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Chapter 20: Problem 28

A de motor has coils with a resistance of \(16 \Omega\) and is connected to an emf of \(120.0 \mathrm{V}\). When the motor operates at full speed, the back emf is \(72 \mathrm{V}\). (a) What is the current in the motor when it first starts up? (b) What is the current when the motor is at full speed? (c) If the current is \(4.0 \mathrm{A}\) with the motor operating at less than full speed, what is the back emf at that time?

Short Answer

Expert verified
Answer: The initial current is 7.5 A, the full-speed current is 3.0 A, and the back emf at a current of 4.0 A is 56 V.

Step by step solution

01

(a) Find initial current in the motor

To find the current in the motor when it first starts up, we need to remember that the back emf is zero at this point. That's because the motor isn't spinning and isn't generating any opposing voltage. Using Ohm's law (\(V = IR\)) and the applied emf (\(120.0 \mathrm{V}\)): \(I = \frac{V}{R} = \frac{120.0}{16} = 7.5 \mathrm{A}\) So, the initial current in the motor when it first starts up is \(7.5 \mathrm{A}\).
02

(b) Find current at full speed

When the motor is running at full speed, the back emf is given as \(72 \mathrm{V}\). The total voltage across the coil will be the difference between the applied emf and the back emf: \(V_{coil} = \varepsilon - \varepsilon_b = 120.0 - 72 = 48 \mathrm{V}\) Now we can use Ohm's law to find the current at full speed: \(I_{full} = \frac{V_{coil}}{R} = \frac{48}{16} = 3.0 \mathrm{A}\) The current in the motor at full speed is \(3.0 \mathrm{A}\).
03

(c) Find back emf at \(4.0 \mathrm{A}\) current

With a given current of \(4.0 \mathrm{A}\) at less than full speed, we can use Ohm's law to find the voltage across the coil: \(V_{coil} = IR = 4.0 \times 16 = 64 \mathrm{V}\) Now let's find the back emf using the equation mentioned earlier: \(\varepsilon_b = \varepsilon - V_{coil} = 120.0 - 64 = 56 \mathrm{V}\) So, the back emf when the motor is operating at less than full speed with a current of \(4.0 \mathrm{A}\) is \(56 \mathrm{V}\).

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Most popular questions from this chapter

A coil has an inductance of \(0.15 \mathrm{H}\) and a resistance of 33 \(\Omega\). The coil is connected to a 6.0 - \(V\) battery. After a long time elapses, the current in the coil is no longer changing. (a) What is the current in the coil? (b) What is the energy stored in the coil? (c) What is the rate of energy dissipation in the coil? (d) What is the induced emf in the coil?
In this problem, you derive the expression for the self inductance of a long solenoid [Eq. \((20-15 a)] .\) The solenoid has \(n\) turns per unit length, length \(\ell,\) and radius \(r\) Assume that the current flowing in the solenoid is \(I\) (a) Write an expression for the magnetic field inside the solenoid in terms of \(n, \ell, r, I,\) and universal constants. (b) Assume that all of the field lines cut through each turn of the solenoid. In other words, assume the field is uniform right out to the ends of the solenoid-a good approximation if the solenoid is tightly wound and sufficiently long. Write an expression for the magnetic flux through one turn. (c) What is the total flux linkage through all turns of the solenoid? (d) Use the definition of self-inductance [Eq. \((20-14)]\) to find the self inductance of the solenoid.
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The component of the external magnetic field along the central axis of a 50 -turn coil of radius \(5.0 \mathrm{cm}\) increases from 0 to 1.8 T in 3.6 s. (a) If the resistance of the coil is \(2.8 \Omega,\) what is the magnitude of the induced current in the coil? (b) What is the direction of the current if the axial component of the field points away from the viewer?
The windings of an electromagnet have inductance \(L=8.0 \mathrm{H}\) and resistance \(R=2.0 \Omega . \quad A \quad 100.0-V \quad d c\) power supply is connected to the windings by closing switch \(S_{2} .\) (a) What is the current in the windings? (b) The electromagnet is to be shut off. Before disconnecting the power supply by opening switch \(S_{2},\) a shunt resistor with resistance \(20.0 \Omega\) is connected in parallel across the windings. Why is the shunt resistor needed? Why must it be connected before the power supply is disconnected? (c) What is the maximum power dissipated in the shunt resistor? The shunt resistor must be chosen so that it can handle at least this much power without damage. (d) When the power supply is disconnected by opening switch \(S_{2},\) how long does it take for the current in the windings to drop to \(0.10 \mathrm{A} ?\) (e) Would a larger shunt resistor dissipate the energy stored in the electromagnet faster? Explain.
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