/*! 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 50 A dc motor with its rotor and fi... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2 \Omega. When the motor is running at full load on a \(120 - \mathrm { V }\) line, the emf in the rotor is 105\(\mathrm { V }\) . (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

Short Answer

Expert verified
(a) 4.69 A; (b) 562.5 W; (c) 492.19 W.

Step by step solution

01

Identify the Known Values

We are given:- Internal resistance of the motor, \( R = 3.2 \, \Omega \).- Supply voltage, \( V = 120 \, \mathrm{V} \).- Emf in the rotor, \( E = 105 \, \mathrm{V} \).
02

Calculate the Current Drawn By the Motor

Using Ohm's Law, we know that the total voltage across the motor's resistance is the supply voltage minus the rotor emf:\[V - E = I \times R\]Substitute the known values:\[120 - 105 = I \times 3.2\]Solving for \( I \):\[I = \frac{15}{3.2} \approx 4.6875 \, \mathrm{A}\]
03

Calculate the Power Delivered to the Motor

Power delivered to the motor can be calculated using:\[P = V \times I\]Substitute the known values:\[P = 120 \times 4.6875 \approx 562.5 \, \mathrm{W}\]
04

Calculate the Mechanical Power Developed by the Motor

The mechanical power developed is the power associated with the rotor emf:\[P_{mech} = E \times I\]Substitute the known values:\[P_{mech} = 105 \times 4.6875 \approx 492.1875 \, \mathrm{W}\]

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 91Ó°ÊÓ!

Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in electrical engineering that relates voltage, current, and resistance in an electrical circuit. It can be expressed with the formula:\[ V = I \times R \]where:
  • \(V\) represents voltage in volts (V)
  • \(I\) is the current in amperes (A)
  • \(R\) is the resistance in ohms (\(\Omega\))
In the context of a DC motor exercise, Ohm's Law helps calculate the current drawn from the power supply. By rearranging the formula, the current \(I\) can be found as:\[ I = \frac{V - E}{R} \]Here, \(V\) is the supply voltage, \(E\) is the electromotive force (emf) induced in the motor, and \(R\) is the internal resistance.
Understanding these relationships is crucial because it allows us to determine how much current flows in the circuit when a particular voltage is applied. It also helps identify potential issues like increased resistance that may affect motor performance.
Power Calculation
Power calculation in a DC motor involves understanding how much energy is consumed and how much is converted into useful mechanical energy. Power can be calculated using the formula:\[ P = V \times I \]This equation calculates the total power delivered to the motor. The total power input depends on the supply voltage \(V\) and the current \(I\).
However, not all supplied power is converted into mechanical power. In any electrical device, some energy is lost as heat due to resistance. Thus, the mechanical power developed by the motor can be determined using the rotor's electromotive force (emf):\[ P_{mech} = E \times I \]where \(E\) is the emf in volts.
These calculations are critical to ensure the motor operates efficiently.
  • High losses may indicate maintenance needs or inefficiencies in design.
  • Knowing both total and mechanical power allows engineers to optimize motor design and energy consumption.
Internal Resistance
Internal resistance in a DC motor refers to the opposition to current flow within the motor itself. It includes the resistance of wires and components like rotor and field coils. This internal resistance is an integral factor when analyzing circuit performance.
In our DC motor task, internal resistance \(R\) affects the voltage drop across the motor, calculated as:\[ V - E = I \times R \]A high internal resistance can lead to significant power loss as heat, reducing the efficiency of energy conversion from electrical to mechanical.
Key aspects of internal resistance to consider include:
  • It can increase with temperature as materials expand and change their conductive properties.
  • Aging can also lead to higher resistance over time due to wear and oxidation.
  • Regular assessments of internal resistance help maintain motor efficiency and performance.
Understanding and managing internal resistance is key to ensuring the prolonged and optimal performance of a DC motor.

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

You wish to hit a target from several meters away with a charged coin having a mass of 4.25\(\mathrm { g }\) and a charge of \(+ 2500 \mu \mathrm { C }\) . The coin is given an initial velocity of 12.8\(\mathrm { m } / \mathrm { s }\) , and a downward, uniform electric field with field strength 27.5\(\mathrm { N } / \mathrm { C }\) exists through-out the region. If you aim directly at the target and fire the coin horizontally, what magnitude and direction of uniform magnetic field are needed in the region for the coin to hit the target?

Paleoclimate. Climatologists can determine the past temperature of the earth by comparing the ratio of the isotope oxygen-18 to the isotope oxygen-16 in air trapped in ancient ice sheets, such as those in Greenland. In one method for separating these isotopes, a sample containing both of them is first singly ionized (one electron is removed) and then accelerated from rest through a potential difference \(V\) . This beam then enters a magnetic field \(B\) at right angles to the field and is bent into a quarter-circle. A particle detector at the end of the path measures the amount of each isotope. (a) Show that the separation \(\Delta r\) of the two isotopes at the detector is given by $$\Delta r = \frac { \sqrt { 2 e V } } { e B } \left( \sqrt { m _ { 18 } } - \sqrt { m _ { 16 } } \right)$$ where \(m _ { 16 }\) and \(m _ { 18 }\) are the masses of the two oxygen isotopes, (b) The measured masses of the two isotopes are \(2.66 \times\) \(10 ^ { - 26 } \mathrm { kg } \left( ^ { 16 } \mathrm { O } \right)\) and \(2.99 \times 10 ^ { - 26 } \mathrm { kg } ( 8 \mathrm { O } ) .\) If the magnetic field is \(0.050 \mathrm { T } ,\) what must be the accelerating potential \(V\) so that these two isotopes will be separated by 4.00\(\mathrm { cm }\) at the detector?

A particle of charge \(q > 0\) is moving at speed \(v\) in the \(+ z\) -direction through a region of uniform magnetic field \(\vec { \boldsymbol { B } }\) . The magnetic force on the particle is \(\vec { \boldsymbol { F } } = F _ { 0 } ( 3 \hat { \boldsymbol { \imath } } + 4 \hat { \boldsymbol { J } } ) ,\) where \(F _ { 0 }\) is a positive constant. (a) Determine the components \(B _ { x } , B _ { y } ,\) and \(B _ { z }\) , or at least as many of the three components as is possible from the information given. (b) If it is given in addition that the magnetic field has magnitude \(6 F _ { 0 } / q v ,\) determine as much as you can about the remaining components of \(\vec { B } .\)

An electron experiences a magnetic force of magnitude \(4.60 \times 10^{-15} \mathrm{N}\) when moving at an angle of \(60.0^{\circ}\) with respect to a magnetic field of magnitude \(3.50 \times 10^{-3} \mathrm{T.}\) Find the speed of the electron.

A straight piece of conducting wire with mass \(M\) and length \(L\) is placed on a friction less incline tilted at an angle \(\theta\) from the horizontal (Fig. P27.69). There is a uniform, vertical magnetic field \(\vec { B }\) at all points (produced by an arrangement of magnets not shown in the figure). To keep the wire from sliding down the incline, a voltage source is attached to the ends of the wire. When just the right amount of current flows through the wire, the wire remains at rest. Determine the magnitude and direction of the current in the wire that will cause the wire to remain at rest. Copy the figure and draw the direction of the current on your copy. In addition, show in a free-body diagram all the forces that act on the wire.

See all solutions

Recommended explanations on Physics Textbooks

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.