1
\$\begingroup\$

This is how I approached the problem.

enter image description here

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?

\$\endgroup\$
  • \$\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\$ – helloworld922 Jan 25 '16 at 20:01
  • \$\begingroup\$ @helloworld922 I didn't get you. Will you please elaborate? \$\endgroup\$ – curiousbrain Jan 25 '16 at 20:04
  • \$\begingroup\$ In the very last line, you have \$V_B = -49.37 A\$. \$\endgroup\$ – helloworld922 Jan 25 '16 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 '16 at 20:23
3
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.