/*! 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 15 A \(75.0-\mathrm{kW}, 230-\mathr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 33: Problem 15

A \(75.0-\mathrm{kW}, 230-\mathrm{V}\) shunt generator has a generated emf of \(243.5 \mathrm{~V}\). If the field current is \(12.5 \mathrm{~A}\) at rated output, what is the armature resistance?

Short Answer

Expert verified
The armature resistance is approximately 0.04 ohms.

Step by step solution

01

Understand the Problem Parameters

We are given a shunt generator with the following parameters: - Power output of 75.0 kW - Terminal voltage of 230 V - Generated emf of 243.5 V - Field current of 12.5 A. Our task is to find the armature resistance.
02

Calculate the Output Current

To find the armature resistance, we first need the output current of the generator. We use the formula for power:\[ P = V_t imes I_t \]where \( P = 75.0 ext{ kW} = 75000 ext{ W} \) and \( V_t = 230 ext{ V} \).Solve for \( I_t \):\[ I_t = \frac{75000}{230} \approx 326.09 ext{ A} \]
03

Determine the Armature Current

Next, we find the armature current \( I_a \) using:\[ I_a = I_t + I_f \]where \( I_t = 326.09 ext{ A} \) and \( I_f = 12.5 ext{ A} \).Thus, we have:\[ I_a = 326.09 + 12.5 = 338.59 ext{ A} \]
04

Apply Ohm's Law to Find Armature Resistance

Finally, use Ohm's Law on the armature circuit to find the armature resistance \( R_a \):\[ E = V_t + I_a \times R_a \]Solve for \( R_a \):\[ R_a = \frac{E - V_t}{I_a} = \frac{243.5 - 230}{338.59} \approx 0.04 \text{ ohms} \]

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.

Shunt Generator
A shunt generator is a specific type of direct current (DC) generator where the field windings are connected in parallel with the armature windings. This parallel connection is what gives it the name "shunt" generator. The primary advantage of a shunt generator is its ability to maintain a relatively constant output voltage even as the load varies. This is highly beneficial for applications that require a steady voltage despite changes in the electrical load.
  • The field winding energizes the magnetic field, which is essential for generating electricity.
  • The armature winding is the part where electricity is actually generated.
  • As the generator operates, the generated emf drives the current through the external load.
In short, a shunt generator is an efficient system for producing a stable voltage supply, making it a common choice for various electrical systems like battery charging and auxiliary power applications.
Armature Resistance
Armature resistance is a critical parameter in electrical generators and motors that affects their efficiency and performance. It represents the electrical resistance offered by the armature winding, which is the winding where the generated current flows. Ideally, low armature resistance is preferred as it means less electrical energy is lost as heat, thereby improving efficiency.
  • A high armature resistance leads to larger power losses, hence higher heat generation.
  • Armature resistance is measured in ohms (惟), and it can be influenced by factors such as winding material, size, and temperature.
  • In practical applications, minimizing armature resistance is a key goal to ensure that more of the generated energy is available as useful output power.
For shunt generators, understanding the armature resistance is crucial as it directly impacts the calculation for total emf and fault diagnostics.
Ohm's Law
Ohm's Law is a fundamental principle in electrical engineering that describes the relationship between voltage, current, and resistance in an electrical circuit. It is usually expressed in the equation:\[ V = I \times R \]where \( V \) is the voltage across the resistor, \( I \) is the current through the resistor, and \( R \) is the resistance of the resistor. This law is instrumental in analyzing and solving circuit problems as it enables the calculation of one parameter if the other two are known.
  • Ohm's Law is applicable to most electrical circuits, though it may not hold in non-linear or non-ohmic devices.
  • It allows engineers to design circuits with precise voltage, current, and resistance requirements.
  • In the context of generators, Ohm鈥檚 Law helps find unknown circuit elements like armature resistance by rearranging the formula.
By applying Ohm's Law, we can determine the armature resistance of a shunt generator by knowing its generated emf, terminal voltage, and current.
Generated EMF
Generated electromotive force (emf) is a crucial concept in the working of generators and motors. It refers to the electrical energy produced by the mechanical rotation of coils within a magnetic field. In DC generators like shunt generators, the generated emf is the voltage produced when mechanical energy is converted into electrical energy.
  • Generated emf is measured in volts (V) and often slightly higher than the terminal voltage.
  • This takes into account voltage drops caused by the internal elements like armature resistance.
  • In calculations, knowing the generated emf helps determine the overall efficiency and performance of the generator.
Ultimately, generated emf provides the driving force for current flow in the circuit, making it fundamental to understanding the output capabilities and limitations of a shunt generator.

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

An ac generator produces an output voltage of \(\mathscr{E}=170 \sin 377 t\) volts, where \(t\) is in seconds. What is the frequency of the ac voltage? A sine curve plotted as a function of time is no different from a cosine curve, except for the location of \(t=0\). Since \(\mathscr{E}=2 \pi N A B f \cos 2 \pi f t\), we have \(377 t=2 \pi f t\), from which we find that the frequency \(f=60 \mathrm{~Hz}\).

A motor armature develops a torque of \(100 \mathrm{~N} \cdot \mathrm{m}\) when it draws \(40 \mathrm{~A}\) from the line. Determine the torque developed if the armature current is increased to \(70 \mathrm{~A}\) and the magnetic field strength is reduced to 80 percent of its initial value. The torque developed by the armature of a given motor is proportional to the armature current and to the field strength (see Chapter 30). In other words, the ratio of the torques equals the ratio of the two sets of values of \(|N I A B|\). Using subscripts \(i\) and \(f\) for initial and final values, \(\tau_{f} / \tau_{i}=I_{f} B_{f} / I_{i} B_{i}\), hence, $$ \tau_{f}=(100 \mathrm{~N} \cdot \mathrm{m})\left(\frac{70}{40}\right)(0.80)=0.14 \mathrm{kN} \cdot \mathrm{m} $$

A shunt motor has a field resistance of \(200 \Omega\) and an armature resistance of \(0.50 \Omega\) and is connected to 120-V mains. The motor draws a current of \(4.6\) A when running at full speed. What current will be drawn by the motor if the speed is reduced to 90 percent of full speed by application of a load?

When turning at 1500 rev/min, a certain generator produces \(100.0 \mathrm{~V}\). What must be its frequency in rev/min if it is to produce \(120.0 \mathrm{~V}\) ? Because the amplitude of the emf is proportional to the frequency, we have, for two frequencies \(f_{1}\) and \(f_{2}\), $$ \frac{\mathscr{E}_{1}}{\mathscr{E}_{2}}=\frac{f_{1}}{f_{2}} \quad \text { or } \quad f_{2}=f_{1} \frac{\mathscr{E}_{2}}{\mathscr{E}_{1}}=(1500 \mathrm{rev} / \mathrm{min})\left(\frac{120.0 \mathrm{~V}}{100.0 \mathrm{~V}}\right)=1800 \mathrm{rev} / \mathrm{min} $$

A \(120-\mathrm{V}\) generator is run by a windmill that has blades \(2.0 \mathrm{~m}\) long. The wind, moving at \(12 \mathrm{~m} / \mathrm{s}\), is slowed to \(7.0 \mathrm{~m} / \mathrm{s}\) after passing the windmill. The density of air is \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\). If the system has no losses, what is the largest current the generator can produce? [Hint: How much energy does the wind lose per second?]
See all solutions

Recommended explanations on Physics Textbooks

Kinematics Physics

Read Explanation

Radiation

Read Explanation

Turning Points in Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Modern Physics

Read Explanation

Medical Physics

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.