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A transformer is used by a European traveler while she is visiting the United States. She primarily uses it to run a 1200-watt hair dryer she brought with her. When the hair dryer is plugged in to her hotel room outlet in Los Angeles, she notices that it runs exactly as it does at home. The input voltage and current are \(120 \mathrm{~V}\) and 11.0 A, respectively. (a) Prove that this is not an ideal transformer. (b) What is its efficiency? (c) What is the rate at which heat is generated in the transformer itself?

Short Answer

Expert verified
Not ideal; efficiency is 90.91%; 120 W is lost as heat.

Step by step solution

01

Understanding an Ideal Transformer

An ideal transformer has no power loss, meaning input power equals output power. The input power in the primary coil is given by \( P_{in} = V_{in} \times I_{in} \), where \( V_{in} = 120 \mathrm{~V} \) and \( I_{in} = 11.0 \mathrm{~A} \). Thus, \( P_{in} = 120 \times 11 = 1320 \mathrm{~W} \).
02

Calculate Output Power

The output power is the power consumed by the hair dryer, which is \( P_{out} = 1200 \mathrm{~W} \).
03

Comparing Powers for Non-Ideal Transformer

For an ideal transformer, \( P_{in} \) should equal \( P_{out} \). However, here \( P_{in} = 1320 \mathrm{~W} \) and \( P_{out} = 1200 \mathrm{~W} \) are not equal, proving that this is not an ideal transformer due to energy losses.
04

Calculate Efficiency

Efficiency is calculated using \( \text{Efficiency} = \frac{P_{out}}{P_{in}} \times 100\% \). Substituting the values gives \( \frac{1200}{1320} \times 100\% = 90.91\% \).
05

Calculate Heat Generated in Transformer

The heat generated, or power loss, is \( P_{loss} = P_{in} - P_{out} = 1320 \mathrm{~W} - 1200 \mathrm{~W} = 120 \mathrm{~W} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Non-Ideal Transformers
Transformers are fundamental devices in electrical engineering that allow us to transfer electrical energy between two or more circuits through electromagnetic induction. When we talk about ideal transformers, we mean transformers that have no energy loss. In a perfect world, the power going into a transformer equals the power coming out. However, in reality, all transformers are non-ideal to some extent. They experience energy losses due to factors such as heat, resistance, and inefficiencies in the core material. Non-ideal transformers are characterized by these inherent power losses:
  • Core losses or hysteresis losses occur because the magnetic core doesn't respond perfectly to the electromagnetic fields.
  • Copper losses happen due to the resistance in the wire windings, generating heat.
  • Leakage flux where some of the magnetic field lines do not fully link the primary and secondary coils.
In our exercise, the input power is measured at 1320 W while the output power is only 1200 W, indicating a non-ideal transformer since not all input power is converted to useful output power. Understanding this concept helps us realize why some amount of power is lost in every real-life transformer operation.
Efficiency of Transformers
The efficiency of a transformer is a crucial aspect when determining its performance. It tells us how effectively a transformer converts input electrical power to output electrical power without unnecessary losses. Efficiency is expressed as a percentage, derived using the formula:\[ \text{Efficiency} = \frac{P_{out}}{P_{in}} \times 100\% \]Here, the output power, \( P_{out} \), is the useful power that the transformer delivers, while the input power, \( P_{in} \), is what initially enters the system. If a transformer were perfect, these two values would be equal, resulting in 100% efficiency. In the given scenario, the hair dryer's output power is 1200 W and the input power is 1320 W, leading to an efficiency calculation of:\[ \text{Efficiency} = \frac{1200}{1320} \times 100\% \approx 90.91\% \]This means approximately 91% of the input power is successfully converted to output power, with about 9% lost in the process. Such efficiency helps us determine how suitable a transformer is for reducing power loss in practical applications.
Power Loss in Transformers
Power loss in transformers is an unavoidable reality due to the non-ideal nature of materials and other physical principles. These losses affect the transformer's performance and depend largely on its design and operation. Primarily, power loss can be understood as the difference between the input power and the output power.In our exercise, power loss is calculated by subtracting the output power from the input power:\[ P_{loss} = P_{in} - P_{out} = 1320 \text{ W} - 1200 \text{ W} = 120 \text{ W} \]This means that 120 W of power is lost primarily as heat within the transformer. These losses are essentially due to:
  • Ohmic heating in the windings, often referred to as copper losses.
  • Magnetic losses in the core, also known as core or iron losses.
  • Additional minor losses from stray losses and dielectric heating.
Mitigating these losses involves using better core materials, reducing resistance in windings, and ensuring optimal design to control factors like heat and magnetic inefficiencies. Understanding power loss is essential for engineers looking to enhance transformer designs and improve energy efficiency.

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Most popular questions from this chapter

(a) A square loop of wire with sides of length \(40 \mathrm{~cm}\) is in a uniform magnetic field perpendicular to its area. If the field's strength is initially \(100 \mathrm{mT}\) and it decays to zero in \(0.010 \mathrm{~s}\), what is the magnitude of the average emf induced in the loop? (b) What would be the average emf if the sides of the loop were only \(20 \mathrm{~cm} ?\)

A uniform magnetic field is at right angles to the plane of a wire loop. If the field decreases by \(0.20 \mathrm{~T}\) in \(1.0 \times 10^{-3} \mathrm{~s}\) and the magnitude of the average emf induced in the loop is \(80 \mathrm{~V},\) (a) what is the area of the loop? (b) What would be the value of the average induced emf if the field change was the same but took twice as long to decrease? (c) What would be the value of the average induced emf if the field decrease was twice as much and it also took twice as long to change?

A boy is traveling due north at a constant speed while carrying a metal rod. The rod's length is oriented in the east-west direction and is parallel to the ground. (a) There will be no induced emf when the rod is (1) at the equator, (2) near the Earth's magnetic poles, (3) somewhere between the equator and the poles. Why? (b) Assume that the Earth's magnetic field is \(1.0 \times 10^{-4} \mathrm{~T}\) near the North Pole and \(1.0 \times 10^{-5} \mathrm{~T}\) near the equator. If the boy runs with a speed of \(1.3 \mathrm{~m} / \mathrm{s}\) northward near each location, and the rod is \(1.5 \mathrm{~m}\) long, calculate the induced emf in the rod in each location.

A \(240-\mathrm{V}\) dc motor has an armature whose resistance is \(1.50 \Omega\). When running at its operating speed, it draws a current of 16.0 A. (a) What is the back emf of the motor when it is operating normally? (b) What is the starting current? (Assume that there is no additional resistance in the circuit.) (c) What series resistance would be required to limit the starting current to \(25 \mathrm{~A} ?\)

An ideal transformer steps \(8.0 \mathrm{~V}\) up to \(2000 \mathrm{~V}\), and the 4000-turn secondary coil carries 2.0 A. (a) Find the number of turns in the primary coil. (b) Find the current in the primary coil.

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