Question

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.

Answer 1

Answer:

$10.63952sin(209.43951t)$

10.63952 V

Explanation:

$N_t$ = Number of turns = 200

$\phi_B$ = Magnetic flux = $2.5\times 10^{-4}cos(\omega t)$

$\omega_e$ = Engine angular speed = $1\times 10^{3}\ rpm$

Alternator angular speed is given by

$N_a=2\times \omega_e\\\Rightarrow N_a=2\times 1\times 10^{3}\\\Rightarrow N_a=2\times 10^{3}\ rpm$

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

Induced emf is given by

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

The function is $10.63952sin(209.43951t)$

The induced maximum emf is 10.63952 V