I have another circuit to analyze here. help please.
perhaps u should try another forum. there's an engineering forum!
Node V1 : 15 = V1(1/1 + 1/2) - Vx(1/2)
if you see there is a resistance in serie with a current source between node V2 and ground
right?
yes
then you can remove that resistance of 2 ohm , right?
yes. !
ok then at node V2 -15 - Vx/9 = V2 (1/1) - Vx (1/1)
and node Vx 0 = Vx(1/2 + 1/1 + 1/3) - V1 (1/2) - V2 (1/1)
then you have 3 equations and 3 variables (V1,V2,Vx)
do you understand what i did? ^^
PS I'm not very tidy
it's alright, i'll try to do as u said. thank you so much! :D
i found Vx = 12.857V
question. why not V1-Vx(1/2)-V2-Vx(1/1) ? at node Vx?
if you see in all equations of node i folow these steps : at node VA 1.) the sum of all currents , in current at node (+) out current (-) 2.) it is equal to : sum of: - this node multiplied by admittances(inverse of resistors) wich come to that node - other nodes that are connected to this node for one (or more) resistances multiplied for the admittance between then but with sign - - so on for all resistances
so for node V1there incoming current : +15 this node is V1 so (there are 2 resistances [1 and 2 ohms] connected to this node) V1*(1/1 + 1/2) remember there are 2 resistances connected to V1 that is 1 that connect to V1 with "ground" , if is ground , i won't take that node because it is 0, and resistance of 2 ohms connect to node V1 with Vx so Vx*(1/2) but with negativ sign! then i will have 15 = V1*(1/1 + 1/2 ) - Vx*(1/2)
if you do the same for node Vx you will get this equation 0 = Vx(1/2 + 1/1 + 1/3) - V1 (1/2) - V2 (1/1)
ok, thank you very much. i get it now! thank you for explaining it to me. :)
tell me the solution that you find ;)
i got 0 for Vx.
-5 for V1 and - 1.5 for V2
I found V1= 5.714 V2 = -26.428 Vx = -12.85
i used cramer's to solve for the unknowns, why can't i get it?
what are your matrix?
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