Recent content by stef6987

No need for the KCL, if all currents are going clockwise, you can say I2 = 4<0 + I3, i'll post up my results when i get home if you like? and have you done Q7, i only have that to do and I'm not sure if my transfer function is right, my numbers are ridiculous haha.

haha :P this one was probably the longest, just write all 4 kvl's, because they are ideal the only purpose of the dots is to show the relationship between the transformer voltages and currents, so if you start from the right mesh you can work your way to the left and solve Vs, takes time but eh...

why are you multiplying the voltages by the currents? :S

i got 9.4j - 18.36 volts :S how'd did you do it?

just use KCL, it's pretty straight forward if you do it like that And yeah i got the same results for 4 & 5. have you done 6 yet?

i re did mine and got 0.231<-133.4 degrees or something around their, your answer is rectangle form should be -0.158 - j0.167. because both terms are negative it lies in the third quadrant, so you have to subtract Pi radians or 180 degrees, and that is the answer,

Oh ok i think i follow, i realized i had a small mistake (not sure why i had the extra s ahah) it should be: u(t) - (2/3)*e-(4/3)t i need to find I for t>0, so at t>0 u(t) = 1, is that what you're implying? so my answer should be: 1 - (2/3)*e-(4/3)t

well A = 1, 1/s = u(t) so what else could i do to that? my final solution was: 1/u(t) - (2/3)*e-(4/3)t/(s+(4/3))

ahh, the answer was so simple, i used the inverse laplace transformation the equation is in the form: s+4 = A/s + B/(3s+4) A = 1 B = -2 1/s = u(t) B/(3(s+4/3) = -2e-(4/3)t*(3) i'm not to sure do i just multiply the 3 to the equation?

Ok, now that i got that right :P i need to convert it back to the time domain, but no solution in my list seems to match. any ideas?

hmm i think i made a few mistakes, i tried it again and this is what i get : kvl 1: -4/s + i1 + i1/s - i0/s => 4/s = i1 + i1/s - i0/s kvl 2: 2io + io - 1/(s+1) + i0/s - i1/s = 0 Now i rearranged kvl 2 to find i1 and i got: i1/s = 3i0 - 1/(s+1) + i0/s = i0(3 + 1/s) - 1/(s+1) i1 =...

haha :) My final answer was 2.30<46.59 volts, i don't think a little difference would matter much :) how did you go on the rest of the assignment? but... I'm pretty sure my answer is off a bit, the angle should be 46.59 - 180... due to it being in the 3rd quadrant.. so my angle would be...

Sorry i just realized you can use your method, you made one mistake, the current through the inductor is i1-2ix

EC203 assignment? ahah i got 46.59 degrees as the final angle, don't use KVL, you know at the 4k resistor the voltage is 16<45, all you need to do is apply a KCL at node v0(t) and just solve. your final answer is very close to mine, try my method and I'm sure your answer will be the same as your...

I just used a simple KCL, seemed like a more simple answer, thankyou so much for your help, it's very appreciated :)