Nodal Analysis 1 equation 2 Unknowns Problem

[eatsleep]

http://imgur.com/W9sWTv7V=IR, that is all I think
3. I have put my ground at the bottom branch. I have tried to write a KCL for the top branch but end up with 0=0 when I substitute in equivalencies. My KCL: (V1+5Ix-0)/20+2+Ix=0. (V1+5Ix)/20=-2-Ix. So 0=0. I am not sure if I am writing my KCL correct also I know I am missing something.

Can I make the voltages sources and the top node for a supernode?

 

[SteamKing]

I think you are being hasty in concluding that you first equation develops into 0 = 0. The algebra doesn't support this conclusion.

 

[rude man]

Your first expression isn't even an equation - where's the = sign?

It says to use nodal analysis. Are you summing currents to zero at each independent node?

(I myself don't ever use KCL. I sum currents at each node to zero - is that KVL? I don't know. Never heard of a supernode either).

 

[eatsleep]

I think you are being hasty in concluding that you first equation develops into 0 = 0. The algebra doesn't support this conclusion.

Is it correct that (V1+5Ix)/20=-2-Ix ? If so that is where 0=0 comes from

 

[SteamKing]

If (V1 + 5Ix)/20 = -2 - Ix,
then
(V1 + 5Ix) = 20*(-2 - Ix)
(V1 + 5Ix) = -40 - 20Ix
V1 + 25Ix = -40

which is not 0 = 0

 

[The Electrician]

My KCL: (V1+5Ix-0)/20+2+Ix=0. (V1+5Ix)/20=-2-Ix.

This is wrong. It should be:

(V1-5Ix-0)/20+2+Ix=0

 

[NascentOxygen]

With the Electrician's correction, you have one equation. Your second equation comes from the branch on the right side, giving its own equation relating V₁ and I_X.