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If I have the following circuit and I apply the KCL to the red surface, the algebric sum of the currents in the surface should equals zero. So ix=-4A. Why isn't that right?

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  • \$\begingroup\$ Because that's not a loop: look at the bridge in the top right corner. \$\endgroup\$
    – pjc50
    Apr 17, 2015 at 9:46
  • \$\begingroup\$ @pjc50 I don't think he is referring to a loop but to a gauss surface. \$\endgroup\$ Apr 17, 2015 at 10:16

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You have four terminals coming out of your closed surface, the current sum across all terminals must be zero.

Assuming positive entering currents for the two top terminals you have \$+i_x\$ and \$-i_x\$, while for the bottom two terminals you have \$-4\$A and \$+4\$A. The sum is \$i_x-i_x+4A-4A=0\$ no matter the value of \$i_x\$.

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  • \$\begingroup\$ You are welcome @sleax. On a stack board you can upvote an answer if you think it is correct (or feel like upvoting it for any reason), and you can mark an answer as accepted if you are the asker. \$\endgroup\$ Apr 17, 2015 at 11:02

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