# Circuit Analysis Polarity Determination

#### SamR

Joined Mar 19, 2019
5,040
Here is the circuit in question.

So, any component adjacent to a battery source will have the polarity of the battery.

What determines the polarity of R2? The negative source is greater, but does the voltage drops across R1 and R3 have any affect on the polarity of R2? It is neccesary for calculating the simultaneous KCL equations to determine current. I know that it is positive toward ground but how is that determined?

#### crutschow

Joined Mar 14, 2008
34,420
What determines the polarity of R2
Since the battery negative terminal is connected to ground through a resistor, there can be no negative voltages anywhere in that array with respect to ground.

#### SamR

Joined Mar 19, 2019
5,040
That would give me

Which is I1 +I3 = I2 and the answer key gives I2 + I3 = I1 and to solve the matrix for the simutaneous equations I need the correct polarites.

#### Irving

Joined Jan 30, 2016
3,884
Errrr, I beg to differ. The 9v battery is + to ground so the non-ground side of R2 is -ve as the 6v battery and its larger series resistors is unlikely to overcome that. Its a moot point anyway if you label it incorrectly you'll have a reverse current through it.

#### WBahn

Joined Mar 31, 2012
30,052
Since the battery negative terminal is connected to ground through a resistor, there can be no negative voltages anywhere in that array with respect to ground.
Huh???

Then what is the voltage at the middle node in the top branch?

Sure looks like it is, by inspection, -9 V with respect to ground.

#### WBahn

Joined Mar 31, 2012
30,052
Here is the circuit in question.
View attachment 314786
So, any component adjacent to a battery source will have the polarity of the battery.
BIG misconception here! The notion of "the polarity of the battery" isn't even well-defined. What does it mean?

Imagine two batteries with there negative terminals tied together (and let's call that 0 V). One battery is 9 V and the other is 9 V. Now put a resistor between the two positive terminals.

How can both sides of the resistor have "the polarity of the battery"?

In this case, the 6 V battery will have a negative current flowing through it, meaning that it is acting like a load (i.e., being charged). Depending on the battery, this may or may not be a bad thing.

View attachment 314788
What determines the polarity of R2? The negative source is greater, but does the voltage drops across R1 and R3 have any affect on the polarity of R2? It is neccesary for calculating the simultaneous KCL equations to determine current. I know that it is positive toward ground but how is that determined?
The best (i.e., surest) way to determine polarity is to analyze the circuit. Pick a polarity for the voltage across R2 and/or the current through R2 and assign a variable accordingly. Doesn't matter which polarity you pick -- use a coin flip if you want. Then solve for that variable by analyzing the circuit. If you guessed right, the value will come out positive. If you guess wrong, it will come out negative. In either case, you now know both the magnitude and the polarity of that quantity for that component.

Sometimes you can reason out what the polarity must be, but other times you can't (or at least not easily).

In this case, you can determine it fairly easily.

Remove R2 from the circuit. You now have a simple series circuit. Going around that circuit, you see that you have a total of 15 V across 11.3 kΩ, with the current flowing clockwise. That means that the voltage drop across the 1.2 kΩ resistor is going to be about 1.5 V with the left side being more positive than the right side. The right side is a -9 V, so the left side is going to be at about -7.5 V.

That means that, when you put R2 back into the circuit, current will flow from right-to-left through it.

You can choose any analysis technique you want -- nodal, mesh, superposition, Thevenin, Norton, whatever. They all better give the same result.

For this circuit, nodal analysis is the most obvious choice. Let's call Vo the node on the left side of R2. Then writing the node equation for Vo, we have:

(Vo / R2) + ((Vo - -9V) / R1) + ((Vo - 6V) / (R3 + R4)) = 0

Now just solve for Vo

#### SamR

Joined Mar 19, 2019
5,040
For this circuit, nodal analysis is the most obvious choice.
It was given as a Branch Circuit Analysis problem.

Remove R2 from the circuit. You now have a simple series circuit. Going around that circuit, you see that you have a total of 15 V across 11.3 kΩ,
K, got it, Thanks! And that gives me I2 + I3 = I1!

#### crutschow

Joined Mar 14, 2008
34,420
(Head slap). Yes I missed the other battery in the circuit.

#### WBahn

Joined Mar 31, 2012
30,052
It was given as a Branch Circuit Analysis problem.

K, got it, Thanks! And that gives me I2 + I3 = I1!
Depends on how you define I2, I3, and I1. This is a meaningless equation unless you describe where these currents are, including their directions.

As you defined them (Post #3), it is I1 + I3 = I2 (just as you said).

