Question

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

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

Step-by-step solution

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 \(E_{2}\).

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: \(\omega_{1} = \frac{2\pi}{60}n_{1}\) \(\omega_{2} = \frac{2\pi}{60}n_{2}\)

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 \(E_{2}\): \(E_{2} = E_{1} \cdot \frac{\omega_{2}}{\omega_{1}}\)

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.