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A motor in normal operation carries a direct current of 0.850 A when connected to a \(120-\mathrm{V}\) power supply. The resistance of the motor windings is 11.8 \Omega. While in normal operation, (a) what is the back emf generated by the motor? (b) At what rate is internal energy produced in the windings? (c) What If? Suppose that a malfunction stops the motor shaft from rotating. At what rate will internal energy be produced in the windings in this case? (Most motors have a thermal switch that will turn off the motor to prevent overheating when this occurs.)

Short Answer

Expert verified
The back emf is 110 Volts, the internal energy production rate is 8.49 Watts and when the motor stops, it jumps to 1216.19 Watts.

Step by step solution

01

Calculate the back emf

The back emf (Electromotive force) can be calculated using Kirchhoff's loop rule. According to the rule, the sum of emfs in any closed loop of a network is equal to the sum of the potential drops in that loop. In other words, \(E - I*R - ε = 0\), where E is the source voltage, I is the current, R is the resistance, and ε is the back emf. In this case, we know E = 120 Volts, I = 0.850 A and R = 11.8 Ω. Solving for ε gives \(ε = E - I*R = 120 Volts - 0.850 A * 11.8 Ω = 110 Volts\).
02

Calculate the rate of internal energy production

The rate of production of internal energy can be found using the formula of electric power, which equals the current squared times the resistance, \(P = I^2 * R\). Substituting the known values gives \(P = (0.850 A)^2 * 11.8 Ω = 8.49 Watts\).
03

Calculate the rate of internal energy production when the motor stops

When the motor stops, the back emf will be zero. Now the full voltage will be across the resistance R. The new rate of internal energy production will then be \(P' = I^2 * R = (0.850 A^2) * 11.8 Ω = 8.49 Watts\), but the current I will be now equal to \(I = E / R = 120 Volts / 11.8 Ω = 10.17 A\). Then, \(P' = (10.17 A)^2 * 11.8 Ω = 1216.19 Watts\).

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Key Concepts

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

Back EMF
The concept of back EMF, or back electromotive force, is crucial in understanding how electric motors work. Back EMF is a voltage generated by the motor itself when it is running. This happens due to the interaction between the motor's moving parts and the magnetic field created by its coils. In simpler terms, when a motor is spinning, it generates its own voltage that acts in opposition to the supplied voltage. This is why it is called "back" EMF.

When the motor is in operation, this back EMF reduces the effective voltage across the motor's windings. To find the back EMF, you subtract the voltage drop across the motor's resistance, calculated by multiplying the current and resistance, from the supply voltage. Here, back EMF can be mathematically represented as \[\varepsilon = E - I \cdot R\]where \(E\) is the supplied voltage, \(I\) is the current, and \(R\) is the resistance. Understanding this concept is key because it determines how much of the supplied electrical energy is effectively used to produce mechanical work.
Kirchhoff's Loop Rule
Kirchhoff's Loop Rule is a fundamental principle used in electrical circuit analysis. It helps in understanding how voltages are distributed in a circuit. This rule states that the total sum of electric potentials (voltage) around any closed loop in a circuit is equal to the sum of the potential drops across each element within that loop.

In mathematical terms, it's written as \[\sum{V} = 0\]This means that all the supplied voltage in a circuit loop is consumed by the components' resistances and back EMFs. For the motor example, the rule is applied by rearranging the equation to find back EMF, where all values must appropriately add to zero. This can help balance and verify calculations when trying to understand electric circuits, particularly with motors and other complex components.
Electric Power
Electric power is a measure of how quickly electrical energy is transferred by an electric circuit. The power in a circuit can be calculated by using the equation \[P = I^2 \cdot R\]where \(P\) is the power in watts, \(I\) is the current in amperes, and \(R\) is the resistance in ohms.

In terms of an electric motor, this formula is used to determine the rate at which internal energy is produced in the motor windings. This power is mainly converted into heat due to the resistance in the wires. Knowing how to calculate this power helps in assessing whether the motor operates effectively and also guards against potential overheating.

For example, when a motor is operating properly, the internal energy rate calculated was 8.49 Watts. However, with a malfunction where the motor stops, the motor draws more current, leading to an increased rate of 1216.19 Watts, which could potentially cause overheating unless a safety mechanism is present to prevent it.

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Most popular questions from this chapter

A 25 -turn circular coil of wire has diameter \(1.00 \mathrm{m}\). It is placed with its axis along the direction of the Earth's magnetic field of \(50.0 \mu \mathrm{T},\) and then in \(0.200 \mathrm{s}\) it is flipped \(180^{\circ} .\) An average emf of what magnitude is generated in the coil?

A rectangular coil of 60 turns, dimensions \(0.100 \mathrm{m}\) by \(0.200 \mathrm{m}\) and total resistance \(10.0 \Omega,\) rotates with angular speed 30.0 rad/s about the \(y\) axis in a region where a \(1.00-\mathrm{T}\) magnetic field is directed along the \(x\) axis. The rotation is initiated so that the plane of the coil is perpendicular to the direction of \(\mathbf{B}\) at \(t=0 .\) Calculate (a) the maximum induced emf in the coil, (b) the maximum rate of change of magnetic flux through the coil, (c) the induced emf at \(t=0.0500 \mathrm{s},\) and \((\mathrm{d})\) the torque exerted by the magnetic field on the coil at the instant when the emf is a maximum.

Very large magnetic fields can be produced using a procedure called flux compression. A metallic cylindrical tube of radius \(R\) is placed coaxially in a long solenoid of somewhat larger radius. The space between the tube and the solenoid is filled with a highly explosive material. When the explosive is set off, it collapses the tube to a cylinder of radius \(r

A coil of area \(0.100 \mathrm{m}^{2}\) is rotating at \(60.0 \mathrm{rev} / \mathrm{s}\) with the axis of rotation perpendicular to a \(0.200-\mathrm{T}\) magnetic field. (a) If the coil has 1000 turns, what is the maximum emf generated in it? (b) What is the orientation of the coil with respect to the magnetic field when the maximum induced voltage occurs?

In a 250 -turn automobile alternator, the magnetic flux in each turn is \(\Phi_{B}=\left(2.50 \times 10^{-4} \mathrm{Wb}\right) \cos (\omega t),\) where \(\omega\) is the angular speed of the alternator. The alternator is geared to rotate three times for each engine revolution. When the engine is running at an angular speed of 1000 rev/min, determine (a) the induced emf in the alternator as a function of time and (b) the maximum emf in the alternator.

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