Maximum Amplitude, 50hx vs 60hz generator

[texas2787]

Homework Statement

We have an AC generator with output current I. I, as a function of time, is a sine wave whose frequency, f, is the same as the rotational frequency of the generator. So,

I(t) = I$$_{0}$$sin(2πft), where I$$_{0}$$ is the max. amplitude of the current.

In terms of magnetic flux, $$\Phi$$$$_{B}$$:

I = - (Constant * Change in $$\Phi$$$$_{B}$$) / time

When the generator revolves at 50 revs/second, the maximum amplitude I$$_{0}$$ = 1000 A. If you increase the spead of the generator to 60 revs/sec, what is the maximum amplitude I$$_{0}$$?

Homework Equations

Given above

The Attempt at a Solution

This question really stumped me. It's difficult to bring in additional equations since we don't know more about the generator -- it's obviously a concept question. I suspect that the maximum amplitude will increase proportionally.

We know that f(old) = 50 Hz, since it will be the same as the rotatation frequency. Thus, the period for this sine curve will be 1/50 second, as the amplitude goes from maximum to maximum. When f(new) = 60 Hz, the period will condense to 1/60 second.

Thinking about flux, there is no change in magnetic field (B) -- though the $$\Theta$$ between B and the area of the coil will be changing faster as rotations speed up. So change in fluz would be greater over a given unit of time (?). Thus, from the second equation, we know that I must also increase (?).

At this point, I simply guess that the change will be proportional -- since rotation is increasing in speed by 20%, so will maximum amplitude ?


Thanks for your help!

 

[anathema]

You are correct in thinking that the maximum amplitude will increase proportionally. This is because the change in magnetic flux, as expressed by the second equation, is directly proportional to the frequency of the generator. As the frequency increases, so does the change in magnetic flux, leading to an increase in the maximum amplitude of the current. Therefore, if the frequency increases by 20%, the maximum amplitude will also increase by 20%. In this case, the maximum amplitude I_{0} would be 1200 A.

I hope this helps. Keep up the good work!

 

[Moetasim]

I would approach this problem by first analyzing the equations provided. We know that the maximum amplitude of the current, I_{0}, is directly proportional to the frequency, f, and the rotational speed of the generator. This can be seen in the equation I(t) = I_{0}sin(2πft), where f is the frequency of the sine wave and also the rotational frequency of the generator.

Since we are given the maximum amplitude at 50 revs/second, we can use this information to find the constant in the equation I = - (Constant * Change in \Phi_{B}) / time. This can be done by rearranging the equation to solve for the constant:

Constant = - (I * time) / Change in \Phi_{B}

Substituting the values given in the problem, we get:

Constant = - (1000 A * 1/50 s) / 0

Since there is no change in magnetic flux, the constant becomes infinity. This means that the maximum amplitude of the current is directly proportional to the frequency, and as the frequency increases, the maximum amplitude will also increase proportionally.

Therefore, if we increase the rotational speed of the generator to 60 revs/second, the maximum amplitude I_{0} will also increase proportionally. So, the new maximum amplitude will be 1200 A (1000 A * 60 Hz / 50 Hz).

In conclusion, the maximum amplitude of the current is directly proportional to the frequency of the generator, and as the rotational speed increases, the maximum amplitude will also increase proportionally.