I am a bit stuck on the conversion process, would appreciate it if someone could offer some advice. I have attached the word document as I do not know how to use the symbols correctly.
The second step of your working out contains an error -- you can't write sin(a+b) as sin(a) + sin(b). You have to use the sum/difference formulas.
I apologise for the new thread, this is actually a reply to the conversion of equation thread I started. My replies are for some reason not appearing. I dont know whether there is a massive queue or I am doing something wrong. Poster Someonesdad replied to my original post and highlighted my mistake; I have followed the advice and had another bash. Could someone please check if I have done it right this time. I dont need answers, just tips on where I went wrong if applicable. I have attached my working out on a word document. Thanks you in advance.
For some reason, the system decided to auto-moderate your posts, rendering them invisible. I have merged your new thread here, where it belongs. We apologize for the inconvenience.
I don't have time at the moment to check your work, but here's a tip. When doing such derivations, you can use a tool like python/numpy/matplotlib to plot the functions you derive. Since you're trying to simplify an identity, your plots of the equations should all be the same. If not, it can identify where the error is. The nice thing is that the plotting routines are pretty simple to write: from pylab import * t = arange(0, 4*pi, 4*pi/1000) plot(t, sin(4*pi*t)*cos(8*pi*t - 1.37) show() I'm just making up an example because I'm in a hurry. The point is it just takes a few lines of code to see your result.
I don't have a copy of your textbook, but I think that Boylestad must say something more, especially by page 778. You are discussing power in a purely capacitive circuit. So the current leads the voltage by ∏/2. You can substitute this value into you working, rather than try to introduce RMS values in order to simplify. Your starting point equations are the instantaneous values for current and voltage and their product is the instantaneous power. I have plotted these equations along with their product. This shows that the power wave has twice the frequency of the original. Half the peaks are positive and half are negative. The average value of power over a complete cycle is zero. For half the current or voltage period the capacitor draws power from the supplies and for the other half it returns it to the supply. You should also do this exercise with pure resistive pure inductive circuits to compare. go well