I have a question regarding using KCL / KVL to analyse the circuit below to find all currents and voltages.

My question is : When performing KCL at the top centered node, what way should I assume that Iz is flowing? Should I assume Iz is flowing with the arrow (out of the node) that is given in the question, or assume it goes into the node based on the voltage drop of -2V over the series resistor?

What is confusing me is when I work out my Iz current, I get a positive value flowing into the resistor, but that doesn't seem correct with a voltage drop of -2V?

I am happy to post my full solutions if anybody would be happy to check them?

Thanks for any replies.

Thanks for the great answers everyone.

Yeah I think my mistake was assuming they were pure resistors for some reason but now I see how I could actually get negative voltage drops over the 'black boxes'. I think that was what I was struggling with.

Anyway here my solutions for all unknown voltages and currents for the same circuit below:

enter image description here

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    \$\begingroup\$ "I am happy to post my full solutions if anybody would be happy to check them?", just do it. Go full Nike on it. \$\endgroup\$ Oct 14, 2018 at 15:16
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    \$\begingroup\$ You are assuming that the elements in the diagram are resistors. I don't think that's true; I think they are just black-box, arbitrary elements. I know that resistors are drawn differently in different countries, so telling us where the book was published might help. \$\endgroup\$ Oct 14, 2018 at 16:09
  • \$\begingroup\$ Nothing in the problem statement says those elements are resistors. \$\endgroup\$
    – The Photon
    Oct 14, 2018 at 16:14
  • \$\begingroup\$ @The Photon It's a pretty good assumption that they are resistors. The only other possibility is impedances, and there's nothing to indicate that might be the case - no reference to frequency, no squiggly lines, no lower case letters... \$\endgroup\$
    – Chu
    Oct 14, 2018 at 16:34
  • \$\begingroup\$ @Chu, they could be batteries, ports of n-port devices, diodes, anything at all. The question is about KVL and KCL in the abstract, not about the properties of any particular device. \$\endgroup\$
    – The Photon
    Oct 14, 2018 at 16:36

2 Answers 2


If the diagram is to be believed....

By simple inspection Iz must be 2 amps and, it must be flowing in the direction shown by the arrow and, if it produces -2 volts across the component top-right then that component has a resistance of negative 1 ohm.

Also, by simple inspection Iy must also be 2 amps and therefore the middle vertical component must be positive 2 ohms to produce 4 volts across it.

This makes Ix equal to 4 amps.

And this ties in with the middle left current of 1.5 amps and the bottom left current of 2.5 amps joining to make Ix = 4 amps.

So, I think the diagram is to be believed.

When performing KCL at the top centered node, what way should I assume that Iz is flowing?

You can assume it flows in any direction and, after doing the analysis it either is positive or negative with negative implying it was flowing in the opposite direction you initially assumed.


It's specified fully enough that all you have to do is a bit of head-scratching and arithmetic. Ignore the negative value of that one resistor -- you can do that sort of thing just fine on paper; just remember that it's a bit hard to order negative-valued resistors from DigiKey.

  • \$\begingroup\$ @Chu the resistance must be negative to get 2 A in the direction of Iz (which you get from KCL), and -2 V as indicated. \$\endgroup\$
    – The Photon
    Oct 14, 2018 at 16:41

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