Recent content by MrPacane

Oh, I just realized I made a mistake! At synchronous speed the slip is zero (slip is equal to 1 at starting because the mechanical speed is zero). Torque is produced by the Lorentz forces (due to moving charges in a magnetic field). In an induction motor, there are moving charges in the...

At synchronous speed, the slip is equal to 1 and so the torque is 0. If there's no torque, I think the motor's speed will decrease because of air friction. In order to have torque (and keep the motor running at a constant speed), there needs to be a small difference between the synchronous speed...

Under no load, does an induction motor run exactly at its synchronous speed (i.e. relative field speed) or at a speed that is close to the synchronous speed but slightly less? Thanks for your time.

Homework Statement A 25-turn coil has a radius of 1 cm and it is carrying a current of 1 A. The coil is in a constant B-field of 1 T (directed along the coil's axis). The coil is moving perpendicularly to the B-field at a speed of 1 m/s. a) Is there an Electromotive force induced in the...

Ok guys, I'm sorry for wasting your time, the prof said it's a mistake. The question should have been: Which is quiet easy to prove. Problem "solved".

Yes, I'm pretty sure... Here's a screenshot of the question: http://imageshack.us/a/img689/3365/21602169.jpg EDIT: I just sent an e-mail to my prof asking him to clarify this... I'll get back to you as soon as I get his answer.

Here's the exact question: What am I missing? Again, thanks for your time :) !

Alright, thanks for your reply. But then, does anyone know how to show that Ex_even = Ex_odd = 0.5 * Ex ?

Hi everyone, In my signals assignment, I'm asked to show that, for a continuous time, real-valued signal x(t): Ex_even = Ex_odd = 0.5 * Ex So here's what I've done: Ex_even = ∫|(x(t) + x(-t))/2|²dt Ex_even = 0.5 * ∫|(x(t)² + 2x(t)x(-t) + x(-t)²)/2|dt Ex_even = 0.5 * [ 0.5 *...