/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 A generating station is producin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 22: Problem 67

A generating station is producing \(1.2 \times 10^{6} \mathrm{W}\) of power that is to be sent to a small town located \(7.0 \mathrm{km}\) away. Each of the two wires that comprise the transmission line has a resistance per kilometer of \(5.0 \times\) \(10^{-2} \Omega / \mathrm{km} .\) (a) Find the power used to heat the wires if the power is transmitted at \(1200 \mathrm{V} .\) (b) \(\mathrm{A} 100: 1\) step-up transformer is used to raise the voltage before the power is transmitted. How much power is now used to heat the wires?

Short Answer

Expert verified
(a) 700 kW. (b) 70 W.

Step by step solution

01

Calculate the Current for Transmission at 1200 V

To find the current, use the formula for electrical power: \( P = VI \), where \( P \) is the power, \( V \) is voltage, and \( I \) is current. For part (a), we know that \( P = 1.2 \times 10^{6} \text{ W} \) and \( V = 1200 \text{ V} \). Solving for \( I \), we have \[ I = \frac{P}{V} = \frac{1.2 \times 10^{6}}{1200} = 1000 \text{ A}. \]
02

Determine the Total Resistance of the Wires

The resistance per kilometer of the transmission line is given as \( 5.0 \times 10^{-2} \Omega/\text{km} \). The total distance is \( 7.0 \text{ km} \). Since there are two wires, the total resistance \( R \) is \[ R = 2 \times 7.0 \times 5.0 \times 10^{-2} = 0.7 \Omega. \]
03

Calculate Power Loss due to Heating (Before Step-Up Transformer)

The power loss in the wires can be calculated using the formula for power dissipation in resistors: \( P_{ ext{loss}} = I^2 R \). We have \( I = 1000 \text{ A} \) and \( R = 0.7 \Omega \). So, the power loss is \[ P_{ ext{loss}} = 1000^2 \times 0.7 = 700,000 \text{ W} = 700 \text{ kW}. \]
04

Calculate the New Voltage with the Step-Up Transformer

A 100:1 step-up transformer increases the transmission voltage. If the original voltage is \( V = 1200 \text{ V} \), then the new voltage after stepping up is \[ V' = 100 \times 1200 = 120,000 \text{ V}. \]
05

Calculate New Current After Voltage Increase

With the new transmission voltage, the current \( I' \) can be found again using \( P = V'I' \). With \( P = 1.2 \times 10^{6} \text{ W} \), we have \[ I' = \frac{P}{V'} = \frac{1.2 \times 10^{6}}{120,000} = 10 \text{ A}. \]
06

Recalculate Power Loss in Wires with Step-Up Transformer

Using the new current \( I' = 10 \text{ A} \), the power loss in the wires after the voltage step-up is \[ P'_{ ext{loss}} = (I')^2 R = 10^2 \times 0.7 = 70 \text{ W}. \]
07

