At a recent exam I had this very simple problem concerning Kirchhoffs Laws: problem statement

I am concerned with problem c

Now my understanding of this is:

I can combine both nodes, and since all currents in the node need to cancel out, \$i_c\$ should be 2A. Now my professor marked this problem as wrong and told me it should be 8A instead.

Am I doing something fundamentally wrong here?

Update: The professor reviewed the answer sheet because some other answers were wrong and did not mention the 8A again, and accepted the solution 🤷‍♂️

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    \$\begingroup\$ Unless I'm missing something I think you're correct with 2A. 0 = -(-1)-(Ic)+(4)-(3) -> Ic = 1+4-3 -> Ic = 2. Did your prof explain how he got 8A? \$\endgroup\$ – brhans Aug 21 '19 at 17:03
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    \$\begingroup\$ The only easy way I get 8A for Ic is by changing the sign or direction on the 3A. \$\endgroup\$ – brhans Aug 21 '19 at 17:08
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    \$\begingroup\$ I like to write the equation with outflowing currents on the left and inflowing currents on the right. These must be equal to each other. In (c), I'd write: \$-1\:\text{A}+3\:\text{A}+i_c=4\:\text{A}\$ and then solve that for \$i_c=2\$. I've no idea what your professor is doing. \$\endgroup\$ – jonk Aug 21 '19 at 17:08
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    \$\begingroup\$ I also must be blind because for me 2A is the correct answer. Or maybe it should be -3A? \$\endgroup\$ – G36 Aug 21 '19 at 17:09
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    \$\begingroup\$ how did you obtain the image? If this is what was on the questio. then professor is wrong but if the arrow 3A would flip he would be right. Maybe its sloppy drawing? \$\endgroup\$ – joojaa Aug 22 '19 at 5:49

Professor is wrong. The correct answer is 2A. I solved it algebraically just as every one else, 4=3+Ic+-1, Ic = 2. A DC simulation in Spice yields 2A as as well.

  • \$\begingroup\$ Which version of Spice did you use, and what did the model look like? \$\endgroup\$ – richard1941 Aug 30 '19 at 16:06

I agree, it should be 2A. I don't know how it could be 8A

enter image description here

  • \$\begingroup\$ Easily: reverse the 3A current direction and you're about to get 4A+3A-(-1A). \$\endgroup\$ – Crowley Aug 22 '19 at 12:54

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