I've spent all day trying to figure out where I am going wrong with this problem:

enter image description here My mesh analysis attempt must have something fundamentally wrong with it as my results are giving me crazy numbers:

enter image description here

I have also atttempted Nodal but I can't post it in this question as I have reached my image limit. Any help would be massively appreciated. I've spent pretty much all day on this one problem and I'm sure that I'm 90% of the way there. Thanks


1 Answer 1


Bad news is that your \$I_3\$ equation is wrong, good one is you shouldn't write it at all.

Writing \$14\,I_3-4\,I_2=0\$ your are just considering voltage across current generator \$I_3\$ naugh.
But this is not (generally) true, this voltage will be some unknown value.

What you can do is just skip it so far. You only have two unknowns \$I_1\$ and \$I_2\$, then you only have to solve their system, substituting known -6A whenever you find a reference to \$I_3\$.

So, your \$I_2\$ equation will become:

\$-2\,I_1+9\,I_2-4\,I_3=0 \quad\Rightarrow\quad -2\,I_1+9\,I_2-4\times(-6)=0\$

which can be used to build this 2x2 system $$ \left\{\begin{align} 2\,I_1-2\,I_2&=10 \\ -2\,I_1+9\,I_2&=-24 \end{align}\right. $$

Generally a circuit with M meshes and C current generator branches is solved with an M-C unknowns system. E.g. 3-1=2 in your sketch.

Afterwards, once solved the above, you have all the three currents, the circuit is solved. If you happen to need voltage across current generator you just use Ohm's law.

  • \$\begingroup\$ So just to clarify, in regards to the I2 equation all I'd need to do is replace I3 for -6A? ie -2I +9I2 -4(-0.6) = 0? \$\endgroup\$
    – ChrisD91
    Aug 24, 2016 at 20:37
  • \$\begingroup\$ See my edits above :) \$\endgroup\$
    – carloc
    Aug 24, 2016 at 20:56
  • \$\begingroup\$ Excellent, I'm getting much more sensical answers now. I knew that it was something simple! Thank you \$\endgroup\$
    – ChrisD91
    Aug 24, 2016 at 21:02

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