# Confused about signs of currents in circuit

#### tjohnson

Joined Dec 23, 2014
611
I had a problem in my physics class today involving Kirchhoff's current law in a circuit. I made a quick drawing showing the circuit arrangement in the problem:

My question isn't about how to calculate the actual values of I1, I2, and I3, but how to figure out their signs. Because there is a battery on each side of the circuit, I expected that I1 and I3 would both be negative, since current flows from the batteries in the direction opposite than that shown by the arrows in the drawing. But according to my physics book, I1 is negative and I3 is positive (and I2 is also positive, which I expected). How can this be true?

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#### WBahn

Joined Mar 31, 2012
28,135
Imagine removing the resistor in the middle (the one that I2 is flowing through). If the batteries have different voltages, then a current will flow from one battery to the other, right? So current flowing out of one battery will be positive while current flowing out of the other will be negative. All this means is that one battery is acting as a source (the one with positive current flowing out of it) and the other will be acting as a load (it is being "charged"). The same in the circuit above. If the middle resistor is small enough, then both batteries will deliver current to it. But if it is large enough (and if the two batteries have different voltages), then one battery will deliver power to the circuit and some of the power will be dissipated in the resistors and the rest will be absorbed by the other battery. What that battery actually does with it will depend on the type of battery -- some will store the energy for later use and some will explode.

#### tjohnson

Joined Dec 23, 2014
611
Imagine removing the resistor in the middle (the one that I2 is flowing through). If the batteries have different voltages, then a current will flow from one battery to the other, right? So current flowing out of one battery will be positive while current flowing out of the other will be negative. All this means is that one battery is acting as a source (the one with positive current flowing out of it) and the other will be acting as a load (it is being "charged"). The same in the circuit above. If the middle resistor is small enough, then both batteries will deliver current to it. But if it is large enough (and if the two batteries have different voltages), then one battery will deliver power to the circuit and some of the power will be dissipated in the resistors and the rest will be absorbed by the other battery. What that battery actually does with it will depend on the type of battery -- some will store the energy for later use and some will explode.
Thanks, that makes sense. In the problem in my book, the battery on the left had a voltage almost twice as high as the one on the right, and the resistors all had small values. So since the battery on the left has a higher voltage, current will flow from left to right in the circuit, which makes I1 negative and I3 positive.

#### WBahn

Joined Mar 31, 2012
28,135
A good question to consider is what value the middle resistor needs to be in order for neither battery to be acting as a load?

The next question is to come up with a general relationship between the resistors to make this happen.

The next question is to come up with a general relationship between all of the components, including arbitrary battery voltages, to make this happen.

#### WBahn

Joined Mar 31, 2012
28,135
Thanks, that makes sense. In the problem in my book, the battery on the left had a voltage almost twice as high as the one on the right, and the resistors all had small values. So since the battery on the left has a higher voltage, current will flow from left to right in the circuit, which makes I1 negative and I3 positive.
I should have been a bit more careful -- if the middle resistor is small enough RELATIVE TO the other two resistors.

#### MikeML

Joined Oct 2, 2009
5,444
To illustrate Wbahn's point, look at this sim with made-up values. I placed the resistors so that the positive direction of current through them is shown by the arrows. Note the sign and magnitude of the three currents I(R1), I(R2), and I(R3) as R2 is varied logarithmically from 10Ω to 100KΩ. Note that I(R2) is always positive. I(R1) is always negative, but I(R3) changes sign when R2=1.34K

#### studiot

Joined Nov 9, 2007
4,998
Please note that it is not possible for all three of I1,I2 and I3 to be in the directions of the red arrows in your diagram, unless they are all zero.

Kirchoff's current law (the law of conservation of charge) says that

The sum of the currents entering a node = the sum of the current leaving

Since you have no currents entering the sum of those leaving must be zero so at least one must be negative (that is in the other direction).

For this reason I would advise always labelling your currents so that there is at least one current entering and one leaving every node.

#### kubeek

Joined Sep 20, 2005
5,776
For this reason I would advise always labelling your currents so that there is at least one current entering and one leaving every node.
Not entirely sure, but I think there could exist a circuit where this condition cannot be cannot be met for all nodes.

#### MikeML

Joined Oct 2, 2009
5,444
Not entirely sure, but I think there could exist a circuit where this condition cannot be cannot be met for all nodes.
Like the one I posted in the sim...

The whole point of this exercise is that frequently when first setting up the analysis, you do not know which directions the currents are actually flowing. It doesn't matter if you guess wrong, because when you solve it, the signs will show you the actual directions.

I use these interchangeably:
The sum of currents into a node is zero.
The sum of currents out of a node is zero.
The currents in to a node equals the current out of a node...

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#### studiot

Joined Nov 9, 2007
4,998
Mike, it's a nice sim, but like I said one of your currents is always negative in it, because you have shown all three leaving the junction.

What the OP seemed to be saying was (as any well motivated learner should do)

"He would like to get a feel for which way the currents actually flow"

You and I can probabaly do this most times by inspection, as a result of experience.
This is a valuable skill that does not come from blind application of a set of equations.
The problem with blind application is that a machine (calculator, computer) designed to output an answer will always do that whether that answer is rubbish or correct.
An experienced operator can tell if it is rubbish.

