Recent content by JosefMTD

Ya, this is a single slit enveloped a grating

Ya, this is a single slit enveloped a grating

Hello! I have done this experiment and managed to find this data: http://puu.sh/bO8TJ/ac3520ec8b.png Am I right for saying that the position that reaches the value of 5 amps are the central bright and the position 220-ish as the 1st bright order and the position 260-ish as the 2nd bright...

Sorry for not using the latex: Calculating total impedance in a phase {\frac{1}{Z_p}} = {\frac{1}{Z_L}} + {\frac{1}{Z_R}} {\frac{1}{Z_p}} = {\frac{1}{754 j\Omega}} + {\frac{1}{3R}} {\frac{1}{Z_p}} = {\frac{-j}{754 \Omega}} + {\frac{1}{3R}} {\frac{1}{Z_p}} = {\frac{754\Omega - 3Rj}{754 j\Omega *...

Homework Statement f = 60 Hz pf = 0.8 (lagging) Homework Equations Z(L) = 2 (pi) * f * L Given z = x + iy tan(angle) = y/x pf = cos(angle) The Attempt at a Solution I changed the Wye-connected resistors to an equivalent Delta-connected resistors: The new resistors become 3R in the...

I do agree I need to read the basics again. "OK. That's fine" I'll assume that equations I brought up is correct then. Thank you all for helping me to solve this problem. ;D

OH OH OH OH OH LOL LOL Yeah, I messed up XD I realized where I did wrong XD Thank you guys for bearing with me ;D (V2 - V1)/5 ohm = 2 A V2 - V1 = 10 V V2 = 20 V V3 = 0 V Voltage across the constant current source = 20 V with the active sign convention, the power on constant current source is...

Because the current generated from the constant voltage source would always make it 2A to the opposite side of constant current source. Ka boom.

OMG. @256bits V2 = 0 V (V1 - V2)/5 ohm = 2A [KCL] (10 V - V2) = 10 V V2 = 10 V - 10 V = 0 V No voltage across current source. Do we use the generator-load convention to make the 20 W on voltage source to be generating power not absorbing because it's positive? Thank you guys so much :D...

1. 10V 2. With using KCL on the node between current source and resistor, I got 10 V across the resistor I don't know why, but I never had problems in solving these kinds of problems, I think I forgot all of the basics because I had a too long vacation :P

Oh, I see. V1 = 10V V2 according to KCL ∑I = 0, V2-V1/5Ω - 2A = 0 V2 - 10V = 10V, V2 = 20V V3 according to KVL V2-V3 = V3-V0 20V = 2V3 V3 = 10V Damn, I'm confused :confused:

I'm taking reference from the previous question which is shown up there, dunno, I figured it should be the same, or it shouldn't? :o

Because of the active sign convention. I did it like I did on the previous question: http://puu.sh/7h4PK.png

Well, thank you very much. I must have fail to listen carefully during class while this happens :P But then, Power on resistance = (2A)(2A)(5Ω) = (10 V)(10 V)/(5Ω) = 20 W Power on voltage source = (10V)(2A) = 20 W Power on current source = (-10V)(2A) = -20 W It adds up to 20W, but...

I think it's a constant current source. It's always of the same value and stays on the same direction, does it override all other current source though? I have no idea, I may have missed this during class :-\