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I am doing the nodal analysis of this circuit:

enter image description here

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

but what about i6?

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.

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

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  • \$\begingroup\$ You are right!!!!!!!!!!! I can see this clearly if I draw the current source over R3. Now I see the whole thing. BRILLIANT! THANKS!!!!!!!!!!! \$\endgroup\$ – SpaceDog May 31 '18 at 17:33
  • \$\begingroup\$ 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. \$\endgroup\$ – laptop2d May 31 '18 at 17:33
  • \$\begingroup\$ Thanks for a good question and an attempt at a solution, most people just throw the question out and expect a solution \$\endgroup\$ – laptop2d May 31 '18 at 17:34

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