Mesh current formulas gives two different results for i2. The formula derived from the second loop provides a current of 1A which is the correct answer. However, when I write my equation using the first loop I get a current of 5A. Why does the first loop provide an incorrect current value? My work is below. I also provided a screen shot of multisim verifying the correct i2 value should be 1A.

My work

Multisim verifying i2=1A

multisim results


There is voltage across the current source. Therefore, your KVL equation for the first loop is not correct.

It is logical to find this voltage with the solved \$i_2\$ in your second loop,

\$V_{2A}=12 + 4\times (i_1-i_2)=12 + 4\times 1=16 \ \rm V\$

  • \$\begingroup\$ Does that mean you can't write a mesh equation anytime there is a current source in the loop? \$\endgroup\$ Jan 6 at 3:08
  • \$\begingroup\$ @KoreyLombardi Yes. This is the basic idea of the "supermesh". \$\endgroup\$ Jan 6 at 3:10
  • \$\begingroup\$ Awesome, thank you for the quick answer. \$\endgroup\$ Jan 6 at 3:11
  • 1
    \$\begingroup\$ @KoreyLombardi You don't need a supermesh concept. Just FYI. You can just create an unknown voltage variable for the current source voltage and solve things using regular mesh ideas. But supermesh also works. I just think it's an extra, unnecessary concept that can be avoided. \$\endgroup\$
    – jonk
    Jan 6 at 3:50

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