# Homework Help: Frequency of AC circuit with inductor

1. Dec 10, 2008

### etagg

1. The problem statement, all variables and given/known data

The voltage across and inductor reaches its maximum value 22ms before the current supplied by the generator reaches its maximum value. What is the lowest possible frequency at which the generator operates?

2. Relevant equations

I used the equation V=V_(max)sin(wt+90), where w is the angular speed equal to 2pi(f), where f is the frequency of the generator.
I also used the equation I=I_(max)sin(w(t+22ms))=I_(max)

3. The attempt at a solution

I recognized that the voltage across an inductor has its maximum value when sin(wt+90)=1, or wt+90=90. This obviously means that wt=0, and i could not get a frequency from that.

I then looked at when the current is maximal, or when w(t+22ms)=90, and again i got stuck.

I think im missing something simple.

2. Dec 10, 2008

### jdstokes

Do you know that the voltage across an inductor leads the current by 90 degrees? What does that tell you about the period of oscillation

3. Dec 10, 2008

### etagg

Well i knew that the voltage across and inductor leads by 90 degrees, which is why i used the equation V=V_(max)sin(wt+90).

If the voltage leads the current by 90 degrees, it means that when the voltage is a maximum the current is zero. Im not sure how that would help me?

4. Dec 10, 2008

### jdstokes

Well, it tells you that 90 degrees of rotation takes 22 ms. How long does it take for a complete revolution? What is the corresponding frequency?

5. Dec 10, 2008

### etagg

ohhh so then the full rotation is 4*22s= 88s and then divided by two pi is the frequency? Thanks so much for your help jdstokes

6. Dec 10, 2008

### jdstokes

Actually, frequency is just 1 over the period which you found to be 88 s. Angular frequency is the number of radias per second, ie 2*pi*frequency. But you have the idea.