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Consider the circuit below from the book Electrical Engineering, Principles and Applications (p. 95). According to that book, we get from Kirchhoff's voltage law that $$ -v_1+0.5v_x+v_2=0 $$

However, I do not understand why $0.5v_x$ and $v_2$ have the same sign (since the current travels from both the reference node and $v_1$ towards $v_2$). Can someone explain what convention has been used to determine the signs in the equation?

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Can someone explain what convention has been used to determine the signs in the equation?

Sure. First note that the equation is for the loop clockwise (the signs would be reversed if counter-clockwise) that includes the voltage across $R_2$ ($v_1$) the voltage across the dependent voltage source, and the voltage across independent current source ($v_2$).

Now, before going 'round the loop, I would add a "plus" sign at the top of $R_2$ and at the top of current source. Why? This sign denotes the reference polarity, i.e., $v_1$ is positive when the top of $R_2$ is more positive than the bottom and similarly for the current source.

The rule for KVL is this: as you go around the loop, if you enter the "plus" marked end of the component, the voltage variable has a positive sign else it has a negative sign.

Starting at the circuit common node and going 'round the loop clockwise, we don't enter the "plus" end of $R_2$ so the voltage $v_1$ gets a negative sign. However, we do enter the "plus" end for both the voltage source and current source so those voltage variables get a plus sign.

That's really all there is to it.

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    $\begingroup$ That made it very clear, thanks a lot. $\endgroup$
    – Logi
    Sep 18, 2022 at 16:31
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Kirchoff's Voltage law deals with the sum of voltages across a closed loop, and the region between the two nodes 1 and 2 is not one. In between the two nodes (with potentials $v_1$ and $v_2$), this simple sign convention can be used (each of the paths are traversed from left ($a$) to right ($b$)):

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The fourth representation applies to your question; with $\Delta V = V_b - V_a = -\varepsilon$, in the circuit given: $$\Delta V = v_2-v_1 = -0.5v_x$$ which can be rearranged to obtain $-v_1 + 0.5v_x + v_2 = 0$

Hope this helps.

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  • $\begingroup$ Funny, in France we use: $ \triangle V=V_{a}-V_{b}$. In the end, it does not change anything, but I am curious, where are you from? $\endgroup$
    – Shaktyai
    Sep 20, 2022 at 9:39
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According to the diagram the potential $v_1$ is greater than the potential $v_2$ by 0.5 $v_x$. So $$v_1-v_2=0.5\ v_x\ \ \ \ \ \text{that is}\ \ \ \ \ -v_1+v_2+0.5v_x=0$$ We have not appealed to Kirchhoff's voltage law, as we have not needed to consider a complete loop, but just the potential difference between two points. The potentials $v_1$ and $v_2$ could be with respect to any point, that is we could take the zero of potential to be anywhere in the circuit: the argument would still work.

However, the 'Earth' symbol is no doubt supposed to tell us that the bottom line of the circuit is to be taken as at zero potential. In that case we can use Kirchhoff's voltage law if we want to ... If we go anticlockwise round the bottom left hand loop, starting at the bottom right hand corner, we gain in potential by $v_2$ as we go up through the current source, then by 0.5 $v_x$ as we go through the voltage source, but we then drop in potential by $v_1$ as we go back down to the bottom line via $R_3$. So $$+v_2+0.5v_x-v_1=0.$$ In fact, we don't even need to take the bottom line as at zero potential in order to apply KVL. If we give it potential $v_0$, then we can use exactly the same procedure as in the last paragraph and write $$(v_2-v_0) +0.5v_x - (v_1-v_0)=0$$ which simplifies to the same relationship. So we have a case where the KVL result is trivially true: it can be established without the use of the law, by the argument I gave originally.

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  • $\begingroup$ Yes, I see your point. The book has several examples like this, and they choose to use KVL for all of them. But that definitely seems like a detour considering what you are writing here. $\endgroup$
    – Logi
    Sep 18, 2022 at 16:52
  • $\begingroup$ @Logi I've extended my answer to include the KVL approach as an alternative. $\endgroup$
    – Philip Wood
    Sep 20, 2022 at 7:42
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Problem here is with sign convention in closed loop. We know that when current pass through resistor or any passive component, there is energy loss or voltage drop. Now this voltage drop, in case of resistance have resistivity, increases with increase in length of resistance. The end of resistor connected with energy source or battery is at higher potential or from direction current pass through the resistor, the tail of current passing is at high potential. Because current is already drops as it pass through resistive medium.

Now battery or source have already terminal defined or direction of current shown in case of current source. Current flows from positive terminal of voltage source or arrow head of current source. Now take any direction to sum voltages in closed loop and that sum equals to zero is KVL. To sum voltages in loop, encountering positive terminal is considered as negative, because current goes into negative terminal of source. Follow direction of current, but one can make own direction just keep in mind current flow through negative of element is positive.

Here we make $R_2$ as source, so current is flowing in clockwise direction. So, $v_1$ is voltage drop along $R_2$, active element has voltage 0.5 $v_x$ but added as negative because counterclock current enters to positive terminal of it. Similarly adding current source $v_2$. Thus,

$v_1-0.5\ v_x-v_2=0$, or $-v_1+0.5\ v_x+v_2=0$

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