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Chapter 27: Problem 48

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) 562.8 W. (c) 492.45 W.

Step by step solution

01

Understanding the Motor Circuit

The total voltage supplied to the motor is 120 V. The motor's internal resistance is 3.2 惟, and when running at full load, the EMF (back electromotive force) of the rotor is 105 V. The line voltage is divided between this EMF and the voltage drop across the internal resistance.
02

Calculating the Current (a)

To find the current, use Ohm's Law. The voltage on the internal resistance is the difference between the line voltage and the EMF: \[ V_{resistance} = V_{line} - E_{back} = 120 ext{ V} - 105 ext{ V} = 15 ext{ V} \]Then, using Ohm's Law, calculate the current: \[ I = \frac{V_{resistance}}{R} = \frac{15 ext{ V}}{3.2 \Omega} = 4.69 \text{ A} \]
03

Determining the Power Delivered to the Motor (b)

The power delivered to the motor is the total power input from the line. It can be calculated using the formula for electric power: \[ P_{input} = V_{line} \times I \]Substitute the known values:\[ P_{input} = 120 ext{ V} \times 4.69 ext{ A} = 562.8 ext{ W} \]
04

Calculating the Mechanical Power Developed by the Motor (c)

Mechanical power developed by the motor is equivalent to the power associated with the EMF, since EMF represents the conversion of electrical to mechanical energy:\[ P_{mechanical} = E_{back} \times I = 105 ext{ V} \times 4.69 ext{ A} = 492.45 \text{ W} \]

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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 a fundamental principle in electrical circuits, providing a simple relationship between voltage, current, and resistance. It states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points. The relationship is defined by the formula: \[ V = I \times R \] where \( V \) is the voltage in volts, \( I \) is the current in amperes, and \( R \) is the resistance in ohms.
  • This law serves as the foundation for calculating how much current flows through a circuit when the voltage and resistance are known.
  • It also helps in determining the necessary resistance to achieve a desired current flow.
In the context of a DC motor, Ohm's Law can be applied to find the current drawn from the line by considering the internal resistance and the back electromotive force (EMF) of the motor.
Electromotive Force (EMF)
Electromotive Force, or EMF, is a term used to describe the electrical potential produced by a cell or generator. Unlike its name implies, it is not a force; instead, it is the energy provided per charge. EMF is essential in driving current through circuits, particularly in motors.
  • In a motor, the back EMF opposes the input voltage, reducing the effective voltage that drives current through the motor.
  • This phenomenon arises because the motor's rotation induces an EMF, which works against the supply voltage.
The presence of back EMF in a DC motor is beneficial as it increases with speed, thus regulating the speed of the motor by decreasing the current as the motor spins faster. To calculate the current using EMF, understand that the effective voltage is the difference between the supplied voltage and the back EMF.
Mechanical Power
Mechanical power represents the rate at which work is done or energy is transferred in a mechanical system. In the context of motors, it is the power that is converted from electrical energy to useful mechanical work. Mechanical power is influenced heavily by the back EMF, which essentially acts as a measure of this conversion.
  • The mechanical power developed by the motor can be calculated using the formula: \[ P_{mechanical} = E_{back} \times I \]
  • Here, \( E_{back} \) represents the back EMF and \( I \) is the current.
Calculating mechanical power gives insight into the efficiency of the motor, allowing for understanding of how effectively electrical energy is turned into mechanical energy. For the exercise, understanding how back EMF contributes to mechanical power helps to comprehend why not all electrical power translates into mechanical output.
Electrical Circuits
Electrical circuits are networks that allow electric current to flow through them to perform work, power devices, or transfer information. They are comprised of different components such as resistors, inductors, capacitors, and active elements like batteries or motors.
  • In series circuits, all components are connected along a single pathway, meaning the same current flows through each component.
  • The total resistance in a series circuit is the sum of individual resistances and affects the total current flow.
When considering a DC motor in a circuit, especially where the rotor and field coils are in series, understanding the circuit configuration is crucial. Knowing how to apply principles such as Kirchhoff's laws and Ohm's Law, alongside understanding EMF, allows for solving complex circuit problems, like calculating total current or electrical power input in motors.

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Most popular questions from this chapter

A particle with negative charge \(q\) and mass \(m=2.58 \times\) 10 \(^{-15} \mathrm{kg}\) is traveling through a region containing a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=-(0.120 \mathrm{T}) \hat{\boldsymbol{k}} .\) At a particular instant of time the velocity of the particle is \(\overrightarrow{\boldsymbol{v}}=\left(1.05 \times 10^{6} \mathrm{m} / \mathrm{s}\right)(-3 \hat{\imath}+4 \hat{\jmath}+\) 12\(\hat{k} )\) and the force \(\vec{F}\) on the particle has a magnitude of 1.25 \(\mathrm{N}\) . (a) Determine the charge \(q\) . (b) Determine the acceleration \(\overrightarrow{\boldsymbol{d}}\) of the particle. (c) Explain why the path of the particle is a helix, and determine the radius of curvature \(R\) of the circular component of the helical path. (d) Determine the cyclotron frequency of the particle. (e) Although helical motion is not periodic in the full sense of the word, the \(x\) - and \(y\) -coordinates do vary in a periodic way. If the coordinates of the particle at \(t=0\) are \((x, y, z)=(R, 0,0),\) determine its coordinates at a time \(t=2 T,\) where \(T\) is the period of the motion in the \(x y\) -plane.

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