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In a 200-turn automobile alternator, the magnetic flux in each turn is ΦB = 2.50 10-4 cos ωt, where ΦB is in webers, ω is the angular speed of the alternator, and t is in seconds. The alternator is geared to rotate three times for each engine revolution. The engine is running at an angular speed of 1.00 103 rev/min.

(a) Determine the induced emf in the alternator as a function of time. (Assume is in V.) =_______.
(b) Determine the maximum emf in the alternator.

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boffeemadrid

Answer:

[tex]10.63952sin(209.43951t)[/tex]

10.63952 V

Explanation:

[tex]N_t[/tex] = Number of turns = 200

[tex]\phi_B[/tex] = Magnetic flux = [tex]2.5\times 10^{-4}cos(\omega t)[/tex]

[tex]\omega_e[/tex] = Engine angular speed = [tex]1\times 10^{3}\ rpm[/tex]

Alternator angular speed is given by

[tex]N_a=2\times \omega_e\\\Rightarrow N_a=2\times 1\times 10^{3}\\\Rightarrow N_a=2\times 10^{3}\ rpm[/tex]

[tex]\omega=N_a\dfrac{2\pi}{60}\\\Rightarrow \omega=2\times 10^{3}\dfrac{2\pi}{60}\\\Rightarrow \omega=209.43951\ rad/s[/tex]

Induced emf is given by

[tex]\epsilon=-N_t\dfrac{d\phi_B}{dt}\\\Rightarrow \epsilon=-200\dfrac{d}{dt}2.54\times 10^{-4}cos(209.43951 t)\\\Rightarrow \epsilon=200\times 2.54\times 10^{-4}\times 209.43951 sin(209.43951t)\\\Rightarrow \epsilon=10.63952sin(209.43951t)[/tex]

The function is [tex]10.63952sin(209.43951t)[/tex]

The induced maximum emf is 10.63952 V

onyebuchinnaji

The induced emf in the alternator as a function of time is 0.05cos(314.2t).

The maximum emf in the alternator is 0.05 V.

Induced emf in the coil

The emf induced in the coil is calculated as follows;

emf = NdФ/dt

where;

  • N is number of turns
  • Ф is magnetic flux = BA

The alternator angular speed

ωa = 3ω

ωa = 3 x 1 x 10³ x 2π x  1/60 s = 314.2 rad/s

emf = 200 x 2.5 x 10⁻⁴ x cos(314.2t)

emf = 0.05cos(314.2t)

Thus, the maximum emf in the alternator is 0.05 V.

Learn more about induced emf here: https://brainly.com/question/13744192

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