# Phasor voltage

1. Apr 8, 2005

### EvLer

Hello,
now we are in phasors and I am missing something in this problem:
given a circuit of indep current source (5cos(wt)), R, L (with vL(t)) all in series, no values for them,
although voltage across inductor vL(t) = 3cos(wt + theta), we need to find theta.

So, by using impedance for L: Z = jwL, vL(t) = Z*I,
so I got vL(t) = jwL5 /_ 0 which is also = 3/_theta
And now what? The answer is 90 degrees, but how do I get it?

Thanks a lot.

Last edited: Apr 8, 2005
2. Apr 9, 2005

### Theelectricchild

The reason its 90 degrees is simple... the voltage is indeed ZI, and I has a phase of zero--- but you would do well to convince yourself that the impedance Z is purely imaginary, so you have a phase for your impedance equivalent to the arctan of Im[Z]/Re[Z] where Re[Z] = 0, and thus you have an arctan of infinity, which is undoubtedly equal to 90 degrees.

Therefore you have V = ZI which yields a phase of 0 + 90, the angular frequency and the amplitudes of Z and I are immaterial for this problem.

3. Apr 9, 2005

### Theelectricchild

By the by, do you intro EE students still happen to make use of Nilsson and Riedel by any chance?

4. Apr 9, 2005

### EvLer

Huh???
I guess, not.

5. Apr 9, 2005

### Theelectricchild

Oh sorry, it's just a textbook that I found really great for my intro EE courses.

6. Apr 11, 2005

### EvLer

Thanks, I might check it out as complementary source. We are using DeCarlo and Lin, since DeCarlo is actually a professor here, I doubt they will change our texbook any time soon