I just started circuit analysis today and learn't about Kirchhoff's Laws.

Its pretty straightforward, but in all the examples my book gives they already specify the direction of the current and the end of the branch with the higher voltage.

For example,

enter image description here

there arrows everywhere indicating the direction of the current at each node and +/- signs indicating the end of the branch with a higher voltage.

Obviously, when actually designing stuff the direction of the current and the side of the branch with the higher voltage wouldn't be indicated.

So what I'm asking is:

1) Can Kirchhoff's Law be used to figure out the direction of the current and side with higher voltage?

2) If not, do techniques exist to find the directions?

Please note that I'm not talking about convention. (i.e. currents flowing into the node are positive and currents flowing out are negative etc.) You can't just have a convention because you can't differentiate which direction the currents or voltages are going.

  • \$\begingroup\$ Use the loop current analysis. Convert all current sources to the equivalent voltage sources. Then recognize the independent loops and label the loop current. A loop is a closed path and a loop current is the current that flows in the closed path following the loop. If voltage source of a loop tends to force the loop current in the same direction as the loop current, then this current will get a (+) sign in the equation. If the voltage source tends to oppose the current this will have a (-) mark. \$\endgroup\$
    – GR Tech
    Commented Mar 16, 2014 at 6:48

4 Answers 4


When analyzing a circuit, you can put the arrows in either direction according to whim, a flipped coin, or Tarot cards.

After applying Kirchoff's laws to compute all the voltages and currents, you'll find some variables have negative values. Those correspond to arrows you drew backwards. Fix those, and then you know the directions of currents in all branches of the circuit.

It is perfectly normal for an experienced engineer to get a few initially backwards, when multiple different voltage sources are pushing in opposite directions. You can only guess, and let algebra tell you the net result.

  • \$\begingroup\$ You can't actually draw a reference direction or polarity 'backwards' and there's no reason to worry about guessing the 'correct' direction or polarity in the first place. There's no more better evidence for this than that the absolute polarity and direction is given by the algebra regardless of one's choice of reference polarity or direction. See my comment to william beaty's correct answer. \$\endgroup\$ Commented Mar 16, 2014 at 12:43

The +- signs indicate the connection of a voltmeter. I.e. if the voltage polarity of Va doesn't match the little +- sign, then that's a negative voltage.

Similiarly with the little arrows, they indicate the connection of an ammeter.

For getting answers in tests/homework, those little arrows are critical in determining whether to write +2.35mA or -2.35mA ! (And regarding "true direction" of current, just remember that your DVM measures conventional current, and it cannot tell whether you're trying to measure an electron beam in a CRT or the two contrary ion flows in a plating tank.)

  • \$\begingroup\$ I know what they are, I'm wondering if its possible to solve for things without knowing them. \$\endgroup\$
    – dfg
    Commented Mar 16, 2014 at 3:25
  • 1
    \$\begingroup\$ You just assume they are in one direction and then if you are wrong, your answer has a negative sign. \$\endgroup\$
    – Eric
    Commented Mar 16, 2014 at 3:30
  • \$\begingroup\$ > without knowing them ...No, because they aren't showing the answer to the posted problem, they aren't showing the current in the circuit. Instead they're showing the connection of an ammeter. If you don't know which way the textbook's invisible ammeter is connected, then your answer on the homework may be marked wrong for having the wrong polarity. (It's just a coincidence that the arrows in your posted circuit are all in the direction of the actual current, so all your homework answers will have positive milliamperes.) \$\endgroup\$
    – wbeaty
    Commented Mar 16, 2014 at 3:37
  • 1
    \$\begingroup\$ This is the correct answer. One must distinguish between reference polarity and reference direction versus absolute polarity and absolute direction. Consider placing two voltmeters across the 6k resistor, one with the red lead on the topmost terminal and one with the red lead on the bottommost terminal. The voltmeters read equal and opposite voltages. Which one is 'correct'? Both! Both voltmeters give you the same information - the magnitude of the voltage and the terminal that is more positive than the other, i.e., the absolute polarity. (cont.) \$\endgroup\$ Commented Mar 16, 2014 at 12:34
  • 1
    \$\begingroup\$ The bottom line is that one cannot choose a 'wrong' polarity or direction because all one is doing is creating a circuit variable. If I label the topmost terminal of the 6k resistor with a plus sign and call that \$v_{6k}\$ and someone else labels the bottommost terminal of the 6k resistor with a plus sign and calls that \$v^*_{6k}\$, neither is 'correct' or 'wrong'. We just have \$v_{6k} = -v^*_{6k}\$. The algebra takes care of the signs and doesn't rely on a 'correct' choice of reference polarities or directions. \$\endgroup\$ Commented Mar 16, 2014 at 12:39

The answer to (1) is yes. So I don't need to answer (2).

