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doodle
doodle is offline
#8
Jan18-07, 12:30 AM
P: 161
Quote Quote by VinnyCee View Post
[tex]V_4\,=\,V_a\,=\,V_3[/tex]
As I have said earlier, to claim that V3 = V4 is incorrect. Read my earlier post.

Quote Quote by VinnyCee View Post
Just what you said above. Now, I use [itex]i\,=\,\frac{V}{R}[/itex] to get the new currents in green.

KCL: [tex]I_3\,+\,I_4\,+\,I_5\,+\,\frac{V_b}{4K\Omega}\,=\,0[/tex]
Well yes, that is right. But you do know that the above expression gives only 1 nodal equation, that is, the nodal equation for node 3. You would have to do the same thing for the other 3 nodes.

Quote Quote by VinnyCee View Post
[tex]I_3\,=\,\frac{(5\,V)}{4000\Omega}\,=\,0.00125\,A\,=\,1.25\,mA[/tex]
This is not right- you're forgetting V3.

Quote Quote by VinnyCee View Post
[tex]I_4\,=\,\frac{V_2\,-\,V_3}{2000\Omega}[/tex]

[tex]I_5\,=\,\frac{V_1\,-\,V_3}{4000\Omega}[/tex]
These are correct.

And yes, combine these terms into an equation. Note that Vb = V2. Also, at the moment, let's forget about V2 = V4-2. Now write down again the nodal equation for node 3.