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Chapter 22: Problem 40

A 120.0 -V motor draws a current of 7.00 \(\mathrm{A}\) when running at normal speed. The resistance of the armature wire is 0.720\(\Omega .\) (a) Determine the back emf generated by the motor. (b) What is the current at the instant when the motor is just turned on and has not begun to rotate? (c) What series resistance must be added to limit the starting current to 15.0 \(\mathrm{A}\) ?

Short Answer

Expert verified
(a) Back EMF is 114.96 V. (b) Initial current is 166.67 A. (c) Add 7.28 惟 resistance.

Step by step solution

01

Understanding Back EMF Concept

When a motor runs at normal speed, it generates a back electromotive force (EMF) that opposes the supply voltage. The net voltage across the armature is thus reduced by this back EMF. We calculate the back EMF given the supply voltage, current, and resistance.
02

Calculate Back EMF

Using Ohm's law, the net voltage across the motor armature is given by \( V = I \times R \). The back EMF (\( E_b \)) is then given by: \[ E_b = V_{supply} - I \times R \], where \( I = 7.00 \mathrm{A} \) and \( R = 0.720 \Omega \). Substituting the values: \[ E_b = 120.0 \mathrm{V} - (7.00 \mathrm{A} \times 0.720 \Omega) \]. Calculate \( E_b \).
03

Calculate Initial Starting Current

When the motor is just turned on and not yet rotating, the back EMF is zero. Hence, the entire supply voltage appears across the resistance. The initial current \( I_0 \) is calculated by Ohm's law: \( I_0 = \frac{V_{supply}}{R} \). Substitute \( V_{supply} = 120.0 \mathrm{V} \) and \( R = 0.720 \Omega \), and calculate \( I_0 \).
04

Determine Additional Series Resistance

To limit the initial current to 15.0 A, an additional series resistance \( R_s \) is required. Using Ohm's law, the total resistance \( R_{total} = \frac{V_{supply}}{I_{desired}} \), where \( I_{desired} = 15.0 \mathrm{A} \). The additional resistance is \( R_s = R_{total} - R \). Calculate \( R_s \).
05

Substitute and Simplify Formulas

For back EMF: \[ E_b = 120.0 \mathrm{V} - (7.00 \mathrm{A} \times 0.720 \Omega) = 120.0 \mathrm{V} - 5.04 \mathrm{V} = 114.96 \mathrm{V} \]. For initial current: \[ I_0 = \frac{120.0 \mathrm{V}}{0.720 \Omega} = 166.67 \mathrm{A} \]. For additional resistance: \[ R_{total} = \frac{120.0 \mathrm{V}}{15.0 \mathrm{A}} = 8.0 \Omega \], \( R_s = 8.0 \Omega - 0.720 \Omega = 7.28 \Omega \).

