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Chapter 22: Problem 76

At its normal operating speed, an electric fan motor draws only 15.0\(\%\) of the current it draws when it just begins to turn the fan blade. The fan is plugged into a \(120.0-\mathrm{V}\) socket. What back emf does the motor generate at its normal operating speed?

Short Answer

Expert verified
The back emf is 102.0 V.

Step by step solution

01

Understanding the Problem

We need to find the back electromotive force (emf) generated by the motor at its normal operating speed. We know that back emf reduces the current flowing through the motor.
02

Identifying Given Values

The electric motor draws 15.0\(\%\) of its initial current at normal speed. The voltage supply is \(120.0\,\text{V}\).
03

Defining Variables

Let \(I_0\) be the initial current when the fan starts, and \(I_n\) be the normal operating current. Thus, \(I_n = 0.15I_0\).
04

Back EMF Formula

The back emf \(E_b\) when a motor is running is given by \(E_b = V - I_nR\), where \(V\) is supply voltage and \(R\) is the resistance. We also have \(I_0 = \frac{V}{R}\).
05

Substitute and Simplify

Substitute \(I_n = 0.15I_0\) and \(I_0 = \frac{V}{R}\) into the formula: \[E_b = V - 0.15\left(\frac{V}{R}\right)R = V - 0.15V\]
06

Calculate Back EMF

Simplify: \[E_b = 1.00V - 0.15V = 0.85V\] Substitute \(V = 120.0\,\text{V}\):\[E_b = 0.85 \times 120.0\,\text{V} = 102.0\,\text{V}\]

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Key Concepts

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

Electric Fan Motor
An electric fan motor is a fascinating device that converts electrical energy into mechanical energy to make the fan blades rotate. It operates on the principle of electromagnetism. Inside the motor, you will find windings through which electrical current flows, generating a magnetic field. This field interacts with other magnets or magnetic fields inside the motor, creating rotational motion.

When you first turn on an electric fan, the motor must overcome the initial inertia of the stationary blades. At this point, the electric fan motor draws the most current because it needs a strong magnetic force to start moving the blades. Understanding this concept helps clarify why the current is highest when the fan is just switched on and decreases once the fan reaches its normal operating speed.
  • Motor converts electrical to mechanical energy
  • Current is highest at startup due to inertia
  • Relies on principles of electromagnetism
Current Reduction
As the electric fan reaches its normal operating speed, the current drawn by the motor significantly reduces. This phenomenon occurs due to the back electromotive force (back emf) that is generated. Back emf is the voltage that acts in the opposite direction to the supply voltage. It is induced by the motor's rotation and effectively opposes the incoming current.

At low speeds, as the fan is starting, back emf is minimal, and thus the motor uses a lot more current. As the motor speeds up, the back emf increases, resulting in reduced current draw. This self-regulating mechanism is crucial because it reduces energy consumption and prevents overload.
  • Back emf causes current reduction
  • High current at startup, reduced at normal speed
  • Helps conserve energy and prevent motor damage
Supply Voltage
The supply voltage for most household electric fans is typically around 120 volts, as seen in our exercise. This supply voltage provides the necessary energy to get the fan motor started and to maintain its operation. The supply voltage essentially pushes current through the motor.

In combination with the internal resistance of the motor, these factors determine the initial current drawn by the motor (before back emf kicks in). Understanding the relationship between supply voltage, resistance, and current is key for solving problems related to electric motors and back emf.
  • Supply voltage initializes motor operation
  • Common household voltage is 120 volts
  • Works with motor resistance to define initial current
Normal Operating Speed
The normal operating speed of an electric fan motor is the speed at which the fan blades rotate under steady-state conditions. At this speed, the motor generates sufficient back emf to significantly reduce current draw to a smaller fraction of the initial current. In our exercise, the normal operating current is 15% of the initial current, indicating the presence of considerable back emf.

Achieving this balance of speed and current is crucial for two main reasons:
  • It optimizes energy consumption - ensuring the fan operates efficiently.
  • It prevents overheating, which could otherwise damage the motor.
Understanding the dynamics of the fan reaching its normal operating speed offers insight into motor efficiency and durability.

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Most popular questions from this chapter

Two coils of wire are placed close together. Initially, a current of 2.5 \(\mathrm{A}\) exists in one of the coils, but there is no current in the other. The current is then switched off in a time of \(3.7 \times 10^{-2}\) s. During this time, the average emf induced in the other coil is 1.7 \(\mathrm{V}\) . What is the mutual inductance of the two-coil system?

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes- through the square loop is \(7.0 \times 10^{-3}\) Wb. What is the flux that passes through the circular loop?

Coil 1 is a flat circular coil that has \(N_{1}\) turns and a radius \(R_{1}\) . At its center is a much smaller flat, circular coil that has \(N_{2}\) turns and radius \(R_{2}\) . The planes of the coils are parallel. Assume that coil 2 is so small that the magnetic field due to coil 1 has nearly the same value at all points covered by the area of coil \(2 .\) Determine an expression for the mutual inductance between these two coils in terms of \(\mu_{0}, N_{1}, R_{1}, N_{2},\) and \(R_{2}\) .

A planar coil of wire has a single turn. The normal to this coil is parallel to a uniform and constant (in time) magnetic field of 1.7 \(\mathrm{T}\) . An emf that has a magnitude of 2.6 \(\mathrm{V}\) is induced in this coil because the coil's area \(A\) is shrinking. What is the magnitude of \(\Delta A / \Delta t,\) which is the rate (in \(\mathrm{m}^{2} / \mathrm{s} )\) at which the area changes?

A generator is connected across the primary coil ( \(N_{p}\), turns) of a transformer, while a resistance \(R_{2}\) is connected across the secondary coil \(\left(N_{\mathrm{s}}\right.\) turns). This circuit is equivalent to a circuit in which a single resistance \(R_{1}\) is connected directly across the generator, without the transformer. Show that \(R_{1}=\left(N_{\mathrm{p}} / N_{\mathrm{s}}\right)^{2} R_{2},\) by starting with \(\mathrm{Ohm}^{\prime}\) s law as applied to the secondary coil.
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