/*! 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 48 A motor is designed to operate o... [FREE SOLUTION] | 91Ó°ÊÓ

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A motor is designed to operate on \(117 \mathrm{V}\) and draws a current of \(12.2 \mathrm{A}\) when \(\mathrm{it}\) first starts up. At its normal operating speed, the motor draws a current of 2.30 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third of the normal speed.

Short Answer

Expert verified
(a) 9.59 Ω, (b) 94.91 V, (c) 8.89 A.

Step by step solution

01

Calculate Armature Coil Resistance (Part a)

To find the resistance of the armature coil, use Ohm's Law. At startup, the motor draws the full voltage of 117 V with a current of 12.2 A. Ohm's Law is given by:\[ V = IR \]Solving for resistance \( R \):\[ R = \frac{V}{I} = \frac{117 \text{ V}}{12.2 \text{ A}} \approx 9.59 \text{ } \Omega \]Thus, the resistance of the armature coil is approximately 9.59 \( \Omega \).
02

Calculate Back EMF at Normal Speed (Part b)

At normal operating speed, the motor draws a current of 2.30 A. The effective voltage across the motor when running at normal speed, accounting for back EMF \( E \), is given by:\[ V - E = IR \]Substitute in the known values:\[ 117 - E = 2.30 \times 9.59 \]Solve for \( E \):\[ E = 117 - (2.30 \times 9.59) \approx 94.91 \text{ V} \]Thus, the back EMF at normal speed is approximately 94.91 V.
03

Calculate Current at One-Third Normal Speed (Part c)

At one-third normal speed, the back EMF is also one-third of its value at normal speed. Calculate this new back EMF:\[ E_{1/3} = \frac{1}{3} \times 94.91 \approx 31.64 \text{ V} \]Using Ohm's Law with the reduced back EMF:\[ 117 - 31.64 = I \times 9.59 \]Solve for current \( I \):\[ I = \frac{117 - 31.64}{9.59} \approx 8.89 \text{ A} \]Hence, the current drawn by the motor at one-third of normal speed is approximately 8.89 A.

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Key Concepts

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

Ohm's Law
Ohm's Law is one of the foundational concepts in physics, particularly when dealing with electrical circuits. It describes the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit. The law is typically expressed as:
  • \[ V = IR \]
This relationship suggests that the voltage across a resistor is directly proportional to the current flowing through it, with the resistance being the constant of proportionality. In practical terms, knowing two of the three variables allows us to calculate the third. For example, in an electric motor operating on a known voltage, if we can measure the current, we can readily find the resistance of the armature coil using this simple formula. This is exactly what is done in the given exercise to determine the armature resistance.
Electric Motors
Electric motors are devices that convert electrical energy into mechanical energy. They are central to a wide range of applications, from small household appliances to large industrial machines. The basic working principle of an electric motor involves the interaction between a magnetic field and electric current, which produces a force (or torque) that turns the motor's shaft.
In the context of the exercise, when a motor first starts, it draws significant current, termed as the start-up current, because there is no back EMF developed yet to oppose the incoming voltage. Understanding how electric motors work includes recognizing the roles of different components, such as:
  • Armature Coil: The part of the motor that carries current and experiences a magnetic field.
  • Stator: The stationary part of the motor providing the magnetic field.
  • Commutator: Used to reverse the direction of current flow through the armature coil.
Recognizing these components helps in understanding the motor's operation, including factors like start-up current and normal operating speed.
Back EMF
Back Electro-Motive Force (EMF) is a critical concept in the operation of electric motors. When a motor rotates, its movement induces a voltage in the opposite direction to the supply voltage, known as back EMF. This phenomenon is governed by Faraday's Law of electromagnetic induction.
When a motor reaches its normal operating speed, this back EMF significantly reduces the net voltage across the motor's armature, thereby decreasing the current it draws. The significance of back EMF can be seen in its stabilizing effect, allowing motors to automatically adjust the current they draw based on their speed. If the motor speed increases, back EMF increases, reducing the current; similarly, if speed decreases, back EMF decreases, increasing the current. This balancing act is essential for efficient motor operation.
  • Key Formula: \[ V - E = IR \]
  • Where: E is the back EMF.
Back EMF, therefore, ensures the motor doesn't draw excessive current once it's up to speed.
Armature Resistance
Armature resistance is the inherent resistance of the wire that makes up the windings in the armature coil of an electric motor. This resistance plays a vital role in determining how much current the motor draws under various operating conditions.
In terms of practical measurement as seen in the exercise, armature resistance can be calculated during startup using Ohm's Law, where the full potential difference is across this resistance due to the absence of back EMF:
  • \[ R = \frac{V}{I} \]
  • Where V is the initial voltage and I is the startup current.
Armature resistance impacts the motor's efficiency and heat generation. Lower resistance is generally preferable as it minimizes power losses within the motor. However, in an actual design, there is always a balance needed between minimizing resistance and designing a practical and manufacturable motor. Understanding armature resistance is crucial for diagnosing electric motor performance issues and ensuring optimum functionality in different scenarios.

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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 uniform magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that an emf of \(0.80 \mathrm{V}\) and a current of \(3.2 \mathrm{A}\) are induced in the coil. The wire is then re-formed into a single-turn square coil, which is used in the same magnetic field (again perpendicular to the plane of the coil and with a magnitude changing at the same rate). What emf and current are induced in the square coil?

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 long, current-carrying solenoid with an air core has 1750 turns per meter of length and a radius of \(0.0180 \mathrm{m}\). A coil of 125 turns is wrapped tightly around the outside of the solenoid, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system?

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 \(15.0 \mathrm{A} ?\)

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