This is how I approached the problem.

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I however do not see any wrong with my approach. But I am not getting the correct answer. Is there anything to do with grounding? Or am I missing something else?

  • \$\begingroup\$ How did you get an amperage for \$V_B\$? That's definitely wrong. I haven't looked through the rest yet to see what else is wrong. \$\endgroup\$ Jan 25, 2016 at 20:01
  • \$\begingroup\$ @helloworld922 I didn't get you. Will you please elaborate? \$\endgroup\$ Jan 25, 2016 at 20:04
  • \$\begingroup\$ In the very last line, you have \$V_B = -49.37 A\$. \$\endgroup\$ Jan 25, 2016 at 20:06
  • \$\begingroup\$ In your first equation for node B, you have \$I_3 + I_4 + I_5 = 0\$ but your diagram shows \$I_3\$ flowing in to node B, while the other currents flow out. So that is a sign error. \$\endgroup\$
    – The Photon
    Jan 25, 2016 at 20:23

1 Answer 1


You need to be consistent with your signs and assumptions.

In writing the KCL equation for node A, you defined it as

\begin{gather} I_1 + I_2 + I_3 = 0 \end{gather}

where \$I_1\$, \$I_2\$, and \$I_3\$ are all leaving the node (consistent with the arrows in your diagram). However, then you immediately define

\begin{gather} I_3 = \frac{V_B - V_A}{2k\Omega} \end{gather}

This is the negative of the convention you just established. I would go through and double-check all of your signs and make sure they are consistent with each other. I can see similar issues in the KCL equations for the other nodes as well. You get lucky a few times by having a double negative cancel out your mistake, but this is just getting the correct answer for the wrong reasons.


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