/*! 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 44 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
The current drawn by the motor from the line is 4.6875 A, the power delivered to the motor is 562.5 W, and the mechanical power developed by the motor is 492.1875 W.

Step by step solution

01

Calculate the Current Drawn by the Motor from the Line

We can apply Ohm's law here, which states that voltage is equal to the current times resistance (V = I * R). Given that we have the total voltage (120 V), which is the sum of the voltages across the field coil and the rotor, and also the resistance (3.2 Ω), we can rearrange the equation to calculate for the current. The total voltage is 120 V (voltage supplied from the line) and the sum of the internal voltage in the field coil and the electromotive force (emf) in the rotor. The internal voltage drop in the field coil can be calculated as total voltage minus emf in the rotor. So, current I = (Total Voltage - Emf) / Resistance = (120 V - 105 V) / 3.2 Ω = 4.6875 A.
02

Calculate the Power Delivered to the Motor

For DC circuits, the electrical power (P) is given by the product of the voltage (V) and the current (I) (P = V * I). We know the total voltage supplied to the motor (120 V) and we've calculated the current in the step 1 (4.6875 A). So, Power = Voltage * Current = 120 V * 4.6875 A = 562.5 W.
03

Calculate the Mechanical Power Developed by the Motor

The mechanical power developed by the motor is given by the product of the emf in the rotor and the current through the motor. We know the emf in the rotor (105 V) and we've calculated the current in the step 1 (4.6875 A). So, mechanical power = emf * current = 105 V * 4.6875 A = 492.1875 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. It is usually expressed by the formula \(V = I \times R\), where \(V\) represents voltage, \(I\) is the current, and \(R\) is the resistance. This law is crucial for understanding how electric circuits operate, as it allows us to calculate one property if the other two are known.

In the context of a DC motor, understanding Ohm's Law helps determine how much current is drawn from the power source. Knowing this helps us avoid issues like overheating or insufficient power delivery. When dealing with a circuit with known resistance, adjusting the voltage will directly affect the current.

This concept is particularly helpful for troubleshooting and optimizing electrical systems by ensuring each component receives the appropriate current without surpassing its limits.
Electromotive Force (emf)
Electromotive force, commonly abbreviated as emf, is a measure of the energy supplied by a source of electric current. It's essentially the voltage generated by a battery or a power supply when no current is flowing. In context, it represents the maximum potential difference a source can provide.

For a DC motor, the emf relates to the internal voltage that helps in producing mechanical rotation. Here, the sum of the emf and the internal resistive voltage drop is used to calculate the effective voltage applied across the motor. Understanding emf helps in maintaining efficient operation of the motor. If the emf is too low, the motor may underperform. Conversely, too high an emf may lead to overheating due to excessive power consumption.
Mechanical Power
Mechanical power in motors is the usable energy output, converted from electrical energy. This conversion is essential for the motor to perform work, such as turning a shaft or moving a load.

For the DC motor example here, mechanical power is calculated by multiplying the electromotive force (emf) by the current flowing through the motor: \(\text{Mechanical Power} = \text{emf} \times \text{Current}\).

This calculation shows how effectively a motor transforms incoming electrical energy into mechanical work. High mechanical power indicates efficient motor performance, whereas a large discrepancy between input electrical and output mechanical power might suggest inefficiencies, such as energy lost as heat.
Internal Resistance
Internal resistance refers to the natural opposition to current flow within the components of a circuit, such as the coils in a DC motor. This resistance affects the overall efficiency of any electrical device, as it causes a voltage drop across its elements.

In the case of the DC motor discussed, the resistance amounts to \(3.2 \Omega\). This internal resistance dictates how much voltage is "lost" inside the motor, impacting the overall voltage available to produce mechanical work. Accurately knowing and managing internal resistance ensures that components function correctly and helps in predicting how changes in external voltage will affect overall current and performance.
Current Calculation
Calculating current within a circuit is vital to ensure proper operation and safety. Based on Ohm's Law, the current can be determined when the total voltage and resistance are known.

For the DC motor scenario, the process involves using Ohm's law to determine the current by rearranging it to \(I = \frac{V}{R}\). Here, the total voltage is the supply voltage minus the electromagnetic force, reflecting the internal resistive losses: \(I = \frac{120 \mathrm{~V} - 105 \mathrm{~V}}{3.2 \Omega} = 4.6875 \, ext{A}\).

This calculation is crucial to assess how much current the motor requires from the power source, helping in system design and efficiency optimization by ensuring the motor's demand is suitably matched to its supply.
Power Calculation
Power in an electrical circuit represents the rate at which energy is used or produced. In terms of a DC motor, power calculations help understand both the input and output energies.Electrical power can be calculated using the formula \(P = V \times I\), where \(V\) is voltage and \(I\) is current. In the given example, it measures how much electrical energy is delivered to the motor; \(P = 120 \, \mathrm{V} \times 4.6875 \, \mathrm{A} = 562.5 \, \mathrm{W}\).

These calculations are significant in determining efficiency. For a well-functioning system, the input electrical power should closely match the useful mechanical power, minimizing losses and optimizing performance.

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 electron is moving in the \(x y\) -plane. If at time \(t\) a magnetic field \(B=0.200 \mathrm{~T}\) in the \(+z\) -direction exerts a force on the electron equal to \(F=5.50 \times 10^{-18} \mathrm{~N}\) in the \(-y\) -direction, what is the velocity (magnitude and direction) of the electron at this instant?

An electron traveling from the sun as part of the solar wind strikes the earth's magnetosphere at latitude \(80.0^{\circ} \mathrm{N}\) in a region where the magnetic field has a strength of \(15.0 \mu \mathrm{T}\) and is directed toward the earth's center. The electron has a speed of \(400 \mathrm{~km} / \mathrm{s}\) and is directed toward the earth's axis parallel to the equator. Magnetic forces send the electron on a helical trajectory. (a) What is the radius of this helix? (b) With what speed does the electron approach the surface of the earth? (c) If you are looking downward toward the earth from space, is the electron's motion clockwise or counterclockwise? (d) What is the frequency of the motion? (e) This electron strikes the ionosphere, where it is further accelerated by an electric field with strength \(20.0 \mathrm{mV} / \mathrm{m}\) directed northward parallel to the earth's surface. What is the electron's new speed after it has been deflected \(100 \mathrm{~km}\) southward by this field? (f) By what factor has its kinetic energy been increased by the electric field?

A straight, \(2.5 \mathrm{~m}\) wire carries a typical household current of \(1.5 \mathrm{~A}\) (in one direction) at a location where the earth's magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is \(155 \mathrm{~V} / \mathrm{m}\) and the magnetic field is \(0.0315 \mathrm{~T}\). The ions next enter a uniform magnetic field of magnitude \(0.0175 \mathrm{~T}\) that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is \(17.5 \mathrm{~cm},\) what is their mass?

A flat circular coil carrying a current of 8.80 A has a magnetic dipole moment of \(0.194 \mathrm{~A} \cdot \mathrm{m}^{2}\) to the left. Its area vector \(A\) is \(4.0 \mathrm{~cm}^{2}\) to the left. (a) How many turns does the coil have? (b) An observer is on the coil's axis to the left of the coil and is looking toward the coil. Does the observer see a clockwise or counterclockwise current? (c) If a huge \(45.0 \mathrm{~T}\) external magnetic field directed out of the paper is applied to the coil, what torque (magnitude and direction) results?

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.