# KCL applied to a closed surface

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?

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

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 $i_x$.