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Chapter 7: Problem 30

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) 492.5 W.

Step by step solution

01

Understand the Circuit

The motor is connected in series, which means the same current flows through the internal resistance and the rest of the motor. We know the total voltage supplied is 120 V, the internal resistance is 3.2 惟, and the emf across the rotor is 105 V.
02

Calculate the Voltage Across the Internal Resistance

Find the voltage drop across the internal resistance using the equation: \(V = V_{total} - ext{emf}\). Substituting the given values, we have:\[V = 120 ext{ V} - 105 ext{ V} = 15 ext{ V}\].
03

Calculate the Current Drawn by the Motor

Use Ohm's Law, \(V = IR\), to find the current \(I\) drawn from the line. Rearrange the equation to solve for \(I\):\[I = \frac{V}{R} = \frac{15 ext{ V}}{3.2 \, \Omega} \approx 4.69 \,A\].
04

Calculate the Power Delivered to the Motor

The total power \(P\) delivered to the motor is given by the formula:\[P = V_{total} \times I\]. Substituting the known values:\[P = 120 ext{ V} \times 4.69 \, ext{A} \approx 562.8 \, ext{W}\].
05

Calculate the Mechanical Power Developed

The mechanical power developed by the motor is the power associated with the back emf (rotor emf). Use the formula:\[P_{mech} = ext{emf} \times I\]. Using the given values:\[P_{mech} = 105 ext{ V} \times 4.69 \, ext{A} \approx 492.5 \, ext{W}\].

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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 used to understand electrical circuits and devices like the DC motor in our example. This law forms the foundation for analyzing electrical circuits by relating voltage (V), current (I), and resistance (R). The law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points and is inversely proportional to the resistance between them. It is expressed through the formula:
  • \( V = I \times R \)
In our exercise, after finding the voltage drop across the internal resistance of the motor (15 V), Ohm's Law was used to calculate the current, leading to the understanding that 4.69 A is the amount drawn from the line. This connection between voltage, current, and resistance allows for precise calculations essential in the design and function of circuits.
Electrical Resistance
Electrical resistance is a measure of the opposition to the flow of electric current through a conductor. In simpler terms, it determines how easily current can pass through a material. Resistance is influenced by factors such as the material's properties and temperature. In circuits, resistance is vital because it affects how much current will flow for a given voltage.
  • Resistance is measured in ohms (\( \Omega \)).
  • Higher resistance means less current for the same voltage.
  • Resistance in components is crucial to controlling and distributing electricity efficiently.
In the context of our DC motor, the internal resistance of 3.2 \( \Omega \) significantly influences how much current circulates. Understanding the role of resistance helps determine the motor's efficiency and performance when using energy from the power supply. It essentially dictates the energy losses within electrical devices.
Mechanical Power
Mechanical power in the context of an electric motor refers to the amount of power being converted from electrical to mechanical energy. Understanding this conversion helps comprehend how motors work in moving or performing tasks with force and speed. The mechanical power output is a vital factor in assessing the efficiency and capability of a motor.
  • It is calculated using the formula: \( P_{mech} = \text{emf} \times I \).
  • In our example, the back emf (105 V) is crucial in determining mechanical output, as this is the actual voltage converted to produce work.
In the case of our DC motor, with a calculated mechanical power of around 492.5 W, it implies that despite the overall richer power input, only a portion is effectively utilized for doing mechanical work. This illustrates the power losses due to internal resistances and inefficiencies. The more mechanical power a motor can develop, the more effective it is at performing its designed tasks.

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Most popular questions from this chapter

A particle with charge \(q\) is moving with speed \(v\) in the \(-y\) -direction. It is moving in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=\) \(B_{x} \hat{\imath}+B_{y} \hat{J}+B_{z} \hat{k} .\) (a) What are the components of the force \(\vec{F}\) exerted on the particle by the magnetic field? (b) If \(q>0\), what must the signs of the components of \(\overrightarrow{\boldsymbol{B}}\) be if the components of \(\overrightarrow{\boldsymbol{F}}\) are all nonnegative? (c) If \(q<0\) and \(B_{x}=B_{y}=B_{z}>0\), find the direction of \(\overrightarrow{\boldsymbol{F}}\) and find the magnitude of \(\overrightarrow{\boldsymbol{F}}\) in terms of \(|q|, v\), and \(B_{x}\)

A particle of charge \(q>0\) is moving at speed \(v\) in the \(+z\) -direction through a region of uniform magnetic field \(\overrightarrow{\boldsymbol{B}}\). The magnetic force on the particle is \(\overrightarrow{\boldsymbol{F}}=F_{0}(3 \hat{\boldsymbol{\imath}}+4 \hat{\boldsymbol{j}})\), 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 \(\overrightarrow{\boldsymbol{B}}\).

Torque on a Current Loop in a Nonuniform Magnetic Field. In Section \(7.7\) the expression for the torque on a current loop was derived assuming that the magnetic field \(\vec{B}\) was uniform. But what if \(\vec{B}\) is not uniform? Figure P7.49 shows a square loop of wire that lies in the \(x y\)-plane. The loop has comers at \((0,0)\), \((0, L),(L, 0)\), and \((L, L)\) and carries a constant current \(I\) in the clockwise direction. The magnetic field has no z-component but has both \(x\) - and \(y\)-components: \(\overrightarrow{\boldsymbol{B}}=\left(B_{0} y / L\right) \hat{i}+\left(B_{0} x / L\right) \hat{j}\), where \(B_{0}\) is a positive constant. (a) Sketch the magnetic field lines in the \(x y-\) plane. (b) Find the magnitude and direction of the magnetic force exerted on each of the sides of the loop by integrating Eq. (7.20). (c) If the loop is free to rotate about the \(x\)-axis, find the magnitude and direction of the magnetic torque on the loop. (d) Repeat part (c) for the case in which the loop is free to rotate about the \(y\)-axis. (e) Is Eq. (7.26), \(\overrightarrow{\boldsymbol{\tau}}=\overrightarrow{\boldsymbol{\mu}} \times \overrightarrow{\boldsymbol{B}}\), an appropriate description of the torque on this loop? Why or why not?

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at \(1.50 \mathrm{~km} / \mathrm{s}\) in the \(+x\)-direction experiences a force of \(2.25 \times 10^{-16} \mathrm{~N}\) in the \(+y\)-direction, and an electron moving at \(4.75 \mathrm{~km} / \mathrm{s}\) in the \(-z\)-direction experiences a force of \(8.50 \times\) \(10^{-16} \mathrm{~N}\) in the \(+y\)-direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the \(-y\)-direction at \(3.20 \mathrm{~km} / \mathrm{s} ?\)

A particle having mass \(m\), charge \(q\) enters a cylinder region having uniform magnetic field \(B\) in the inward direction as shown. If the particle is deviated by \(60^{\circ}\) as it emerges out of the field then what is the time spent by it in the field.
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