# Nodal analysis: how can I represent the current i2 in terms of voltage over R2?

I am doing the nodal analysis of this circuit:

I have successfully wrote the current equations for node A, using nodal analysis, that I found it to be:

i3 = i5 + i7 or

(Vc-Va)/R3 = (Va-Vb)/R5 + i7


where i7 is the current source value of 1A.

That will give me

-9Va + 3Vb + 5Vc = 400


So far, so good...

Now analyzing node C I have a doubt.

The currents for node C would be:

i1 = i3 + i6


where

i1 = (V1-Vc)/R1 = (12-Vc)/20

i3 = (Vc-Va)/R3 = (Vc-Va)/80


I know that i2 = i6 + i7 resulting on i6 = i2-i7 or substituting the value for i7,

i6 = i2 - 1


The problem here is, how can I declare i2 in terms of voltages and resistors? If the current source was not there I would say i2 = Vc/R2 but with the current source there I don't now.

Any help is welcome. Thanks.

• i6 = Vc/R2 - i7. Ohm's law must hold for R2! – Chu May 31 '18 at 17:22
• duh! Obviously! Thanks. Please convert this comment to an answer, so I can accept. – SpaceDog May 31 '18 at 17:24

The problem here is, how can I declare i2 in terms of voltages and resistors? If the current source was not there I would say $i_2 = V_c/R_2$ but with the current source there I don't now.

The problem is, i6 doesn't really exist, C is the same node as the both the nodes on either side of the current you have drawn for i6. Since the voltage for node C is the same as the node between R2 and the current source, the resistance for i6 would be zero and technically the current would be infinite.

A better node equation for node C would be this: $0 = i_1 -i_2 -i_3 + i_7$ And you need to write C on the node between R2 and the current source

The voltage equations for R2 is $i_2 = V_c/R_2$ so you are correct, you can take any resistor and the two nodes around it and find the current through the resistor

$\frac{V_A- V_B}{R} = i_R$

In this case Va would be Node C and the other side is ground, which you have correctly identified.

• You are right!!!!!!!!!!! I can see this clearly if I draw the current source over R3. Now I see the whole thing. BRILLIANT! THANKS!!!!!!!!!!! – SpaceDog May 31 '18 at 17:33
• Yes it is, but it is but there are also other currents to consider here, it is important to make sure you identify nodes correctly. – laptop2d May 31 '18 at 17:33
• Thanks for a good question and an attempt at a solution, most people just throw the question out and expect a solution – laptop2d May 31 '18 at 17:34