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Chapter 20: Problem 80

The alternator in an automobile generates an emf of amplitude $12.6 \mathrm{V}\( when the engine idles at \)1200 \mathrm{rpm}$. What is the amplitude of the emf when the car is being driven on the highway with the engine at 2800 rpm?

Short Answer

Expert verified
Answer: The amplitude of the emf generated at 2800 rpm is approximately 29.4 V.

Step by step solution

01

Understand the information given

We are given the initial emf amplitude \(E_{1} = 12.6 V\), the initial engine speed \(n_{1} = 1200 rpm\), and the final engine speed \(n_{2} = 2800 rpm\). We need to find the final emf amplitude \(E_{2}\).
02

Convert engine speeds to angular velocities

Angular velocity (ω) can be found by multiplying the engine speed (n) by the conversion factor (2π/60), where 2π represents a full circle, and 60 converts minutes to seconds: \(\omega = \frac{2\pi}{60}n\) Calculate the angular velocities ω₁ and ω₂: \(\omega_{1} = \frac{2\pi}{60}n_{1}\) \(\omega_{2} = \frac{2\pi}{60}n_{2}\)
03

Use the proportionality relationship to solve for the final emf amplitude

Since the emf generated by the alternator is directly proportional to its angular velocity, we can write: \(\frac{E_{1}}{\omega_{1}} = \frac{E_{2}}{\omega_{2}}\) Now, solve for \(E_{2}\): \(E_{2} = E_{1} \cdot \frac{\omega_{2}}{\omega_{1}}\)
04

Calculate the final emf amplitude

Now, plug in the given values and the calculated angular velocities to find the final emf amplitude: \(E_{2} = 12.6 V \cdot \frac{\frac{2\pi}{60} \cdot 2800}{\frac{2\pi}{60} \cdot 1200}\) Simplify and solve for \(E_{2}\): \(E_{2} = 12.6 V \cdot \frac{2800}{1200}\) \(E_{2} \approx 29.4 V\) The amplitude of the emf when the car is being driven on the highway with the engine at 2800 rpm is approximately 29.4 V.

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Most popular questions from this chapter

A transformer with 1800 turns on the primary and 300 turns on the secondary is used in an electric slot car racing set to reduce the input voltage amplitude of \(170 \mathrm{V}\) from the wall output. The current in the secondary coil is of amplitude \(3.2 \mathrm{A}\). What is the voltage amplitude across the secondary coil and the current amplitude in the primary coil?
An ideal solenoid ( \(N_{1}\) turns, length \(L_{1},\) radius \(r_{1}\) ) is placed inside another ideal solenoid ( \(N_{2}\) turns, length \(L_{2}>L_{1}\), radius \(r_{2}>r_{1}\) ) such that the axes of the two coincide. (a) What is the mutual inductance? (b) If the current in the outer solenoid is changing at a rate \(\Delta I_{2} / \Delta t,\) what is the magnitude of the induced emf in the inner solenoid?
The current in a \(0.080-\mathrm{H}\) solenoid increases from \(20.0 \mathrm{mA}\) to \(160.0 \mathrm{mA}\) in \(7.0 \mathrm{s} .\) Find the average emf in the solenoid during that time interval.
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?
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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