Hi all.
I am stuck on the above question.
I managed to do part a), but not sure whether my solution for part b) is correct.
My attempt: (Fixed URL)
http://s10.postimg.org/a9y8ti8mh/New_Doc_1_1.jpg
Last edited:
Oops, wrong link, haha. Will fix it shortly.Your link appears to be to your question, not your attempted answer?
b) (i)What does the negative sign imply for the ratio of a voltage across a resistor to the current through it?
Awesome, my company internet blocked that URL and called it "adult and pornography".
I wrote: The magnitude of the current is solely given by i(2), as there is no current entering the second loop. The negative sign is due to the passive sign convention?Why not just post it here?
What about my question on the negative sign?
Hmm, that actually sounds like a better idea. I will try this tomorrow, and report back.Hi,
I think instead of using KVL and crunching through the math as you had done, you could very well use the result from part a.
Take Z1=1/(j*w*C) ; Z2=j*w*L; Z3=Z1;
Expand the matrix form equation as
V1=Z11*I1 + Z12*I2;
V2=Z21*I1 + Z22*I2;
Also use the fact that V2=-I2*R;
With this express I2 as a function of I1, and from expression of V1 you can get Zin as V1/I1.
Turns out this is indeed the best way to solve it IMO, worked a treat. Thanks.Hi,
I think instead of using KVL and crunching through the math as you had done, you could very well use the result from part a.
Take Z1=1/(j*w*C) ; Z2=j*w*L; Z3=Z1;
Expand the matrix form equation as
V1=Z11*I1 + Z12*I2;
V2=Z21*I1 + Z22*I2;
Also use the fact that V2=-I2*R;
With this express I2 as a function of I1, and from expression of V1 you can get Zin as V1/I1.
Hmm, this is what I got? Not sure if it is correct??Hi,
I think instead of using KVL and crunching through the math as you had done, you could very well use the result from part a.
Take Z1=1/(j*w*C) ; Z2=j*w*L; Z3=Z1;
Expand the matrix form equation as
V1=Z11*I1 + Z12*I2;
V2=Z21*I1 + Z22*I2;
Also use the fact that V2=-I2*R;
With this express I2 as a function of I1, and from expression of V1 you can get Zin as V1/I1.
Hi,You're making finding Zin much harder than it needs to be.
Forget about two-ports; it's just a simple ladder network. To find Zin, start at the right side. You've got R in series with 1/jωC. Then that combination is in parallel with jωL, and finally add 1/jωC in series, simplify and you have Zin.
Hi Electrician, the original poster has vanished and I suspect his assignment is past due now, but part C was interesting, thought you may be interested in an alternative approach (of no use to the original poster - he was set an algebra grind).What happened to part C? Even more algebra is involved, but the final answer is amazing simple.
by Jake Hertz
by Jake Hertz
by Jake Hertz