Conclusion: Compare Power Loss Before and After Step-Up Transformer

Without the transformer, the power loss due to heating the wires is \( 700 \text{ kW} \). With the transformer, the power loss reduces drastically to \( 70 \text{ W} \). The step-up transformer significantly reduces power loss.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Power Loss Calculation
Electrical power transmission often leads to power loss due to heating in the conducting wires. This power loss is given by the formula for power dissipation in resistors: \( P_{\text{loss}} = I^2 R \), where \( I \) is the current flowing through the wires and \( R \) is the total resistance of the wires.
When power is transmitted at a low voltage, a high current is needed to transport the same amount of power. This high current results in a greater power loss because the power loss is directly proportional to the square of the current.
For example, in the problem where power is transmitted at 1200 V, the current \( I \) is 1000 A. Calculating power loss, we end up with a significant 700 kW loss due to the high squared value of the current multiplied by the resistance. Hence, in power transmission, reducing the current significantly cuts down on power losses.
Resistance in Wires
Resistance in transmission wires is an essential factor in determining how much power is lost during electric power transmission. The resistance is given by the product of the resistance per unit length and the total length of wire used.
In our exercise, the wire resistance per kilometer is provided as \( 5.0 \times 10^{-2} \ \Omega/\text{km} \). If each wire is 7 km long, and there are two wires, the total resistance \( R \) can be calculated as follows:
  • Resistance of each wire \( = 7 \times 5.0 \times 10^{-2} = 0.35 \ \Omega \)
  • Total resistance for two wires \( = 2 \times 0.35 = 0.7 \ \Omega \)
The calculated resistance allows you to evaluate the power loss due to resistance when an electric current flows through the wires. Understanding wire resistance helps in designing power systems that minimize energy losses, thereby making the transmission more efficient.
Step-Up Transformer Effect
Step-up transformers play a vital role in reducing power losses in electrical transmission. A transformer increases the voltage and reduces the current that needs to travel through the wires. With the relationship \( P = VI \), when the voltage \( V \) is increased, the current \( I \) decreases if the power \( P \) remains constant.
In this problem, a 100:1 step-up transformer raises the voltage from 1200 V to 120,000 V. As a result, the current decreases from 1000 A to 10 A.
Remember that power loss is proportional to the square of the current \( I^2 \), so reducing the current drastically reduces power loss. With the transformer, the power loss in the wires plunges from 700 kW to just 70 W.
Therefore, step-up transformers are crucial for efficient long-distance power transmission because they help cut down on power losses by minimizing the current flowing through the transmission wires, which in turn reduces heating losses.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Mutual induction can be used as the basis for a metal detector. A typical setup uses two large coils that are parallel to each other and have a common axis. Because of mutual induction, the ac generator connected to the primary coil causes an emf of \(0.46 \mathrm{V}\) to be induced in the secondary coil. When someone without metal objects walks through the coils, the mutual inductance and, thus, the induced emf do not change much. But when a person carrying a handgun walks through, the mutual inductance increases. The change in emf can be used to trigger an alarm. If the mutual inductance increases by a factor of three, find the new value of the induced emf.

A circular coil (950 turns, radius \(=0.060 \mathrm{m}\) ) is rotating in a uniform magnetic field. At \(t=0\) s, the normal to the coil is perpendicular to the magnetic field. At \(t=0.010\) s, the normal makes an angle of \(\phi=45^{\circ}\) with the field because the coil has made one-eighth of a revolution. An average emf of magnitude \(0.065 \mathrm{V}\) is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

A circular coil of radius 0.11 m contains a single timm and is located in a constant magnetic field of magnitude \(0.27 \mathrm{T} .\) The magnetic field has the same direction as the normal to the plane of the coil. The radius increases to \(0.30 \mathrm{m}\) in a time of \(0.080 \mathrm{s} .\) Concepts: (i) Why is there an emf induced in the coil? (ii) Does the magnitude of the induced emf depend on whether the area is increasing or decreasing? Explain. (iii) What determines the amount of current induced in the coil? (iv) If the coil is cut so it is no longer one continuous piece, are there an induced emf and an induced current? Explain. Calculations: (a) Determine the magnitude of the emf induced in the coil. (b) The coil has a resistance of \(0.70 \Omega\). Find the magnitude of the induced current.

A rectangular loop of wire with sides 0.20 and \(0.35 \mathrm{m}\) lies in a plane perpendicular to a constant magnetic field (see part \(a\) of the drawing). The magnetic field has a magnitude of \(0.65 \mathrm{T}\) and is directed parallel to the normal of the loop's surface. In a time of 0.18 s, one-half of the loop is then folded back onto the other half, as indicated in part \(b\) of the drawing. Determine the magnitude of the average emf induced in the loop.

A \(120.0-\mathrm{V}\) motor draws a current of \(7.00 \mathrm{A}\) when running at normal speed. The resistance of the armature wire is \(0.720 \Omega .\) (a) Determine the back emf generated by the motor. (b) What is the current at the instant when the motor is just turned on and has not begun to rotate? (c) What series resistance must be added to limit the starting current to \(15.0 \mathrm{A} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Work Energy and Power

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Electric Charge Field and Potential

Read Explanation

Waves Physics

Read Explanation

Radiation

Read Explanation

Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.