0

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

Korey Lombardi
  • 3
  • 1

1 Answers1

2

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\$

Muyun Benjamin Li
  • 56
  • 4
  • Does that mean you can't write a mesh equation anytime there is a current source in the loop? – Korey Lombardi Jan 06 '21 at 03:08
  • @KoreyLombardi Yes. This is the basic idea of the "supermesh". – Muyun Benjamin Li Jan 06 '21 at 03:10
  • Awesome, thank you for the quick answer. – Korey Lombardi Jan 06 '21 at 03:11
  • 1
    @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. – jonk Jan 06 '21 at 03:50