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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 generated at normal speed is 102.0 V.

Step by step solution

01

Understand the Problem

The electric fan motor draws a lower current at its normal operating speed due to the generated back emf. The back emf is a voltage that opposes the supply voltage, hence reducing the current. We need to calculate this back emf given the initial and reduced current levels, as well as the supply voltage.
02

Define the Known Values

The supply voltage is given as \( V_s = 120.0 \) V. The initial current (when the fan starts) is the full current, say \( I_0 \). At normal speed, the fan draws \( 15.0\% \) of this current, so the current is reduced to \( I = 0.15 \times I_0 \).
03

Apply Ohm's Law

Ohm's Law is given by \( V = IR \). When the fan starts, the emf is zero, hence the supply voltage \( V_s = I_0R \), which gives \( I_0 = \frac{V_s}{R} \). At normal speed, the back emf reduces the effective voltage across the motor from \( V_s \) to \( V_s - \text{back emf} \).
04

Calculate Effective Voltage at Normal Speed

At normal speed, let the current be \( I \), where \( I = 0.15I_0 \). Using Ohm's Law, the voltage across the motor is \( V = IR \). Since \( I = 0.15I_0 \), \( V = (0.15I_0)R \).
05

Relate Effective Voltage to Supply Voltage and Back Emf

We relate effective voltage to the back emf with the equation: \[ V_s = V + \text{back emf} \].Thus, \[ \text{back emf} = V_s - IR = V_s - (0.15I_0R). \]
06

Substitute Variables and Solve for Back Emf

Since \( I_0 = \frac{V_s}{R} \), we have \( I_0R = V_s \). Substitute this into the equation for back emf as follows:\[ \text{back emf} = V_s - 0.15 \times V_s = 0.85 \times V_s. \]
07

Final Calculation

Substitute the given supply voltage \( V_s = 120.0 \) V into the equation:\[ \text{back emf} = 0.85 \times 120.0 \text{ V} = 102.0 \text{ V}. \]

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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 an essential component that powers the operation of an electric fan, making it ideal for home cooling and ventilation. It works to convert electrical energy into mechanical energy, which ultimately rotates the fan blades and circulates air.
Electric motors used in fans are commonly designed using components like rotors, stators, and bearings. When the motor receives an electric current, it produces a magnetic field that turns the rotor. This movement is then transferred to the fan blades, causing them to spin and create airflow.
A significant aspect of electric motors is their efficiency. As the fan reaches its optimum speed, the motor operates more efficiently, generating a back electromotive force (emf) which plays a vital role in reducing the current drawn from the power source.
Ohm's Law
Ohm's Law is a fundamental principle in the field of electronics and physics. It relates the voltage (V), current (I), and resistance (R) in an electrical circuit with the equation:
  • \( V = IR \)
This equation implies that the voltage across a conductor is directly proportional to the current it carries, given the resistance stays constant. In practical terms, if you know any two of these quantities, you can easily calculate the third.
When an electric fan motor starts, it draws a higher current because no back emf is generated yet. Here, the supply voltage equals the product of the initial current and resistance, \( V = I_0 R \).
As the motor speeds up and creates back emf, it transforms the effective voltage. This new effective voltage is precisely why the current withdraws to a stable level, as covered by Ohm's Law.
Current Reduction
Current reduction in an electric fan motor occurs as the fan reaches its normal operating speed. Initially, when the fan starts, it consumes more electrical current. However, as the fan stabilizes at its intended speed, the motor generates back emf, which in turn, decreases the amount of current needed.
The process is critical because excessive current usage can lead to overheating and possible wear of the motor. By designing electric motors to capitalize on back emf, manufacturers ensure that the fan does not take in more current than necessary. For this fan motor, the normal operational current is just 15% of the initial startup current.
This reduction safeguards the motor while also promoting cost savings on electricity consumption. The decrease in current to a safe level is integral to the longevity and efficiency of the motor.
Supply Voltage
Supply voltage refers to the voltage provided by a power source to an electrical device. In the case of our electric fan, it is the voltage from the power socket that energizes the motor.
For this particular fan motor, the supply voltage is specified as 120.0 volts. This value acts as the driving force that powers the fan when it is plugged into the socket. At the start, the full supply voltage pushes a high current through the motor, as no opposing back emf is present yet.
Once the fan reaches its average operational speed, the supply voltage works alongside the back emf. The back emf effectively reduces the portion of the supply voltage being used for current production within the motor.
By understanding supply voltage, one can better comprehend how it and back emf collaborate to maintain an ideal current for electrical devices like fan motors.

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Most popular questions from this chapter

Reconfiguring a Transformer. You and your team are exploring an abandoned science facility on the coast of western Antarctica when a large storm hits, and it is clear that you will be stuck there for a few days. You and the others search for supplies and find a generator and a tank of fuel. Having electrical power would allow you to keep your communication devices operational and make your stay more comfortable. A team member gets the generator running, but there is a complication: The output of the generator is \(50 \mathrm{Hz}\) at \(440 \mathrm{VAC}(\mathrm{RMS}),\) and your devices require \(60 \mathrm{Hz}, 110 \mathrm{VAC}(\mathrm{RMS})\) The electrical power in many European countries runs on \(240 \mathrm{V}\) at \(50 \mathrm{Hz},\) so a few of your team members have converters. However, \(440 \mathrm{V}\) is still too high to use them. You search and eventually find a large transformer that, according to a worn tag on its case, is designed to step down from \(5000 \mathrm{V}\) to \(880 \mathrm{V}\). The tag also indicates that its primary coil has 1500 turns, but you cannot read the number of turns in the secondary coil. (a) How many turns should its secondary coil have? (b) It will be a difficult job, but you can change the number of primary turns by cutting some of them out. How many turns should you leave on the primary coil so that, with the primary connected to \(440 \mathrm{V},\) the secondary outputs \(240 \mathrm{V}\) (so that you can use the \(240 \mathrm{V}\) to \(110 \mathrm{V}\) converters)? (c) You find that the current at the source (i.e., that connected to the primary) is limited to a maximum of \(20.0 \mathrm{A}\). What is the maximum current limit through the secondary coil? (d) What is the maximum average power available at the secondary coil?

A magnetic field is passing through a loop of wire whose area is \(0.018 \mathrm{m}^{2}\). The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of \(0.20 \mathrm{T} / \mathrm{s}\) (a) Determine the magnitude of the emf induced in the loop. (b) Suppose that the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in \(\mathrm{m}^{2} / \mathrm{s}\) ) should the area be changed at the instant when \(B=1.8 \mathrm{T}\) if the induced emf is to be zero? Explain whether the area is to be enlarged or shrunk.

In some places, insect "zappers," with their blue lights, are a familiar sight on a summer's night. These devices use a high voltage to electrocute insects. One such device uses an ac voltage of \(4320 \mathrm{V},\) which is obtained from a standard \(120.0-\mathrm{V}\) outlet by means of a transformer. If the primary coil has 21 turns, how many turns are in the secondary coil?

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 \(\left.\mathrm{m}^{2} / \mathrm{s}\right)\) at which the area changes?

A generating station is producing \(1.2 \times 10^{6} \mathrm{W}\) of power that is to be sent to a small town located \(7.0 \mathrm{km}\) away. Each of the two wires that comprise the transmission line has a resistance per kilometer of \(5.0 \times\) \(10^{-2} \Omega / \mathrm{km} .\) (a) Find the power used to heat the wires if the power is transmitted at \(1200 \mathrm{V} .\) (b) \(\mathrm{A} 100: 1\) step-up transformer is used to raise the voltage before the power is transmitted. How much power is now used to heat the wires?
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