I am trying to analyze the circuit given below:

enter image description here

The reference node is in the bottom with 0 V.

At nodes A and B writing down Kirchhoff’s first law equation trying to express each current in terms of the voltage across the branch.

My equation for node A looks like this:

$$ \frac{V_{G_1}-V_A}{R_1} - \frac{V_A-V_{reference}}{R_2} - \frac{V_A-V_{reference}}{R_3} + \frac{V_A-V_B}{R_4} + I_{G_2} = 0 $$

For node B:

$$ -I_{G_3} - \frac{V_B-V_{reference}}{R_5} - \frac{V_B-V_A}{R_4} + I_{G_2} = 0 $$

As of now I only have two equations and three unknowns. Therefore I am not sure if my solution attempt is correct.

  • \$\begingroup\$ VB - VA = VG2 is your last equation. With this, is the set solvable ? \$\endgroup\$
    – AJN
    Oct 4, 2020 at 17:06
  • \$\begingroup\$ @AJN thanks and how about I_G2? \$\endgroup\$
    – ytho
    Oct 4, 2020 at 17:09
  • \$\begingroup\$ You are right. You can use Super Node concept in such situations. \$\endgroup\$
    – AJN
    Oct 4, 2020 at 17:11
  • \$\begingroup\$ You are right, that would be much easier but the assignment states we must use the node analysis \$\endgroup\$
    – ytho
    Oct 4, 2020 at 17:12
  • \$\begingroup\$ Since IG2 is common to both equations, you may be able to eliminate it by subtraction. \$\endgroup\$
    – AJN
    Oct 4, 2020 at 17:13

1 Answer 1


You have done a good job drawing arrows to represent your currents. When you apply Kirchoff's current law you are just saying all current into a node = all current out of a node.

Below I'll write the node A equation and will arbitrarily assign current into the node as positive:

$$ \frac{V_{G_1}-V_A}{R_1} - \frac{V_A-V_{reference}}{R_2} - \frac{V_A-V_{reference}}{R_3} - \frac{V_A-V_B}{R_4} - I_{G_2} = 0 $$

Equally as valid - i can arbitrarily assign current into the node as negative which would result in the following equation:

$$ -\frac{V_{G_1}-V_A}{R_1} + \frac{V_A-V_{reference}}{R_2} + \frac{V_A-V_{reference}}{R_3} + \frac{V_A-V_B}{R_4} + I_{G_2} = 0 $$

Clearly, these are both equivalent.

EDIT: Using your figure and adding an arrow for IG2. I will arbitrarily decide that current into node B is negative and write a KCL equation:

$$-I_{R_4} - I_{G_2} - I_{R_5} - I_{G_3} = 0$$

enter image description here

  • \$\begingroup\$ It surely does help to clarify things. So for node B: \$-I_{G_3} - \frac{V_A-V_B}{R_4} - \frac{V_B-V_{reference}}{R_5} + I_{G_2} = 0\$ correct? \$\endgroup\$
    – ytho
    Oct 4, 2020 at 18:15
  • \$\begingroup\$ You need to draw an arrow for IG2 first. If you draw it going out of node B and you put a + sign on your third term then your above equation will be correct. \$\endgroup\$ Oct 4, 2020 at 18:17
  • \$\begingroup\$ I suppose I can not draw it going out of node B since it is already going out of node A according to the first equation. Am I right? So changing the equation for nod B to \$-I_{G_3} - \frac{V_A-V_B}{R_4} - \frac{V_B-V_{reference}}{R_5} - I_{G_2} = 0\$ would be the right solution? \$\endgroup\$
    – ytho
    Oct 4, 2020 at 18:21
  • \$\begingroup\$ Yes, once you decide on the direction of the arrow you need to keep it while writing both node A and B equations. Your equation just above will be correct if you change the sign on the 3rd term to + (because as you have written that term it is a current leaving node B). \$\endgroup\$ Oct 4, 2020 at 18:22
  • \$\begingroup\$ Isn't current \$I_{R_5}\$ entering node B because of the arrow that was already there? \$\endgroup\$
    – ytho
    Oct 4, 2020 at 18:33

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