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Chapter 20: Problem 24

The starter motor in an automobile has a resistance of \(0.40 \Omega\) in its armature windings. The motor operates on \(12 \mathrm{~V}\) and has a back emf of \(10 \mathrm{~V}\) when running at normal operating speed. How much current does the motor draw (a) when running at its operating speed, (b) when running at half its final rotational speed, and (c) when starting up?

Short Answer

Expert verified
(a) 5 A, (b) 17.5 A, (c) 30 A.

Step by step solution

01

Understand Ohm's Law

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points. The formula for this is given by \( I = \frac{V}{R} \), where \( I \) is the current, \( V \) is the voltage, and \( R \) is the resistance.
02

Determine Current at Normal Operating Speed

When running at normal operating speed, the back emf is 10 V. The effective voltage across the motor is: \( V = 12 V - 10 V = 2 V \). Using Ohm's Law, we calculate the current: \( I = \frac{2 V}{0.40 \Omega} = 5 \) A.
03

Find Current at Half of Final Speed

At half the final speed, the back emf is half as well, i.e., \( \frac{10}{2} = 5 \) V. The effective voltage becomes: \( V = 12 V - 5 V = 7 V \). Calculate the current: \( I = \frac{7 V}{0.40 \Omega} = 17.5 \) A.
04

Calculate Current When Starting Up

When the motor is starting, the back emf is 0 because the motor hasn't moved yet. Therefore, the total voltage is applied across the resistance: \( V = 12 V \). Calculate the starting current: \( I = \frac{12 V}{0.40 \Omega} = 30 \) A.

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Key Concepts

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

Electric Current
Electric current is the rate at which electric charge flows through a conductor. It's kind of like a river of electricity moving through wires or other materials. The unit for measuring electric current is the ampere, often just called an "amp." This is abbreviated as A. Electric current is crucial in circuits because it determines how fast or how much electricity flows through.
It's driven by a voltage, making the electrons move. Think of electric current as the flow of water through a pipe. The more current there is, the more electrons are flowing. In the example of a starter motor, current determines how quickly energy is delivered to turn the motor.
  • Measured in amperes (A)
  • Driven by voltage
  • Flows in a loop or circuit
Resistance
Resistance is an essential concept in understanding how electricity flows through materials. It represents how much a material opposes or "resists" the flow of electric current. With higher resistance, less current flows through a conductor. Resistance is measured in ohms, symbolized by the Greek letter \( \Omega \).
In our starter motor example, the resistance is given as \(0.40 \Omega\). Knowing the resistance allows us to use Ohm's Law. This helps us calculate the current flow for different voltages or conditions.
  • Measured in ohms (\(\Omega\))
  • More resistance means lower current flow
  • Ohm's Law helps connect resistance, voltage, and current
Back Electromotive Force (emf)
Back electromotive force, or back emf, is a fascinating phenomenon that occurs in electric motors when they operate. As the motor spins, it generates its own voltage, opposing the applied voltage. This happens because the motor, while running, acts like a generator.
Back emf is crucial for controlling the speed of the motor. At the normal operating speed, the back emf reduces the effective voltage, which decreases the current. In our motor problem, the back emf is \(10 \text{ V}\) at full speed, thus reducing the voltage to \(2 \text{ V}\).
  • Created by the motor's movement
  • Opposes the applied voltage
  • Affects motor speed by changing current flow
Voltage
Voltage, often referred to as electric potential difference, is the force that pushes electric current through a conductor. It's like the pressure that pushes water through a hose. Measured in volts (V), voltage is crucial in determining how much work electric charges can do.
In the motor tasks, a \(12 \text{ V}\) battery provides the initial push. However, due to back emf, the effective voltage changes through different operational stages. This adjustment in voltage, from \(12 \text{ V}\) down to \(2 \text{ V}\) when running at full speed, helps us calculate different current values.
  • Measured in volts (V)
  • Drives the electric current
  • Varies due to components like back emf in a circuit

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Most popular questions from this chapter

An ac generator operates at a rotational frequency of \(60 \mathrm{~Hz}\) and produces a maximum emf of \(100 \mathrm{~V}\). Assume that its output at \(t=0\) is zero. What is the instantaneous \(\mathrm{emf}(\mathrm{a})\) at \(t=1 / 240 \mathrm{~s} ?(\mathrm{~b})\) at \(t=1 / 120 \mathrm{~s} ?(\mathrm{c})\) at \(t ?\) (d) How much time elapses between successive 0 -volt outputs? (e) What maximum emf would this generator produce if it were operated, instead, at \(120 \mathrm{~Hz} ?\)

The flux through a loop of wire changes uniformly from \(+40 \mathrm{~Wb}\) to \(-20 \mathrm{~Wb}\) in \(1.5 \mathrm{~ms}\). (a) What is the significance of the negative number attached to the final flux value? (b) What is the average induced emf in the loop? (c) If you wanted to double the average induced emf by changing only the time, what would the new time interval be? (d) If you wanted to double the average induced emf by changing only the final flux value, what would it be?

The transformer in the power supply for a computer's 500-GB external hard drive changes a \(120-\mathrm{V}\) input voltage (from a regular house line) to the 5.0 - \(\mathrm{V}\) output voltage that is required to power the drive. (a) Find the ratio of the number of turns in the primary coil to the number of turns in the secondary coil. (b) If the drive is rated at \(10 \mathrm{~W}\) when running and the transformer is ideal, what is the current in the primary and secondary when the drive is in operation?

A flat coil of copper wire consists of 100 loops and has a total resistance of \(0.500 \Omega\). The coil diameter is \(4.00 \mathrm{~cm}\) and it is in a uniform magnetic field pointing toward you (out the page). The coil orientation is in the plane of the page. It is then pulled to the right (without rotating) until it is completely out of the field. (a) What is the direction of the induced current in the coil: (1) clockwise, (2) counterclockwise, or (3) there is no induced current? (b) During the time the coil leaves the field, an average induced current of \(20.0 \mathrm{~mA}\) is measured. What is the average induced emf in the coil? (c) If the field strength is \(5.50 \mathrm{mT}\), how much time did it take to pull the coil out?

(a) A square loop of wire with sides of length \(40 \mathrm{~cm}\) is in a uniform magnetic field perpendicular to its area. If the field's strength is initially \(100 \mathrm{mT}\) and it decays to zero in \(0.010 \mathrm{~s}\), what is the magnitude of the average emf induced in the loop? (b) What would be the average emf if the sides of the loop were only \(20 \mathrm{~cm} ?\)
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