0
\$\begingroup\$

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

\$\endgroup\$
2
\$\begingroup\$

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\$

\$\endgroup\$
4
  • \$\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

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.