Let's randomize everything. Make the arrows go any way you please. Then apply the Kirchhoff voltage law and Kirchhoff current law.


simulate this circuit – Schematic created using CircuitLab

I've placed arrows indicating whatever I felt at the moment. Didn't matter. But whatever I decide, of course, I must use consistently in setting up the equations. So let's set up the loop equations. I'll always start where the ground is at on this diagram. The tail of an arrow is (+) and the head is (-). As I move around a loop, I use the sign that I first encounter for the sign of the term. So:

\$-12 + 9k\Omega\cdot I_1 - 6k\Omega\cdot I_2 = 0\$

\$-12 + 9k\Omega\cdot I_1 - 3k\Omega\cdot I_3 + 4k\Omega\cdot I_4 = 0\$

\$-12 + 9k\Omega\cdot I_1 - 3k\Omega\cdot I_3 - \left(9k\Omega + 3k\Omega\right)\cdot I_5 = 0\$

Then, you also have these from Kirchhoff's current law. [Again, you MUST observe those arrows, so an arrow going into a node is (+) and an arrow going out of a node is (-)]:

\$I_1 + I_2 + I_3 = 0\$

\$-I_3 - I_4 + I_5 = 0\$

If you arrange the above equations into a matrix equation you get:

\$\left[ \begin{array}{ccccc} 9000 & -6000 & 0 & 0 & 0 \\ 9000 & 0 & -3000 & 4000 & 0 \\ 9000 & 0 & -3000 & 0 & -12000 \\ 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & -1 & -1 & 1 \end{array} \right] % \left[ \begin{array}{c} I_1 \\ I_2 \\ I_3 \\ I_4 \\ I_5 \end{array} \right] = \left[ \begin{array}{c} 12 \\ 12 \\ 12 \\ 0 \\ 0 \end{array} \right]\$

Which solves out as:

\$\begin{array}{l} I_1=0.001 \\ I_2=-0.0005 \\ I_3=-0.0005 \\ I_4=0.000375 \\ I_5=-0.000125 \end{array}\$

The minus signs indicate that the chosen direction for analysis was wrong and should be changed. So change the direction arrows for \$I_2\$, \$I_3\$, and \$I_5\$.

So it really doesn't matter what you think at first. Just pick a direction and stick with it during analysis. When you get done, the answer will tell you whether or not you picked it right.

In other words, "Yes, Kirchhoff's laws will let you figure out the directions."


I know what they are, I'm wondering if its possible to solve for things without knowing them

William Beaty's answer is the correct one and I wish to expand on it some since this question frequently comes up.

When assigning a voltage or current variable for circuit analysis, one must choose a reference polarity for a voltage and a reference direction for a current, just as, when placing a voltmeter or ammeter in a circuit, one must choose an orientation for the leads. You cannot avoid choosing a reference polarity / direction.

Often, students learning the process are 'afraid' of choosing incorrectly but, rest assured, that's impossible. Indeed, two students can choose opposite reference polarities and directions and, assuming no mistakes, both will solve the circuit correctly. Yes, their answers will differ by a sign but both answers give the same information - the magnitude and the absolute polarities / directions.

So, what exactly does the reference polarity / direction imply?

For example, in the circuit diagram, we have the topmost terminal of the 6k resistor labelled positive. Does this mean that we 'believe' the topmost terminal is actually more positive? No!

It means that the voltage we calculate for this variable is the voltage we will measure if we place the red lead on the topmost terminal and the black lead on the bottommost terminal.

Obviously, if we reverse the reference polarity, the sign of the calculated answer changes just as when we reverse the leads of the voltmeter, the sign of the measured voltage changes.

But the choice of reference polarity cannot affect the absolute polarity of the voltage.

Thus, if we place the leads of the voltmeter across a circuit element and measure a positive voltage, we know that the terminal connected to the red lead is more positive than the other terminal.

Conversely, if we measure a negative voltage, we know that the terminal connected to the red lead is less positive than the other terminal, i.e., the terminal connected to the black lead is more positive.

To summarize, by picking a reference polarity / direction (which we must do), we're not 'guessing' at that absolute polarity / direction; the calculated or measured result will tell us that.


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