I am currently studying the textbook "Lectures about the fundamentals of Electrical Engineering" by Adalbert Prechtl and after finishing the theory about Kirchhoff's law, I tried to solve some basic exercises on them, but my answers seem to deviate a lot from the answers the book gives. The question is rather simple, I have to apply Kirchhoff's first law on the nodes A to D. Circuit:

Concerning node A, after considering the direction of current based on the voltage sources' poles my initial answer was: $$ I_6 +I_4-I_5=0$$ My path of thought was: Current from 1st Voltage source, goes through R1 then through R4 and then R5. Current from 2nd Voltage source goes through R4 and then R5, and current from third Voltage source goes Through R6 then R5.

The answer that the book states is: $$ I_6 +I_4+I_5=0$$

I was really suprised from this answer and I am still thinking that I am missing something rather important about Kirchhoff's laws. Then I simulated the circuit in a simple circuit simulator where the flow of current was as my initial answer stated. Any idea on why this is the given answer would really help.

  • \$\begingroup\$ Can you add the step's you took that resulted in your initial answer ? \$\endgroup\$
    – Nick
    Commented Nov 23, 2018 at 10:34
  • \$\begingroup\$ Both answers are fine. Just set the direction of current at every branch. Σ Ι_in = Σ Ι_out. \$\endgroup\$
    – thece
    Commented Nov 23, 2018 at 10:41
  • \$\begingroup\$ @Nick edited my steps. To thece; How can the second answer be correct when, as I see it, it states that current only flows into the node and not out of it? \$\endgroup\$ Commented Nov 23, 2018 at 10:47
  • \$\begingroup\$ Kirchhoffs Laws are the result of philosophers thinking about new phenomena (currents) and then developing constraints (defining directions of current flowing) for the math modeling that make useful predictions. Either define ALL currents at a node as INCOMING, or OUTGOING; then manipulate the signs. Our brains need some consistency in these matters. \$\endgroup\$ Commented Nov 23, 2018 at 15:25

2 Answers 2


Concerning node A, after considering the direction of current based on the voltage sources' poles

Here's the fundamental misconception that you have. You don't need to consider the directions of current; you don't even need to take a slight guess. All you need to do is draw arrows on the circuit and stick with the direction they imply and, if the answer is negative then you have got that direction wrong. But, by the time you find out you will have solved the problem.

  • \$\begingroup\$ Thank you for your answer, I understand what you are saying. So apparently, either way, I'll end up with either the correct answer or the wrong one. But how can you explain the book's answer? \$\endgroup\$ Commented Nov 23, 2018 at 12:03
  • \$\begingroup\$ But how can you explain the book's answer? The book assumes directions of current different to how you have assumed them to be. And also, a negative answer isn't incorrect - it just tells you that the current flow direction is opposite to where you initially placed the arrow. \$\endgroup\$
    – Andy aka
    Commented Nov 23, 2018 at 12:38

Maybe what you are missing is that current has a direction. When you write your equation you assume a given flow. You assume that I6 and I4 are flowing into the node (so you sum them), and that I5 is flowing out of the node (so you subtract it).

Depending on how they actually flow, one or more of your currents will come out as negative. This is not a problem, it's just relative to the direction.

The book does not pick the same assumption, and assumes that they all flow into the node. They will most certainly get at least one negative result.

No matter which one you pick, it's very important to track these signs.

  • \$\begingroup\$ Sorry, but I am totally new to these concepts and I need some clarification. How can someone assume that all currents flow into one node? Isn't that like...counterintuitive? Also, what do you mean by "they will get at least one negative result" ?. Thank you for your answer \$\endgroup\$ Commented Nov 23, 2018 at 10:59
  • \$\begingroup\$ @ΚωνσταντίνοςΖαφείρης Visualize it with water in channels. Your Current is a water flow. Kirchhoffs law says, that all water flowing to point A must also flow away (equals 0). So you sum up all water streams who are connected to A. Now it is up to you, which direction you consider positive. If water that flows towards A is defined as positive, there must also be some channel with water flowing away from you (negative). \$\endgroup\$
    – jwsc
    Commented Nov 23, 2018 at 11:06
  • \$\begingroup\$ @jwsc So based on this way of thinking, I can just sum all the currents that connect to any node and, without considering the direction of flow, my 1st Kirchhoff's equation will be correct? Like the book's answer? \$\endgroup\$ Commented Nov 23, 2018 at 11:10
  • \$\begingroup\$ @ΚωνσταντίνοςΖαφείρης correct. \$\endgroup\$
    – jwsc
    Commented Nov 23, 2018 at 11:48

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