Rotation Rate of a Generator Coil

AI Thread Summary
Doubling the rotation rate of a generator coil results in a peak emf that also doubles, according to Faraday's Law. The equation for induced emf indicates that the maximum voltage is directly proportional to the angular frequency. When the frequency is increased, the induced voltage increases correspondingly. The discussion emphasizes the importance of using mathematical relationships rather than intuition to derive conclusions. Therefore, the peak emf will not remain the same; it will increase as the rotation rate increases.
lampshader
Messages
36
Reaction score
0

Homework Statement


If the rotation rate of a generator coil is doubled, what happens to the peak emf?


Homework Equations





The Attempt at a Solution



I think it stays the same. Am i right?
 
Physics news on Phys.org
lampshader said:

Homework Statement


If the rotation rate of a generator coil is doubled, what happens to the peak emf?


Homework Equations





The Attempt at a Solution



I think it stays the same. Am i right?

no. look at the equation for induced emf...
 
I don't even have a number in my word problem..How can I not use intuition..lol
 
lampshader said:
I don't even have a number in my word problem..How can I not use intuition..lol

by doing it symbolically, and you do have a number: 2

the new \omega is twice that of the old \omega

V_{max} is whenever the trig function on the right is equal to 1. Therefore, we can say this:
V_{oldmax}=\omega BA
V_{newmax} =2\omega BA
Do you see a substitution?
V_{newmax} =2V_{oldmax}
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top