Recent content by newbie991

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...