Homework Statement
[/B]
The photo included provides the problem but the professor stated that the right side source is supposed to be 100t and not 80t. The problem wasn't printed correctly in the book so I assume the solution wasn't either. The solution via the book is v1=2.56cos(100t+139.2°)...
but if the left side of v1 is the same polarity as the top of vs, then why would current flow through the wire at the top of the source and into v1 on the left side? starting at the source you have - then + then + then - then + then -. You have 2 +'s together...shouldn't it always go from + to -...
I appreciate the real world view Don, and the theoretical exercise as well, gneill. Maybe you two can clear up something that's slightly confusing for me. When I took circuit analysis 1 we would go around and label the polarity on components starting with the source. The source in this circuit...
Really?! Oh thank you thank you! I got that the first time I worked this out and couldn't figure out where I was going wrong! I hope these book publishers can get their act together before the 9TH EDITION geeez well thank you sir!
Ok, I have been able to get that. Then I find the current in the loop using ohm's law I = V/R which gives me 0.997sin(20t+4.4°) Amps. Then I use the current through the capacitor which I just found and the Z of the capacitor which is 0.385∠-90° to find the voltage across the capacitor, using...
Ok I see that now. I think I was confusing ideas from transients. My main problem is even when I didnt have the minus sign in there, I wasn't getting the book solution so I don't know if my process is flawed or not. The book solution is Ic=88.7sin(20t-27.5°)mA and Vc=2.31sin(20t+62.5°)Volts with...
Because the polarity of the capacitor in reference to the current flow. If the current flow Ic was in the direction towards the negative terminal of the capacitor then I wouldn't have added the minus sign to the current. Is this a wrong assumption on my part?
My book solution is 5. Where is my mistake? If I input 3 into the formula instead of 4 and run the whole process, I end up with 5.23 which is closer to the book answer. Is that formula for maxima and not minima and that's where the "1 off" error is?
ahhh that makes sense, the minima occurs last because after that, there can be no maxima because the surface the slits are in would be in the way. There are 7 minima starting on each side of the central maxima which means there is one less maxima than there are minima for a total of 6 maxima on...