#### tjohnson

Joined Dec 23, 2014
611
Please note that it is not possible for all three of I1,I2 and I3 to be in the directions of the red arrows in your diagram, unless they are all zero.

Kirchoff's current law (the law of conservation of charge) says that

The sum of the currents entering a node = the sum of the current leaving

Since you have no currents entering the sum of those leaving must be zero so at least one must be negative (that is in the other direction).

For this reason I would advise always labelling your currents so that there is at least one current entering and one leaving every node.
Thanks, I understood that some of the currents would have to be negative because of the direction of the arrows. When I made my first post, I was incorrectly thinking that I2 = -I1 - I3, but now I see that -I1 = I2 + I3.

#### studiot

Joined Nov 9, 2007
4,998
When I made my first post, I was incorrectly thinking that I2 = -I1 - I3, but now I see that -I1 = I2 + I3.
Isn't that the same equation ?

#### tjohnson

Joined Dec 23, 2014
611
Isn't that the same equation ?
Yes, I wasn't thinking. It looks like what I originally did wrong was to use incorrect signs for the currents when I wrote equations for the circuit using Kirchoff's voltage law.

#### studiot

Joined Nov 9, 2007
4,998
That's exactly what I was warning against.
You will be better for the mistake though.

#### WBahn

Joined Mar 31, 2012
28,135
Note that the standardized method of applying KCL is to assume that the currents at each node are always flowing out of the node. We call this Nodal Analysis (or Node Voltage Analysis). By always making this assumption, you are far less likely to make mistakes in establishing your governing equations. It also permits you to very quickly write down those equations by inspection for most, if not all, nodes in the circuit.

#### studiot

Joined Nov 9, 2007
4,998
Note that the standardized method of applying KCL is to assume that the currents at each node are always flowing out of the node.
That's clever, what do you do at the other end of each branch then?
For instance which way does I2 flow out of the bottom node in the given example?
(I know it is inappropriate in this simple case but a couple of extra resistors in the negative legs it would not be.)

#### tjohnson

Joined Dec 23, 2014
611
Note that the standardized method of applying KCL is to assume that the currents at each node are always flowing out of the node.
Sorry for my ignorance, but I'm somewhat confused. I understand that whatever goes into the node must also go out.

In physics class, most of the circuit problems involving KCL were set up as shown by the arrows in the diagram, except that the arrow for I1 pointed right, not left. (Until I had a lesson this past week which made them more complex.) With all of the arrows pointing in the direction of the current flow, I1 = I2 + I3.

Am I understanding correctly that you're saying that the current directions shown by the arrows are the ones you would consider standard?

@studiot: I'm curious what your avatar is? It looks to me like a pig captured in a net.

#### WBahn

Joined Mar 31, 2012
28,135
That's clever, what do you do at the other end of each branch then?
For instance which way does I2 flow out of the bottom node in the given example?
Very simple.

First, you never label a specific current as I2. You express the currents in terms of the node voltages.

So if you have nodes A and B connected by a resistance Rab, then when you are analyzing Node A you express the current in that branch as

(Va - Vb)/Rab

And when you are analyzing Node B you express the current in that branch as

(Vb - Va)/Rab

Because of the branches at a given node will be of this form the left hand side of the equation, say for Node A connected through resistors to four other nodes, can be written down by inspection as:

Va(1/Rab + 1/Rac + 1/Rad + 1/Rae) - Vb(1/Rab) - Vc(1/Rac) - Vd(1/Rad) = ???

The ??? on the right hand side is zero if the other nodes are all "normal" nodes otherwise it is of the form Vx/Rax if Node X is at a known voltage X (say due to a voltage source) or Ix if it is connected to a current source sending current into the node (since the left hand side is the sum of currents leaving the node to other "normal" nodes).

#### WBahn

Joined Mar 31, 2012
28,135
Sorry for my ignorance, but I'm somewhat confused. I understand that whatever goes into the node must also go out.

In physics class, most of the circuit problems involving KCL were set up as shown by the arrows in the diagram, except that the arrow for I1 pointed right, not left. (Until I had a lesson this past week which made them more complex.) With all of the arrows pointing in the direction of the current flow, I1 = I2 + I3.

Am I understanding correctly that you're saying that the current directions shown by the arrows are the ones you would consider standard?

@studiot: I'm curious what your avatar is? It looks to me like a pig captured in a net.
You usually express each node as the sum of currents LEAVING the node -- that's so that the node voltage is positive in the resulting equation. You can do it the other way and it is just as correct.

Imagine a resistor and defining two currents through it, I1 and I2, being in opposite directions. Well, you know that I1 = -I2 and you can use either one in any of your equations as long as you preserve this relationships.

In actuality, what you know is that the actual current, I0, is equal to the algebraic sum of these two. If I0 is defined to be in the same direction as I1, then I0=I1-I2. This is the basis for the standardized method of applying KVL known as Mesh Current Analysis.

#### MikeML

Joined Oct 2, 2009
5,444
I still say:

You can use these interchangeably:
The sum of currents into a node is zero.
The sum of currents out of a node is zero.
The currents in to a node equals the current out of a node...