KVL and KCL *updated*

[Omegachi]

hmm, i am a little confused here trying do my prelabs.

here are the problems
linky

okay, for #3 a), to find the equations for the node voltage for v1,v2,v3, using kcl, do i use the form:

(v1-v3)/3 + (v1-v3)/1 + (7-v1)/2 = 0 for eq1

then find the current on v2 and v3 then resulting in 3 equations?


for problem d) do i use (I1-I2)*r to find the currents?

sorry, i am just a little confused here and i need some help. thanks

 

[Chaotic42]

Man, I wish I could go back to college!

 

[ArmenK]

for 3a the 3 equations are:
(v1-v2)/1 + (v1-v3)/3 + (v1-7)/2 = 0
(v2-v1)/1 + (v2-8)/4 + (v2-0)/5 = 0
(v3-v1)/3 + (8-v2)/4 + (v3-0)/6 = 0

i think

 

[Stojakapimp]

i think you typed that up wrong. It should say (v1-v2)/1. Other than that, it looks like the equation to the first part was right. The only thing that might be wrong is the + signs. one of them might need to be negative. So I guess i'm not much help at all

 

[Stojakapimp]

Originally posted by: ArmenK
for 3a the 3 equations are:
(v1-v2)/1 + (v1-v3)/3 + (v1-7)/2 = 0
(v2-v1)/2 + (v2-8)/4 + (v2-0)/5 = 0
(v3-v1)/3 + (8-v2)/4 + (v2-0)/6 = 0

i think

in your 2nd equation, it should be (v2-v1)/1 and in the 3rd it should be (v3-0)/6

 

[ArmenK]

Originally posted by: Stojakapimp

Originally posted by: ArmenK
for 3a the 3 equations are:
(v1-v2)/1 + (v1-v3)/3 + (v1-7)/2 = 0
(v2-v1)/2 + (v2-8)/4 + (v2-0)/5 = 0
(v3-v1)/3 + (8-v2)/4 + (v2-0)/6 = 0

i think

in your 2nd equation, it should be (v2-v1)/1 and in the 3rd it should be (v3-0)/6

you are correct sir, fixed the typos

 

[Omegachi]

you guys ever used matlab before? i think that program is pretty good, but then the sucky part about the program is that you have to solve for the equations and varibles by hand before you can crunch it into the computer. is there a program that allows you to physically build the ciruit in the program and solves for variables and plots all for you with a click of a button?

 

[ArmenK]

spice ?

 

[Stojakapimp]

yeah, i've used matlab in two classes so far. I haven't really done to much complex stuff with it, but it seems pretty powerful.

And to answer your question about part d, you should just write out 3 different KVL's for those loops. I think that middle wire uses the mesh method, but if you do the loops i don't think you have to worry about that.
Just make sure you assign polarities to the resistors and that your +'s and -'s are correct in your voltage/current relationships

 

[Omegachi]

hmm... i am kinda behind in my circuit analysis, its kinda sad . i will need to work super hard to catch up to all the stuff that is being taught in my current lectures. ( i think we are doing opamps)

 

[Omegachi]

Originally posted by: ArmenK
for 3a the 3 equations are:
(v1-v2)/1 + (v1-v3)/3 + (v1-7)/2 = 0
(v2-v1)/1 + (v2-8)/4 + (v2-0)/5 = 0
(v3-v1)/3 + (8-v2)/4 + (v3-0)/6 = 0

i think

oh oh, why is (v2-0)/5 ? why 0 and not 7v?

 

[Stojakapimp]

Well first off, label voltages across every resistor...such as calling a 4 ohm resistore V4 or something like that. Then assign polarities to those voltages. It doesn't matter at all how you assign those polarities. Now you already have currents and their directions assigned so you don't have to worry about that. So just do KVL and go in the separate loops. For instance, in the loop i1, I believe KVL would give you something like 8V + V4 + V1 + V3. Now remember, those V4, V1, and V3's are different than what is already given on that picture. Now just use ohm's law for V4, V1, and V3. But remember that when you have current flowing into the negative end of the voltage, ohm's law becomes negative. So it would turn out to be 8V - 4i1 - 1i1 - 3i1 = 0
At least I think that's what it is. There's a possibility that it's -8V but i don't think so...but you get the point

 

[Stojakapimp]

Originally posted by: Omegachi

Originally posted by: ArmenK
for 3a the 3 equations are:
(v1-v2)/1 + (v1-v3)/3 + (v1-7)/2 = 0
(v2-v1)/1 + (v2-8)/4 + (v2-0)/5 = 0
(v3-v1)/3 + (8-v2)/4 + (v3-0)/6 = 0

i think

oh oh, why is (v2-0)/5 ? why 0 and not 7v?

Because the other side is grounded

 

[Omegachi]

Originally posted by: Stojakapimp

Originally posted by: Omegachi

Originally posted by: ArmenK
for 3a the 3 equations are:
(v1-v2)/1 + (v1-v3)/3 + (v1-7)/2 = 0
(v2-v1)/1 + (v2-8)/4 + (v2-0)/5 = 0
(v3-v1)/3 + (8-v2)/4 + (v3-0)/6 = 0

i think

oh oh, why is (v2-0)/5 ? why 0 and not 7v?

Because the other side is grounded

did not know that. thanks

 

[Omegachi]

whee! hey guys, i need help again. (hope you guys are still awake.

okay, lets check and see if omegachi did his kvl right.

(I1-I3)3 + 8 + (I1-I2)4 + I1(1) = 0
(I2-I1)4 - 8 + (I2-I3)6 + I2(5) = 0
I3(2) - 7 + (I3-I1)3 = 0

okay, now for the matlab part, how do i go about doing this in the program? (matlab newbie)

* i want to try and finish this lab assignment so i don't have to go to the lab session tomorrow.*

 

[tigerbait]

"what's that horseshoe looking thing? "

actually quote from someone in circuits II class

 

[Omegachi]

morning bump