that is the correct answer :wink:
went over everything again and -11.66 is correct :biggrin:
was just a silly mistake on my part!
thanks a lot for your help Kruum and berkeman, much appreciated :wink:
thanks for the reply :smile:
458 * 0.005 = 2.29 :-p
ok this is exactly what I am entering:
rad mode:
60 sin (458(0.005) + 1/3 ∏)
and I am getting: -11.66159...
ah sorry that's kind of irrelevant that's the peak to peak voltage,
could you tell me what your answer was so i can see if I am on the right track?
or could you tell me where I am going wrong here:
60 Sin (458 (5x10-3) + 1/3 ∏)
Thanks, very interesting forum indeed! :)
Thanks for the reply, I've roughly plotted out this function and I am getting a different answer to the theoretical value.
what would you estimate it to be?
from the forumula I've extracted:
vmax=60 p-p = 120
f=72.892Hz
T=13.71ms
assuming all that...
Homework Statement
v = 60sin(458t + 60º)
Find the voltage of the waveform when t = 5ms
It may be a silly question but I've just started studying ac, just wondering how that + 60º affects the equation?
any help would be appreciated as I've looked everywhere for the method...