Using those definitions,

I3 = (Vo / R2)
I1 = ((Vo - -9V) / R1)
-I2 = ((Vo - 6V) / (R3 + R4))

The answer key is using different directions on it's definitions of these three currents. Unless it indicates what those definitions are, then it becomes very hard to verify with it is correct or not.

#### SamR

Joined Mar 19, 2019
5,040
I'm using a matrix equation to resolve 3 simultaneous equations. First 2 equations are the KVL expressions of I1R +I2R +I3R = E for the two loops and the third equation is the KCL expression of I1 + I2 + I3 = 0 to resolve the I1, I2, and I3 values. I do cheat a bit by once I can state the simultaneous equations in the matrix, I use MS Mathematics to crunch and resolve the matrix. For this problem the matrix is:
(I1) (I2) (I3) (E)
-1.2k 0 -8.2k -9
0 -10.2k 8.2k -6
-1 1 1 0

Which resolves as I1 = 2.03mA, I2 = 1.23mA, I3 = 0.8mA and agrees with the I2 + I3 = I1 so -I1 + I2 + I3 = 0

EDIT: Arghh, posting removed all my formatting but you get the idea I hope...

#### AnalogKid

Joined Aug 1, 2013
11,038
Maybe a bit late, but - To make it a bit more clear what is going on, do three things:

1. R3, R4, and E2 are a simple series leg. Swap the positions of R4 and E2 so both batteries have a direct connection to GND.

2. Combine R3 and R4 into a single resistor.

3. Redraw the circuit with the two batteries stacked vertically, 6 V on top.

From the 6 V + terminal there is 1 resistor to the right.
From the GND point between the batteries there is 1 resistor to the right.
From the 9 V - terminal there is 1 resistor to the right.

The right ends of all three resistors are connected together. The schematic should look like a block figure-8.

ak

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#### SamR

Joined Mar 19, 2019
5,040
Simplification does make sense! Hadn't considered that approach but it is a valid one. Need to try and remember that approach!

#### MrAl

Joined Jun 17, 2014
11,472
Here is the circuit in question.
View attachment 314786
So, any component adjacent to a battery source will have the polarity of the battery.
View attachment 314788
What determines the polarity of R2? The negative source is greater, but does the voltage drops across R1 and R3 have any affect on the polarity of R2? It is neccesary for calculating the simultaneous KCL equations to determine current. I know that it is positive toward ground but how is that determined?
To add to the other good replies here...

I'd like to point out the node which is the junction of R1, R2, and R3.
Make E1 one million times greater than it is in the schematic: E1=9000000 volts.
What is the likely polarity of that one node?
Now instead, make E2 a million times greater than it is in the schematic: E2=6000000 volts.
What is the likely polarity of that one node now?

This illustrates the point that the only way to know what voltage polarity happens to be across a circuit element is to analyze the circuit with a regular technique such as Nodal Analysis. The end result of that analysis will give you the voltage of that one node and then it will be simple to figure out the polarity.

With this circuit though it should be fairly easy to figure out using superposition with the two sources. You end up with two voltage solutions that are comprised mostly of parallel and series resistors forming a voltage divider, and a single voltage source, and the sum is the node voltage.

#### AnalogKid

Joined Aug 1, 2013
11,038
What determines the polarity of R2?
The 10:1 difference in resistor values.

Round number quickie analysis - the 9 V battery is in series with 1 K, while the 6 V battery is in series with 10 K. That's an over 10x difference in battery currents. The 9 V battery polarity will dominate.

k.i.s.s.

ak

#### SamR

Joined Mar 19, 2019
5,040
it should be fairly easy to figure out using superposition with the two sources.
At this point, superposition had not been introduced. I'm just starting the section on supermesh now... AK made a very valid point as to simplifying by combining resistances in the loop as they end up being added together in the KVL equation anyhow. Simplifying by combining them in the schematic makes it much clearer for me to visualize in my thick skull what is going on in the circuit. Also, as WBahn pointed out, removing the R2 resistor from the middle (ala supermesh) also makes it clearer to mentally visualize.

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#### MrAl

Joined Jun 17, 2014
11,472
At this point, superposition had not been introduced. I'm just starting the section on supermesh now... AK made a very valid point as to simplifying by combining resistances in the loop as they end up being added together in the KVL equation anyhow. Simplifying by combining them in the schematic makes it much clearer for me to visualize in my thick skull what is going on in the circuit. Also, as WBahn pointed out, removing the R2 resistor from the middle (ala supermesh) also makes it clearer to mentally visualize.
Ok no problem.
I just wanted to point this out because with this circuit you end up with two voltage dividers which gives us two voltages v1 and v2, and when you add those two together you get the voltage of that one node vn=v1+v2. That seems kind of easy to do.
I guess you are going through some sort of regular coursework so probably best to stick with that.