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The problem provided gives information on a single-phase transformer. Here’s a structured approach to solving each part.

### Given Data:
1. **Apparent Power, S** = 100 kVA
2. **Secondary Voltage, \( V_2 \)** = 400 V
3. **Primary Turns, \( N_1 \)** = 200
4. **Secondary Turns, \( N_2 \)** = 40
5. **Load Power** = 25 kW with a power factor of 0.8
6. Neglecting transformer losses.

We are asked to calculate:
1. \( V_1 \): the primary applied voltage.
2. \( I_1 \) and \( I_2 \): the normal primary and secondary currents.
3. \( I_2 \): the secondary current for a 25 kW load at 0.8 power factor.

---

### Step-by-Step Solution

#### Part (i): Calculating the Primary Applied Voltage \( V_1 \)
We use the **turns ratio** formula for a transformer:

\[
\frac{V_1}{V_2} = \frac{N_1}{N_2}
\]

Rearranging to solve for \( V_1 \):

\[
V_1 = V_2 \times \frac{N_1}{N_2}
\]

Substituting the values:

\[
V_1 = 400 \, \text{V} \times \frac{200}{40} = 400 \times 5 = 2000 \, \text{V}
\]

So, the primary applied voltage \( V_1 \) is **2000 V**.

---

#### Part (ii): Calculating Normal Primary and Secondary Currents \( I_1 \) and \( I_2 \)
Using the **apparent power** equation for both primary and secondary sides:

1. **Secondary Current \( I_2 \)**:

   \[
   S = V_2 \times I_2
   \]

   Solving for \( I_2 \):

   \[
   I_2 = \frac{S}{V_2} = \frac{100,000 \, \text{VA}}{400 \, \text{V}} = 250 \, \text{A}
   \]

2. **Primary Current \( I_1 \)**:

   Using the primary side with the apparent power equation:

   \[
   I_1 = \frac{S}{V_1} = \frac{100,000 \, \text{VA}}{2000 \, \text{V}} = 50 \, \text{A}
   \]

So, the normal primary current \( I_1 \) is **50 A**, and the secondary current \( I_2 \) is **250 A**.

---

#### Part (iii): Calculating Secondary Current \( I_2 \) for a 25 kW Load at 0.8 Power Factor
For this load condition, we first calculate the apparent power \( S \) at 25 kW with a power factor of 0.8:

\[
S = \frac{\text{Real Power}}{\text{Power Factor}} = \frac{25,000 \, \text{W}}{0.8} = 31,250 \, \text{VA}
\]

Then, we calculate \( I_2 \) using this new apparent power:

\[
I_2 = \frac{S}{V_2} = \frac{31,250 \, \text{VA}}{400 \, \text{V}} = 78.125 \, \text{A}
\]

So, the secondary current \( I_2 \) when the load is 25 kW at 0.8 power factor is **78.125 A**.

---

### Summary of Results:
1. **Primary Applied Voltage \( V_1 \)**: 2000 V
2. **Normal Primary Current \( I_1 \)**: 50 A
3. **Normal Secondary Current \( I_2 \)**: 250 A
4. **Secondary Current \( I_2 \) for 25 kW load at 0.8 power factor**: 78.125 A

---

Would you like a detailed explanation of any part of the solution, or do you have further questions? 

Here are some related questions to consider:
1. How would the calculations change if the power factor was different?
2. What happens to the transformer currents if there were losses in the transformer?
3. How does changing the primary or secondary voltage affect the transformer’s operation?
4. What is the significance of the turns ratio in transformer design?
5. How does the transformer’s efficiency affect the current and voltage calculations?

**Tip**: Always verify power factor and load conditions, as they can significantly affect current calculations in transformers.

The following data apply to a single-phase transformer: output: 100 kVA, secondary voltage: 400 V; Primary turns: 200; secondary turns: 40; Neglecting the losses, calculate: (i) the primary applied voltage (ii) the normal primary and secondary currents (iii) the secondary current, when the load is 25 kW at 0.8 power factor.

Calculate the primary voltage, normal primary and secondary currents, and secondary current at a specific load for a single-phase transformer with 100 kVA output and 400 V secondary voltage. Learn to apply transformer turns ratio and apparent power formulas with a 25 kW load at 0.8 power factor. Ideal for engineering students.

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