Basic circuits need help.

[iman00b]

Ive been trying to do some problems for the past 6 hours and got only 2 done and they are so simple but i just cant figure them out. Its been a while since ive taking this class so im in a bit of trouble. I could try to email or AIM the circuit pix. I dont need them solved but want to know what im doing wrong.

 

[So]

post the pics on pics.bbzzdd.com and link them here. There are plenty of EE members here who will be happy to help. So long as you don't ask us to solve them for you (which you don't seem to be)

 

[timosyy]

Originally posted by: So
post the pics on pics.bbzzdd.com and link them here. There are plenty of EE members here who will be happy to help. So long as you don't ask us to solve them for you (which you don't seem to be)

Or imageshack, photobucket, w/e

 

[PowerMacG5]

I took my circuits course last semester, so I can probably help.

But yeah, post it here, so others can help/see the solution.

 

[NaOH]

I past that class 2 years ago and have taken advanced circuits. if you post the circuit I can probably help you. POST!

 

[iman00b]

http://img502.imageshack.us/my.php?image=aa2cc.jpg

oops should have mentioned that im looking for thevenin equiv.

 

[So]

Are you just trying to find Vab?

 

[drinkmorejava]

um, what's the question

 

[NaOH]

alright, to find the load R, think of the sources as open and close circuits. Voltage source would be short and current source would be open circuit. Then find the total resistance in the circuit and that is your thevenin load R.


to find the thevenin circuit voltage, just find the voltage at the open circuit of the complete circuit.

Your gonna have to do a bunch of mesh equations to solve for the voltage across resistor R6. Once you have that, you have the two things necessary for a thevenin equivalent circuit:

Thevenin Open Circuit voltage, and load resistance.

 

[iman00b]

i got the top 3 nodes made currents going away from them = 0. so theres 3 equations and ix is the voltage drop on the 4k resistor divided by the resistor so i can substitute all those into each other and find the voltage at A which is the thevenin equivalent correct?

 

[NaOH]

ix looks like the current going t hrough the 4k resistor.

 

[blustori]

Originally posted by: AMDUALY
alright, to find the load R, think of the sources as open and close circuits. Voltage source would be short and current source would be open circuit. Then find the total resistance in the circuit and that is your thevenin load R.


to find the thevenin circuit voltage, just find the voltage at the open circuit of the complete circuit.

Your gonna have to do a bunch of mesh equations to solve for the voltage across resistor R6. Once you have that, you have the two things necessary for a thevenin equivalent circuit:

Thevenin Open Circuit voltage, and load resistance.

.
Great answer! :thumbsup:

 

[NaOH]

did you get your thev. load to be 11.34 or something

 

[iman00b]

If you label the top 3 nodes left-right 1,2,3 i got:
(V1-90)/15k + V1/10k + (V1-V2)/4K = 0
(V2-V1)/4k + V2/40k + (V2-V3)/5K - 19ix = 0
V3/89K + (V3-V2)/5K + 19ix = 0
these are currents leaving the node = 0
ix = (V1-V2)/4k
Substitute and solve for V3 which is the Vth

^ i dont know if i did the Rth correct cuz i got 44.5k

dammit i just noticed what im doing wrong.

 

[NaOH]

everything looks right, just not sure about that dependent source. Let me try doing mesh and see if i get the same thing as you.

 

[iman00b]

nevermind no i dont kno what im doing wrong lol

 

[NaOH]

alright, these are my mesh equations. You can punch them in your graphing calculator (matrices or else it won't work, I'm sure you learned how to do this) and solve for i3. Then multiply i3 to 89K for the VOC.
I don't have a graphing calculator on me.

90V = 25K(i1) - 10K(ix) Mesh 1

0 = -10K(i1) - 40K(i3) + 54K(ix) Mesh 2

0 = 134K(i3) - 135K(ix) Mesh 3

3 equations 3 unknowns

 

[iman00b]

sorry didnt learn how :[ i could do them by hand?

 

[NaOH]

Goddamn, I forgot what the method is called. Or I could give you a link.

This should help a lot :

http://mathworld.wolfram.com/CramersRule.html

 

[NaOH]

Remember this method, you will use it a lot.

 

[krotchy]

my school taught an incremental method, where you basically just chop and gather pieces of the circuit until you have your circuit made

im going to do this in my head, cuz im to lazy to get a calculator, so here it goes:

So for node 1, you have 90V accross 15K, So you could model this as a 4mA(90V/15kOhm) current source in parrallel with a 15k Resistor.

So now we have the 15k and 10k in parallel with a 4mA current souce, which is the same as the 4 mA current source and a 6k resistor in parrallel 1/(1/10k+1/15k)

Then we rebuild this new "source" as a voltage source in series with a 6k resistor. So we have a 4mA x 6k voltage source = 24V.

So now we have removed R1 and R4 and V1 and can replace that chunk with a Voltage source of 24V and 6K series resitance. Since this is in series with R2, we can model the resistance as 10k

Boom, now we have the block of (24V in series with 10K) in parrallel with R5. And we have removed R1,R2 and R4

Just continue this process until you have removed everything.

 

[iman00b]

I tried doing that but i have no clue what to do with the current controlled current source. I tried solving it again for the 1000th time and still no luck getting an answer. Does anyone know the answer and can show how they did it? Im checking my answer with pspice but never close yet.

 

[mrrman]

I started to do a nodal analysis on it and stoped...what are you trying to find? iX??

 

[NaOH]

Originally posted by: krotchy
my school taught an incremental method, where you basically just chop and gather pieces of the circuit until you have your circuit made

im going to do this in my head, cuz im to lazy to get a calculator, so here it goes:

So for node 1, you have 90V accross 15K, So you could model this as a 4mA(90V/15kOhm) current source in parrallel with a 15k Resistor.

So now we have the 15k and 10k in parallel with a 4mA current souce, which is the same as the 4 mA current source and a 6k resistor in parrallel 1/(1/10k+1/15k)

Then we rebuild this new "source" as a voltage source in series with a 6k resistor. So we have a 4mA x 6k voltage source = 24V.

So now we have removed R1 and R4 and V1 and can replace that chunk with a Voltage source of 24V and 6K series resitance. Since this is in series with R2, we can model the resistance as 10k

Boom, now we have the block of (24V in series with 10K) in parrallel with R5. And we have removed R1,R2 and R4

Just continue this process until you have removed everything.

That's not exactly how you do it in this case.... You don't even know what he's trying to solve for.