Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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-V line, the emf in the rotor is 105 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) 562.8 W; (c) 492.45 W.

Step by step solution

01

Identify Known Values

We are given that the internal resistance \( R \) of the motor is 3.2 \( \Omega \), the line voltage \( V \) is 120 V, and the electromotive force (emf) \( E \) is 105 V.
02

Apply Ohm's Law to Find Current

Use the equation for a circuit with resistance, \( I = \frac{V - E}{R} \). Substitute the known values: \( I = \frac{120 \text{ V} - 105 \text{ V}}{3.2 \Omega} = \frac{15}{3.2} \approx 4.69 \text{ A}. \) Thus, the current drawn by the motor is approximately 4.69 A.
03

Calculate Power Delivered to the Motor

The power delivered to the motor \( P \) can be found using the formula \( P = VI \). Substitute the known values: \( P = 120 \, \text{V} \times 4.69 \text{ A} \approx 562.8 \text{ W}. \) This is the electrical power going into the motor.
04

Determine Mechanical Power Developed by the Motor

The mechanical power developed by the motor can be calculated as \( P_{mech} = EI \). Substitute the known values: \( P_{mech} = 105 \, \text{V} \times 4.69 \, \text{A} \approx 492.45 \, \text{W}. \) This is the output mechanical power of the motor.

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 Vaia!

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 the voltage, current, and resistance in an electrical circuit. In simple terms, Ohm's Law states that the current (\( I \)) flowing through a conductor between two points is directly proportional to the voltage (\( V \)) across the two points and inversely proportional to the resistance (\( R \)) of the conductor. This can be mathematically expressed as:
  • \( I = \frac{V}{R} \)
In the context of DC motors, Ohm's Law helps us determine the current drawn by the motor from the supply line, taking into account any internal resistance and electromotive force present. By rearranging the formula to accommodate these additional factors, we can confirm the motor's current involvement in running efficiently.
Always remember to ensure all units are consistent when applying Ohm's Law and in practical circuit analysis scenarios.
Electromotive force
Electromotive force (emf) is a crucial concept in understanding how DC motors operate. It represents the energy provided by a power source, like a battery or generator, to a charge moving in a circuit. Despite its name, emf is not actually a force but rather a potential difference measured in volts (V). It is the driving factor that moves electrons around a circuit, allowing electric current to flow.
For DC motors, the emf can be thought of as the back electromotive force, which acts against the supply voltage when the motor is in operation. This is because the motor acts like a generator when it is rotating, producing its own voltage as a response to the motion, commonly known as back emf. Understanding emf is key to calculating how much voltage is effectively applied to overcome the internal resistance and perform work.
Power Calculation
Power calculation in electrical circuits is about determining the rate at which energy is transferred or converted. In terms of DC motors, we often calculate two types of power: electrical power supplied to the motor and mechanical power output by the motor.
  • Electrical power (\( P \)) is calculated using the formula \( P = VI \), where \( V \) is voltage and \( I \) is current.
  • For mechanical power, the formula \( P_{mech} = EI \) is used, where \( E \) is the electromotive force.
Accurately calculating these powers helps in assessing the motor efficiency and its operation under a given load. Understanding how to navigate between these types of power is fundamental to ensuring the motor's performance aligns with expectations.
Internal Resistance
Internal resistance is a natural resistance within electrical components that hinders current flow. In a DC motor, internal resistance refers to the cumulative resistance of the wires and components inside the motor.
It plays a vital role when analyzing circuits and can significantly affect motor performance. Increased internal resistance reduces the net voltage available for mechanical work, hence decreasing efficiency.
In the given exercise, identifying the internal resistance is crucial for calculating the current flow and the power delivered to the motor, as it enables us to precisely determine how much of the input energy is actually being converted to perform useful work.
Mechanical Power
Mechanical power in the context of DC motors is the useful power output that transforms electrical energy into mechanical energy. This power reflects how effectively the motor converts electrical inputs to perform work, such as rotating shafts or driving a load.
To find the mechanical power, we use the formula\[ P_{mech} = EI \]where \( E \) is the electromotive force of the motor and \( I \) is the current. High mechanical power indicates efficient conversion, while low values may point to energy losses.
Hence, understanding mechanical power helps in evaluating motor performance, ensuring that the motor meets design specifications and operates effectively in real-world applications.

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

A singly charged ion of \(^7\)Li (an isotope of lithium) has a mass of 1.16 \(\times\) 10\(^{-26}\) kg. It is accelerated through a potential difference of 220 V and then enters a magnetic field with magnitude 0.874 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?

A particle with charge -5.60 nC is moving in a uniform magnetic field \(\overrightarrow{B} =\) -(1.25 T)\(\hat{k}\). The magnetic force on the particle is measured to be \(\overrightarrow{F} =\) -(3.40 \(\times\) 10\(^{-7}\)N)\(\hat{\imath}\) + (7.40 \(\times\) 10\(^{-7}\)N)\(\hat{\jmath}\). (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{v}\) \(\cdot\) \(\overrightarrow{F}\). What is the angle between \(\vec{v}\) and \(\overrightarrow{F}\)?

A particle of charge \(q\) > 0 is moving at speed v in the \(+z\)-direction through a region of uniform magnetic field \(\overrightarrow{B}\). The magnetic force on the particle is \(\overrightarrow{F} =\) \(F_0\)(3\(\hat{\imath}\) + 4 \(\hat{\jmath}\)), 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\(F_0/qv\), determine as much as you can about the remaining components of \(\overrightarrow{B}\).

An electron moves at 1.40 \(\times\) 10\(^6\) m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.40 \(\times\) 10\(^{-2}\) T. (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?

The plane of a 5.0 cm \(\times\) 8.0 cm rectangular loop of wire is parallel to a 0.19-T magnetic field. The loop carries a current of 6.2 A. (a) What torque acts on the loop? (b) What is the magnetic moment of the loop? (c) What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field?

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.

Sign-up for free