Recent content by butchsd

Thanks for all the help :P

Ok, so U would be R*IS = 10*2∠0° = 20∠0° V That makes 3U = 3*20∠0° = 60∠0° V Yeah let's just disregard the thevenin thing I tried. :) Voc = 3U + U = 4U = 4*20∠0° = 80∠0° V = 80cos(ωt) V Zt = Voc / IN = 80∠0° / 1.940∠-14.04° = 14.237∠14.04° Ω ≈ 40+j10 Ω

Ah yes, my mistake. That should be 1.940∠-14.04° A Okay so let's first determine U over R so I can determine 3U: IR = IS - IN = 2∠0° - 1.940∠-14.04° = 2 - 1.882 - j0.471 = 0.118 - j0.471 = 0.486∠-75.9° A U = R*IR = 10 * 0.486∠-75.9° = 4.86∠-75.9° V 3U = 14.58∠-75.9° V Incidentally...

Disregard the multiple norton thing, just spitballing. :) Let's give the mesh analysis a try. So if I solve the mesh analysis for IN I get this: IS: IS = 2.0∠0° A IN: -R(IS-IN) - 3U + ZIN = 0 Ohm's law: -R(IS-IN) - 3R(IS-IN) + ZIN = 0 -4R(IS-IN) + ZIN = 0 Factor into IN and IS...

Hmm. Since I know that UR = 20∠0° A, I know that 3U is 60∠0° A. So then we have a Norton in serial with a Thevenin. Let's play a bit. If I first convert the circuit to two Nortons in series, maybe it'll be better. Thevenin -> Norton for the dependent source and Z: Isc = 60∠0° / j10 =...

Here's what I have (or think I have) so far: An open circuit analysis shows that the current over R is Is: IR = 2.0∠0° A. Voc = IR * R = 2.0∠0° * 10∠0° = 20∠0° A.. Closing the circuit from a to b and doing current division shows: Iab = Is * R / (R+Z) = 2.0∠0° * 10 / (10+j10) = 2.0∠0° *...

Hi! I'm having some difficulty understanding how to go about solving the following problem. I'm a beginner at this and am completely lost. I've seen a few similar problems that could be solved with node/current analysis, but I don't really understand if or how to apply that here. Can someone...