# Homework Help: Why do resistors have polarity?

1. Mar 1, 2016

### goonking

1. The problem statement, all variables and given/known data
When I took intro physics, resistors didn't have polarity.

Now, in electrical circuits class, we start seeing resistors with polarity.

For example, this diagram in our textbook:

My professor insist that this problem had a misprint/typo: the polarity on the 2 ohm resistor should be reversed because "the current flow cannot go backwards when it reaches the 2 ohm resistor".

My questions are

1) why did the textbook bother showing the polarity of the resistors if it wasn't possible to have the arrangement of this diagram?

2) if my professor is wrong, and the book is correct, In this case, is one of the resistors resisting the current flow and the other resistor is "un-resisting" the current flow (ie: "speeding up" the flow)?

2. Relevant equations

3. The attempt at a solution

2. Mar 2, 2016

### cnh1995

Your professor is right. The polarity of resistor simply indicates the direction of current through it, which is from + to -.

3. Mar 2, 2016

### goonking

if the circuit had a bunch of resistors, would the polarities of all the resistors be the same?

4. Mar 2, 2016

### cnh1995

No resistor speeds up the current. Resistors always oppose the current.

5. Mar 2, 2016

### cnh1995

That depends on how they are connected with respect to the sources. Current enters at + terminal and leaves at - terminal.

6. Mar 2, 2016

### goonking

in a series, like the case i posted above. If I add 10 more resistors to that circuit, they would be all have the same polarity, correct?

7. Mar 2, 2016

### robphy

Do the +/- signs indicate the orientation of the measured voltages v1, v2 (akin to the choosing the arrowhead to indicate the positive x-axis)?

8. Mar 2, 2016

### vela

Staff Emeritus
It's not the polarity of the resistors. The signs indicate the how the potential drops are defined.

I wouldn't say the book is wrong or that it's a typo. The voltage v2 is simply going to be negative because of the way it's defined.

9. Mar 2, 2016

### cnh1995

For series circuit, the polarity would be + - + - + - and so on. For parallel resistors, polarity of all the resistors is same.

10. Mar 2, 2016

### robphy

Maybe this is a better way to describe what I think the notation means:

Clarifying earlier posts....
it's not about the polarities of the resistors
but the polarities of the voltmeters whose readouts are then assigned to variables v1, v2, etc

Last edited: Mar 2, 2016
11. Mar 2, 2016

### Staff: Mentor

With this apparently being an introductory textbook for basic electronics, I'd side with your professor in saying it probably is a typo, though it's no big deal if it isn't.

I'm wondering whether you have tried to calculate the magnitude of the current through the resistors here?

12. Mar 2, 2016

### robphy

Author, title, and edition?

Is there a textbook equation (Kirchhoff) using v1 and v2 that goes along with that diagram?
(Are there other diagrams with corresponding equations?)

13. Mar 2, 2016

### CWatters

Yes it could be that or a typo as NascentOxygen suggested.

When you come to analyse more complicated circuits it's not always possible to know at the start which way the current is flowing through all the resistors. The trick is to define what you mean by positive by marking up the circuit and solve the equations to work out the voltages. If some turn out to be negative that means the current is actually flowing the other way.

14. Mar 3, 2016

### milesyoung

Your professor apparently doesn't know basic circuit theory. It's like @vela wrote, the polarities of the resistors aren't those of the actual elements in the circuit. They're the reference polarities of the variables v1 and v2.

Let's say you didn't show a polarity for v1, but you just wrote v1 over the resistor without any reference polarity. If I told you v1 was some positive value, would that mean the left terminal of the resistor was at a higher potential than the right? Would it mean the opposite? The reference polarity makes it unambiguous.

The references might look goofy to you since you have some intuition about the current direction in the circuit, but they're still perfectly valid. As you start learning circuit theory, you'll see a bunch of these types of circuits, where everything seems turned around, precisely to teach you to think in more abstract terms, which will help you avoid mistakes when trying to solve complex circuits.

If you solve the circuit, v1 will be positive and v2 will be negative with the current going clockwise around the circuit, so both resistors will consume power, as passive devices must.

15. Mar 3, 2016

### Staff: Mentor

Once we've dealt with the likely intention of the polarity label on each resistor, would anyone like to guess what exactly is intended by the voltage source we see on the right?

16. Mar 3, 2016

### milesyoung

Can you explain what problem(s) you see with the OP's circuit?

The voltage source on the right has the polarity shown with a value of -8 V. What's the issue?

17. Mar 3, 2016

### nsaspook

To confuse the student.

18. Mar 4, 2016

### goonking

since some of you guys wanted to see the question

from Fundamentals of Electric Circuits (5th ed)

19. Mar 4, 2016

### Staff: Mentor

The textbook answers confirm that the diagram is quite correct, there are no mistakes in the labelling. So when you clip a voltmeter across either resistor by attaching the voltmeter's red probe to the + resistor label and the black probe to the resistor's - label, the voltmeter will give the reading (and associated polarity) as in the textbook answers.

So your professor's claim is disproved.

If you were to swap the + and - signs on either resistor, you will need to amend the textbook's answer by changing its sign.

20. Mar 4, 2016

### goonking

my professor swapped the signs of the 2 ohm resistor, and the answers he got were both positive : 16 V and 8 V

21. Mar 4, 2016

### Staff: Mentor

Sure, he can certainly alter any textbook question or diagram if he wishes.

However, you as a student don't have such liberty to go altering the questions in an exam situation---you must answer the question as presented.

22. Mar 4, 2016

### goonking

so basically, the polarities of the resistors are only important when using Kirchhoff's laws? (separating the circuit into loops)

23. Mar 4, 2016

### CWatters

Resistors don't have polarities. The + and - signs refer to the voltage drop across them in the circuit. If you turn a resistor around they don't move.

24. Mar 4, 2016

### Staff: Mentor

The polarity of the assumed potential across any element is important in circuit analysis.

The polarity is also important when you may be about to connect a meter or to connect another element. Some meters must always be connected so that the red probe is always more positive than the black probe----there could be damage to the meter movement if this is not observed.

As others have pointed out, ordinary resistors are not manufactured to need one end more positive than the other (they would be useless for AC if this were true!). The polarity marked near resistor ends on your circuit indicates the assumed polarity of the potential across that resistor. So where a voltage is designated v2 the + sign indicates the end where that voltage is measured referenced to the - end. (An alternative way to designate assumed polarity is with an arrow across the resistor, the arrowhead serving the same purpose as the + sign.)

25. Mar 4, 2016

### milesyoung

The reference polarities are always arbitrary. As an example, here's your circuit with some references assigned that I just picked out to personal taste (you can pick them completely at random if you so choose):

KVL going clockwise with voltage drops counted positive:
$$-v_a + v_b + v_c + v_d = 0, \quad (1)$$
The element law for the resistor is (passive sign convention):
$$v = Ri$$
The solution to this equation is only correct if you have the reference current going into the positive end of the polarity of the reference voltage. You must never, ever apply it otherwise.

You have then:
$$v_b = R_1 i \quad (2)\\ v_d = R_2 i \quad (3)$$
The solution to the system of $(1),(2),(3)$ always gives correct results. You just have to figure out what $v_a,v_b,v_c,v_d$ are for your particular circuit. In your case, they're:
$$v_a = 32\,\mathrm{V}, v_b = v_1, v_c = 8\,\mathrm{V}, v_d = -v_2$$