### Transformer Voltage and Power Factor Calculation#### Problem Statement:A 460 V:2400 V transformer has a series leakage reactance of \( 37.2 \, \Omega \) as referred to the high-voltage side. A load connected to the low voltage side is observed to be absorbing \( 25 \, \text{kW} \), at unity power factor, and the voltage is measured to be \( 450 \, \text{V} \). Calculate the corresponding voltage and power factor as measured at the high-voltage terminals.---1. **Transformer Details:** - **Primary Voltage (High Voltage Side):** \( 2400 \, \text{V} \) - **Secondary Voltage (Low Voltage Side):** \( 460 \, \text{V} \) - **Series Leakage Reactance (referred to the high-voltage side):** \( 37.2 \, \Omega \)2. **Load Conditions:** - **Power Absorbed by Load:** \( 25 \, \text{kW} \) - **Voltage at Low Voltage Side:** \( 450\, \text{V} \) - **Power Factor of Load:** Unity (1.0)---### Solution Steps:1. **Load Current Calculation:** Given the power (\( P = 25 \, \text{kW} \)) and voltage on the secondary side (\( V_L = 450 \, \text{V} \)) at unity power factor: \( I_L = \frac{P}{V_L} \) \[ I_L = \frac{25000 \, W}{450 \, V} \approx 55.56 \, A \]2. **Refer Load Current to the Primary Side:** The turns ratio (\( a \)) of the transformer is given by: \[ a = \frac{V_1}{V_2} = \frac{2400 \, V}{460 \, V} \approx 5.22 \] The current referred to the high-voltage side (\( I_H \)): \[ I_H = \frac{I_L}{a} = \frac{55.56 \, A}{5.22} \approx 10.64 \, A \]3.

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1. A 460 V:2400 V transformer has a series leakage reactance of 37.2 as referred to the high-voltage side. A load connected to the low voltage side is observed to be absorbing 25 kW, at unity power factor, and the voltage is measured to be 450 V. Calculate the corresponding voltage and power factor as measured at the high-voltage terminals.

1. A 460 V:2400 V transformer has a series leakage reactance of 37.2 as referred to the high-voltage side. A load connected to the low voltage side is observed to be absorbing 25 kW, at unity power factor, and the voltage is measured to be 450 V. Calculate the corresponding voltage and power factor as measured at the high-voltage terminals.

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Concept explainers

Single Phase Transformer

A single-phase transformer is a type of power transformer. It is an electrical device that accepts converts the single phase alternating power into single phase alternating power. In this power conversion the voltage is stepped up or down. If the voltage is stepped up, the current will be stepped down and if the voltage is stepped down, the current will be stepped up.

Question

### Transformer Voltage and Power Factor Calculation

#### Problem Statement:
A 460 V:2400 V transformer has a series leakage reactance of \( 37.2 \, \Omega \) as referred to the high-voltage side. A load connected to the low voltage side is observed to be absorbing \( 25 \, \text{kW} \), at unity power factor, and the voltage is measured to be \( 450 \, \text{V} \). Calculate the corresponding voltage and power factor as measured at the high-voltage terminals.

---

1. **Transformer Details:**
- **Primary Voltage (High Voltage Side):** \( 2400 \, \text{V} \)
- **Secondary Voltage (Low Voltage Side):** \( 460 \, \text{V} \)
- **Series Leakage Reactance (referred to the high-voltage side):** \( 37.2 \, \Omega \)

2. **Load Conditions:**
- **Power Absorbed by Load:** \( 25 \, \text{kW} \)
- **Voltage at Low Voltage Side:** \( 450\, \text{V} \)
- **Power Factor of Load:** Unity (1.0)

---

### Solution Steps:

1. **Load Current Calculation:**
Given the power (\( P = 25 \, \text{kW} \)) and voltage on the secondary side (\( V_L = 450 \, \text{V} \)) at unity power factor:

\( I_L = \frac{P}{V_L} \)

\[
I_L = \frac{25000 \, W}{450 \, V} \approx 55.56 \, A
\]

2. **Refer Load Current to the Primary Side:**
The turns ratio (\( a \)) of the transformer is given by:

\[
a = \frac{V_1}{V_2} = \frac{2400 \, V}{460 \, V} \approx 5.22
\]

The current referred to the high-voltage side (\( I_H \)):

\[
I_H = \frac{I_L}{a} = \frac{55.56 \, A}{5.22} \approx 10.64 \, A
\]

3.

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