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Chapter 22: Problem 36

A \(120.0-\mathrm{V}\) motor draws a current of 7.00 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) 114.96 V, (b) 166.67 A, (c) 7.28 惟 to limit initial current to 15.0 A.

Step by step solution

01

Understand the Problem

We need to calculate the back electromotive force (emf) for a motor running under normal conditions, find the initial current when the motor is turned on, and calculate the series resistance to limit the starting current.
02

Calculate Back EMF (a)

The back emf (\(E_b\)) can be calculated using the formula: \[V = E_b + I \times R\]where \(V\) is the supply voltage, \(I\) is the current, and \(R\) is the resistance. Rearranging this gives:\[E_b = V - I \times R\]Substitute \(V = 120.0 \, V\), \(I = 7.00 \, A\), and \(R = 0.720 \, \Omega\):\[E_b = 120.0 - (7.00 \times 0.720) = 120.0 - 5.04 = 114.96 \, V\]
03

Calculate Instantaneous Current (b)

When the motor is just switched on, the back emf is zero because the motor has not started rotating. Hence, the current \(I\) can be calculated using Ohm's law:\[I = \frac{V}{R}\]Substitute \(V = 120.0 \, V\) and \(R = 0.720 \, \Omega\):\[I = \frac{120.0}{0.720} = 166.67 \, A\]
04

Determine Series Resistance (c)

To limit the starting current \(I_{\text{start}}\) to \(15.0 \, A\), we use the formula:\[V = I_{\text{start}} \times (R + R_s)\]Rearrange this to find the series resistance \(R_s\):\[R_s = \frac{V}{I_{\text{start}}} - R\]Substitute \(V = 120.0 \, V\), \(I_{\text{start}} = 15.0 \, A\), and \(R = 0.720 \, \Omega\):\[R_s = \frac{120.0}{15.0} - 0.720 = 8.0 - 0.720 = 7.28 \, \Omega\]

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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 electronics and physics that helps us understand how current, voltage, and resistance interact in an electrical circuit. The law is usually expressed by the formula:\[ V = I \times R \]where:
  • \(V\) is the voltage across the circuit in volts (V).
  • \(I\) is the current flowing through the circuit in amperes (A).
  • \(R\) is the resistance within the circuit in ohms (惟).
Using Ohm's Law, we can determine how much current will flow through a circuit for a given voltage and resistance. In the context of a motor, this principle helps us understand the initial surge in current when the motor starts but is not yet producing back EMF, as the resistance is the only factor opposing the applied voltage at that instant.
In practical scenarios, adjusting resistance or voltage allows for control over the current, which is crucial in circuits with motors to manage performance and prevent damage.
Series Resistance
Series resistance refers to the resistors placed sequentially in a linear path in an electrical circuit. When resistors are connected in series, their resistances add up to give a total resistance, increasing the opposition to current flow. The formula for total resistance in a series configuration is:\[ R_{ ext{total}} = R_1 + R_2 + R_3 + ext{...} \]In the motor scenario, additional series resistance is crucial to limit the initial current surge. At start-up, a motor draws a very high current due to the absence of back EMF. By calculating the necessary extra resistance, engineers can ensure that the current remains at safe levels when the motor is turned on.
When the calculated additional resistance is added in series with the motor's resistance, it acts as a buffer, ensuring the starting current doesn鈥檛 exceed the safe limit, thus prolonging the life of the motor and associated components.
Current Limitation
Current limitation is essential to ensure the safe and efficient operation of electrical circuits, particularly in systems involving motors. At startup, motors draw a much larger current due to the lack of back EMF. This spike can cause overheating and damage if unmanaged. Limiting the start-up current is achieved by manipulating the circuit parameters such as series resistance or using current limiting devices like resistors.
The purpose of current limitation is to safeguard components and enhance longevity while maintaining performance standards. By understanding the initial demands of the motor and setting a current limit, we protect the system from the excessive current that can trigger circuit breakers or, worse, damage parts.
  • Current limitation reduces the risk of overheating and ensures compliance with the safety ratings of circuit components.
  • This is especially important in applications like motors, where the demand fluctuates dynamically.
Understanding these principles allows for better design and execution of electrical systems, ensuring safety and efficiency.

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Most popular questions from this chapter

Concept Questions The drawing shows a coil of copper wire that consists of two semicircles joined by straight sections of wire. In part \(a\) the coil is lying flat on a horizontal surface. The dashed line also lies in the plane of the horizontal surface. Starting from the orientation in part \(a,\) the smaller semicircle rotates at an angular frequency \(\omega\) about the dashed line, until its plane becomes perpendicular to the horizontal surface, as shown in part \(b\). A uniform magnetic field is constant in time and is directed upward, perpendicular to the horizontal surface. The field completely fills the region occupied by the coil in either part of the drawing. (a) In which part of the drawing, if either, does a greater magnetic flux pass through the coil? Account for your answer. (b) As the shape of the coil changes from that in part \(a\) of the drawing to that in part \(b\), does an induced current flow in the coil, and, if so, in which direction does it flow? Give your reasoning. To describe the flow, imagine that you are above the coil looking down at it. (c) How is the period \(T\) of the rotational motion related to the angular frequency \(\omega\), and in terms of the period, what is the shortest time interval that elapses between parts \(a\) and \(b\) of the drawing? Problem The magnitude of the magnetic field is \(0.35 \mathrm{~T}\). The resistance of the coil is \(0.025 \Omega,\) and the smaller semicircle has a radius of \(0.20 \mathrm{~m} .\) The angular frequency at which the small semicircle rotates is \(1.5 \mathrm{rad} / \mathrm{s} .\) Determine the average current, if any, induced in the coil as the coil changes shape from that in part \(a\) of the drawing to that in \(\operatorname{part} b\)

A vacuum cleaner is plugged into a \(120.0-\mathrm{V}\) socket and uses 3.0 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 loop of wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius \(r=0.20 \mathrm{~m} .\) The normal to the plane of the loop is parallel to a constant magnetic field \(\left(\phi=0^{\circ}\right)\) of magnitude \(0.75 \mathrm{~T}\). What is the change \(\Delta \Phi\) in the magnetic flux that passes through the loop when, starting with the position shown in the drawing, the semicircle is rotated through half a revolution?

Two \(0.68\) -m-long conducting rods are rotating at the same speed in opposite directions, and both are perpendicular to a 4.7-T magnetic field. As the drawing shows, the ends of these rods come to within \(1.0 \mathrm{~mm}\) of each other as they rotate. Moreover, the fixed ends about which the rods are rotating are connected by a wire, so these ends are at the same electric potential. If a potential difference of \(4.5 \times 10^{3} \mathrm{~V}\) is required to cause a \(1.0\) -mm spark in air, what is the angular speed (in \(\mathrm{rad} / \mathrm{s}\) ) of the rods when a spark jumps across the gap?

Electric doorbells found in many homes require \(10.0 \mathrm{~V}\) to operate. To obtain this voltage from the standard 120-V supply, a transformer is used. Is a step-up or a stepdown transformer needed, and what is its turns ratio \(N_{\mathrm{s}} / N_{\mathrm{p}}\) ?
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