i cant read my formula in my notes on how to do this problem here is the problem
the peak voltage of a sine wave is 10v. what is its instantaneous value at 30 degrees and at 40 degrees?
anyone?

o_O

Who did what to who now?

i got the question from my book

you did waht at 30 degrees?

Get the answer from the back of the book...page 327

-- Rogue

i tryed just a glossry and index

You need to know the amplitude and frequency of the wave before you can do the calcs and this is assuming that your offset is zero.

I'd have to pull my old college books out.....and I'm at work right now......I can give you a rough idea of what your answers would be:

At 0, 180 and 360, your instantaneous voltage is 0, right?

At 90° you're 10v peak (assuming 20v p-p?), so at 30° you should be .3 of 10, or 3v. 40° would be 4v. Actually a pretty simple problem.

Doesn't quite work that way Bill...for pete'* sake you are a laser technician! You are right on the quadrant angle values.

You have to know the cycle time to know the phase angle. The top (or bottom) of a sine wave is a parabola based on the amplitude and half-cycle time, therefor it is not a linear relationship for amplitude vs. angle.

Yes, I realize I am telling everything but how to find the answer. I would have to look it up either in a AC circuits or a trig textbook.

Jay

**** it'* been a long time. I got my degree in 92! Gimme a break. I don't have to figure anything any more. I just use an oscilloscope now!