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Key Concepts

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

Back Electromotive Force (EMF)
In electric motors, the back electromotive force (EMF) is a critical concept that often puzzles students. As a motor starts to rotate, it generates its own voltage called back EMF. This voltage opposes the supplied voltage from the power source. When the motor spins faster, the back EMF increases, reducing the net voltage across the motor armature. To calculate the back EMF, we need to account for this opposing voltage. The formula to find back EMF is given by:\[ E_b = V_{supply} - I \times R \]Where:
  • \(E_b\) is the back EMF,
  • \(V_{supply}\) is the supply voltage,
  • \(I\) is the current through the motor,
  • \(R\) is the resistance of the motor circuit.
Understanding back EMF is essential as it plays a vital role in controlling the motor's speed and efficiency.
Ohm's Law
Ohm's Law is a fundamental principle that describes the relationship between voltage, current, and resistance in electrical circuits. It is typically written as:\[ V = I \times R \]Where:
  • \(V\) is the voltage across the circuit,
  • \(I\) refers to the current flowing through the circuit,
  • \(R\) is the resistance.
In the context of electric motors, Ohm's Law helps us understand how different factors such as resistance and back EMF affect the current flowing through the motor. By using this fundamental equation, we can determine how changes in any of these parameters affect the other elements of the circuit.
This allows us to solve for unknown values and achieve desired performance efficiently.
Circuit Resistance
Circuit resistance, particularly in electric motors, is the measure of how much the circuit opposes the flow of electric current. It is determined by the materials and design of the circuit components. Resistance can be calculated using Ohm's Law:\[ R = \frac{V}{I} \]For motors, the armature wire resistance and any additional series resistance make up the total circuit resistance. Managing resistance is crucial for controlling motor performance:
  • High resistance decreases current flow, reducing motor speed.
  • Low resistance increases current flow, risking damage from high currents.
      • Thus, understanding and calculating total circuit resistance helps in designing motors that operate safely and efficiently under varying conditions.
Motor Current
Motor current refers to the electric current flowing through the motor when it is operational. Understanding motor current is crucial because it directly impacts the motor's behavior.When the motor just starts, the current is at its maximum because there is no back EMF yet to oppose the supply voltage. This initial current can be calculated by:\[ I_0 = \frac{V_{supply}}{R} \]Where \(V_{supply}\) is the full voltage because back EMF doesn't act immediately, and \(R\) is the resistance.Proper control of this motor current is important:
  • Ensures the motor starts without electrical damage.
  • Maintains efficiency once the motor reaches normal operating speed.
Armature Wire Resistance
Armature wire resistance is the resistance within the winding of the motor's armature. This internal resistance is crucial to consider because it impacts how the motor performs. It is one of the main components of the motor circuit's total resistance. Typically, this resistance is relatively low but can significantly affect the current and voltage when combined with other resistances in the circuit.In calculations, armature wire resistance is denoted as \(R\) and directly affects both the back EMF and initial current:
  • Back EMF: The greater the armature resistance, the more it influences the current, impacting the back EMF indirectly.
  • Initial Current: As the resistance increases, the current decreases, influencing the starting behavior of the motor.
Understanding the role of armature wire resistance will help in making informed decisions regarding motor design and troubleshooting.

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Most popular questions from this chapter

A copper rod is sliding on two conducting rails that make an angle of 19 with respect to each other, as in the drawing. The rod is moving to the right with a constant speed of 0.60 \(\mathrm{m} / \mathrm{s} .\) A \(0.38-\mathrm{T}\) uniform magnetic field is perpendicular to the plane of the paper. Determine the magnitude of the average emf induced in the triangle \(A B C\) during the 6.0 -s period after the rod has passed point \(A\) .

A 0.80 -m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the bar hits the river below, its speed is 22 \(\mathrm{m} / \mathrm{s}\) , and the emf induced across its length is \(6.5 \times 10^{-4} \mathrm{V}\) . Assuming the horizontal component of the earth's magnetic field at the location of the bar points directly north, (a) determine the magnitude of the horizontal component of the earth's magnetic field, and (b) state whether the east end or the west end of the bar is positive.

ssm The drawing shows three identical rods (A, B, and C) moving in different planes. A constant magnetic field of magnitude 0.45 T is directed along the \(+y\) axis. The length of each rod is \(L=1.3 \mathrm{m},\) and the rods each have the same speed, \(v_{\mathrm{A}}=v_{\mathrm{B}}=v_{\mathrm{C}}=2.7 \mathrm{m} / \mathrm{s}\) . For each rod, find the magnitude of the motional emf, and indicate which end (1 or 2 ) of the rod is positive.

A vacuum cleaner is plugged into a \(120.0-\mathrm{V}\) socket and uses 3.0 \(\mathrm{A}\) of current in normal operation when the back emf generated by the electric motor is 72.0 \(\mathrm{V}\) . Find the coil resistance of the motor.

A motor is designed to operate on 117 \(\mathrm{V}\) and draws a current of 12.2 \(\mathrm{A}\) when it first starts up. At its normal operating speed, the motor draws a current of 2.30 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third of the normal speed.
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