primary side

Now, add the referred values to the primary resistances and reactances:

\[
R_{eq} = R_1 + R_2' = 0.2 + 0.2 = 0.4 \, \Omega
\]
\[
X_{eq} = X_1 + X_2' = 0.75 + 0.8 = 1.55 \, \Omega
\]

### Step 3: Calculate the impedance magnitude

The equivalent impedance magnitude is:

\[
Z_{eq} = \sqrt{R_{eq}^2 + X_{eq}^2} = \sqrt{0.4^2 + 1.55^2} = \sqrt{0.16 + 2.4025} = \sqrt{2.5625} = 1.6 \, \Omega
\]

### Step 4: Full-load current

The full-load current on the primary side is given by:

\[
I_{fl} = \frac{S}{V_1} = \frac{5000}{250} = 20 \, \text{A}
\]

### Step 5: Voltage drop across the equivalent impedance

The voltage drop across the equivalent impedance is:

\[
V_{drop} = I_{fl} \times Z_{eq} = 20 \times 1.6 = 32 \, \text{V}
\]

### Step 6: Effect of power factor

Since the power factor is leading (0.8 leading), the voltage drop is not entirely resistive, so we need to decompose it into real and imaginary components:

- Real component (resistive drop): \( I_{fl} R_{eq} = 20 \times 0.4 = 8 \, \text{V} \)
- Imaginary component (reactive drop): \( I_{fl} X_{eq} = 20 \times 1.55 = 31 \, \text{V} \)

For a leading power factor, the voltage drop due to reactance is subtracted. So the net voltage drop:

\[
V_{net} = (I_{fl} R_{eq}) \cos(\theta) - (I_{fl} X_{eq}) \sin(\theta)
\]

Where:
- \( \cos(\theta) = 0.8 \) (power factor)
- \( \sin(\theta) = 0.6 \) (from \( \sin(\theta) = \sqrt{1 - \cos^2(\theta)} \))

\[
V_{net} = 8 \times 0.8 - 31 \times 0.6 = 6.4 - 18.6 = -12.2 \, \text{V}
\]

### Step 7: Voltage regulation

Finally, the voltage regulation is:

\[
\text{Voltage Regulation} = \frac{V_{net}}{V_1} \times 100 = \frac{-12.2}{250} \times 100 = -4.88\%
\]

### Conclusion:

The voltage regulation of the transformer at full load with a power factor of 0.8 leading is approximately **-4.88%**. The negative sign indicates that the voltage rises under leading power factor conditions.

Answer 2

To determine the voltage regulation of the transformer under full load with a power factor of 0.8 leading, we'll follow these steps:

1. **Given Data:**
   - Primary Voltage \( V_1 = 250 \, V \)
   - Secondary Voltage \( V_2 = 125 \, V \)
   - Apparent Power \( S = 5 \, kVA = 5000 \, VA \)
   - Primary Resistance \( R_1 = 0.2 \, \Omega \)
   - Primary Reactance \( X_1 = 0.75 \, \Omega \)
   - Secondary Resistance \( R_2 = 0.05 \, \Omega \)
   - Secondary Reactance \( X_2 = 0.2 \, \Omega \)
   - Power Factor \( \cos \phi = 0.8 \) (leading)

2. **Full Load Current Calculation:**
   \[
   I_2 = \frac{S}{V_2} = \frac{5000}{125} = 40 \, A
   \]

3. **Referred Impedance to Secondary Side:**
   Referred primary resistance and reactance to the secondary side:
   \[
   R_1' = \left(\frac{V_2}{V_1}\right)^2 R_1 = \left(\frac{125}{250}\right)^2 \times 0.2 \, \Omega = 0.05 \, \Omega
   \]
   \[
   X_1' = \left(\frac{V_2}{V_1}\right)^2 X_1 = \left(\frac{125}{250}\right)^2 \times 0.75 \, \Omega = 0.1875 \, \Omega
   \]

   Total resistance and reactance referred to the secondary side:
   \[
   R_{eq} = R_2 + R_1' = 0.05 + 0.05 = 0.1 \, \Omega
   \]
   \[
   X_{eq} = X_2 + X_1' = 0.2 + 0.1875 = 0.3875 \, \Omega
   \]

4. **Voltage Drop Calculation:**
   The voltage drop across the equivalent impedance:
   \[
   V_{drop} = I_2 \times (R_{eq} \cos \phi + X_{eq} \sin \phi)
   \]

   With \( \cos \phi = 0.8 \) and \( \sin \phi = 0.6 \) (for leading P.F.):
   \[
   V_{drop} = 40 \times (0.1 \times 0.8 - 0.3875 \times 0.6)
   \]
   \[
   V_{drop} = 40 \times (0.08 - 0.2325)
   \]
   \[
   V_{drop} = 40 \times (-0.1525) = -6.1 \, V
   \]

5. **Voltage Regulation Calculation:**
   Voltage regulation is given by:
   \[
   \text{Regulation} = \frac{V_{drop}}{V_2} \times 100\%
   \]
   \[
   \text{Regulation} = \frac{-6.1}{125} \times 100\% = -4.88\%
   \]

### Final Answer:
The voltage regulation of the transformer while supplying full load at a 0.8 leading power factor is \(-4.88\%\). This negative value indicates that the output voltage increases when the transformer is loaded, which is typical for a leading power factor.