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I was wondering how exactly passive convention works with respect to unlabeled resistors. In the problem attached, I realized you only get the answer if you label the resistor terminals one way, regardless of the direction you choose for the current. I was wondering, then, how you know which way to label these resistors. The initial method I used was wrong, which was to go clockwise and label the unlabelled resistor terminals positive to negative in that direction, but that turned out to be incorrect. Is there a general rule on how to label them, and if so what is it? I know it works the other way, (setting them negative to positive in the clockwise direction starting from the top left corner), but I don't understand why, or what dictates that. I am not concerned about the answer since I know the method to get it after trial and error, just the reasoning behind it, as this was a question of mine throughout other problems as well. I truly appreciate any help!

Problem:

enter image description here

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  • \$\begingroup\$ The general rule is: label them how you want. "[...] you only get the answer if you label the resistor terminals one way" that's just impossible. \$\endgroup\$ Commented Jun 6, 2015 at 8:54

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enter image description here

I annotated your circuit a bit. In red the loop current \$I\$, while in green currents and voltage references for each resistor. Please note that I followed no particular rule to do the latter.

First of all, when solving a circuit you need to choose a reference for the voltage. I did so on bottom left corner, just under the \$50V\$ generator.

Now since this circuit has only one loop you would normally need only one equation, plus one additional equation for each controlled source: that makes two for this problem.

Starting from the reference: $$ 50 - 2I + 2V_x -3I-10I-10-5I=0 $$ While the additional equation is: $$ V_x = -10I $$ I am sure you can solve it from here.

What about all the green currents then? Whell I've drawn them to prove that how you choose them does not matter at all. For each resistor you can write its voltage and current with respect to the green reference: $$ I_2=-I\qquad V_2=-2I\\ I_3=I\qquad V_3=-3I\\ I_{10}=-I\qquad V_{10}=-10I\\ I_5=-I\qquad V_5 = 5I $$

easy as pie. Please note that usually (as in please do so) for passive components you use a reference for which a positive IV product means dissipated power, i.e. currents enters at the plus terminal, while for generators and such you do the opposite. That's because that's what usually happens: generators source current from the positive terminal, passives drain currents in the positive terminal... But that's just a convention: look at your 10V generator. If we choose a reference with the same direction as I then it has the same references a passive would have... It is actually dissipating power as a matter of fact.

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  • \$\begingroup\$ the downvoter could explain his concerns regarding this answer instead of just hitting the -1 button \$\endgroup\$ Commented Jun 6, 2015 at 13:23
  • \$\begingroup\$ No, the equations are different. If you use the various I_2, I_3 etc. then you can change some signs but you need the equation to have only a single unknown, so substituting back I should lead to my equation. \$\endgroup\$ Commented Jun 7, 2015 at 12:07
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The reference directions of currents and the reference polarity of voltages across circuit elements is completely arbitrary and independent of each other (for sake of uniformity I'll talk about reference directions for both currents and voltages). You can choose them any way you like. The only thing to be aware is that some basic relationships between current and voltage for the same element will have sometimes a minus sign added in front of them.

For example, Ohm's law will have two different formulations according to which choice you make between these two:

  • The (arbitrarily and independently) chosen reference directions for I and V are associated, i.e. the current's direction "flows" from the terminal marked with "+" to the terminal marked with "-":

\[ V = R \cdot I \]

  • The (arbitrarily and independently) chosen reference directions for I and V are non-associated, i.e. the current's direction "flows" from the terminal marked with "-" to the terminal marked with "+":

\[ V = - \; R \cdot I \]

schematic

simulate this circuit – Schematic created using CircuitLab

EDIT (To address a concern expressed in a comment)

If you are worried that this freedom of choice is (apparently) in contrast with basic circuit laws think twice about it. In circuit theory you need references for (V and I) directions the same way in mechanics you need references when you describe the motion of some cogs and shafts in a machine. These directions are arbitrary because they are just references, not the true directions of the quantities involved. When you choose a voltage reference direction you are saying nothing about its true direction (which you will find out solving equations written using KVL, KVC and the V-I relationship for each element in the circuit).

Consider this analogy: a car is traveling from city A toward city B (the distance A-B being D=500km). The car has traveled for \$t=2\$ hours at a speed of \$v=100km/hour\$. Physics tells you that the distance the car has traveled is \$d=v t=200km\$. Now you want to know the position x of the car, therefore you choose a reference for positions with origin in A and positive values going from A to B. Now you can say that the position of the car is \$x = d = 200km\$ relative to the reference chosen. Had you chosen a different reference, i.e. origin in B and positive values toward B, you would have expressed the position of the car y as \$y=d-D=-300km\$. Note that the position is different just because of the different choice of the reference, but the physical situation is the same.

In circuit theory we choose the best reference to ease the solution of the problem at hand, but this fact is not always clear in this kind of homework exercises, which are meant to make you take acquaintance with the math involved, rather than develop the intuition needed to guess which choice of references is best to avoid messier calculations (e.g. to avoid two minus signs canceling each other).

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If you mean how to determine the + and - signs on the resistors, the process is to select a current direction and label each resistor with a + where that assumed current enters and - where it exits. The initial choice of current direction is arbitrary, but must be consistent with KCL (in this case, as there's only one loop, KCL is trivial). As one of the resistors already has + and - signs to define the dependent voltage source, it would make sense to assume an anticlockwise current direction for this particular exercise.

If you choose clockwise as the current direction, you will need to pencil in +/- on the top/bottom of the 10 Ohm, and Vx will change polarity etc ... too messy!!

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  • \$\begingroup\$ What do you mean by "the choice of current direction must be consistent with KCL"? \$\endgroup\$ Commented Jun 6, 2015 at 8:54
  • \$\begingroup\$ @Vladimir Cravero, in general, for more than one loop, current nominations at a junction must conform with KCL. \$\endgroup\$
    – Chu
    Commented Jun 6, 2015 at 9:14
  • \$\begingroup\$ I am sorry, but I still do not understand. The choice of current direction can either be arbitrary or consistent with something, not both. \$\endgroup\$ Commented Jun 6, 2015 at 9:17
  • \$\begingroup\$ Perhaps my wording is confusing: if I have a node of 3 branches, I can arbitrarily attribute I1 to any branch, I2 to any other branch, but I3 is constrained by KCL \$\endgroup\$
    – Chu
    Commented Jun 6, 2015 at 9:27
  • \$\begingroup\$ What's confusing is that you make no difference between the current reference and the current value... The reference is not constrained by anything, while the value is constrained by whatever reason, including but not limited to KCL. \$\endgroup\$ Commented Jun 6, 2015 at 9:32

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