Different Sign Conventions used with Kirchhoff's Law?

[bagasme]

TL;DR

Standard or Flipped One?

Hello,

As far as I know, the standard sign convention for Kirchhoff's law are:

• For resistor: the potential difference (voltage) is negative if the chosen loop direction is the same as chosen current direction through the resistor, positive otherwise.
• For battery: the voltage is positive if the chosen loop direction is from negative toward positive terminal, positive otherwise.

But in high school, I'd been taught different, flipped convention:

• For resistor: the voltage is positive if the chosen loop direction is the same as chosen current direction through the resistor, negative otherwise.
• For battery: the voltage is negative if the chosen loop direction is from negative toward positive terminal, positive otherwise.

To demonstrate effects of different sign conventions above, I attached sample problem
from this homework forum. The problem image:

By standard sign convention, the equations are:

$$\begin{align}
I_1 + I_2 &= I_3 \nonumber \\
6 - 3I_1 - 3I_3 &=0 \nonumber \\
6 - 6I_2 - 3I_3 &=0 \nonumber
\end{align}$$

Solving the equation, we have $I_1 = \frac 4 5 A, I_2 = \frac 2 5 A, I_3 = \frac 6 5 A$.

By flipped sign convention, the equations are:
$$\begin{align}
I_1 + I_2 &= I_3 \nonumber \\
-6 + 3I_1 - 3I_3 &= 0 \nonumber \\
-6 + 6I_2 + 3I_3 &= 0 \nonumber
\end{align}$$

Solving the equation, we have same solutions as above with standard convention.

I have questions:

1. In what cases both sign conventions produce same solutions, and in what cases different solutions are produced?
2. Which convention is right, the standard or the flipped one?

Regards, Bagas

 

[Dale]

Summary:: Standard or Flipped One?

Which convention is right, the standard or the flipped one?

They are conventions. There is no right or wrong involved. Either works, you just need to be consistent. The only time you will have a problem is if you switch conventions in the middle of a problem.

 

[bagasme]

They are conventions. There is no right or wrong involved. Either works, you just need to be consistent. The only time you will have a problem is if you switch conventions in the middle of a problem.

Hmm...

Are there cases when those conventions produce different results, assuming that there aren't any calculation errors?

This could happen because in my high school, when doing Kirchhoff's Law problems, most of my answers are invalidated only for using standard convention one instead of the flipped one.

 

[Dale]

Are there cases when those conventions produce different results, assuming that there aren't any calculation errors?

No

This could happen because in my high school, when doing Kirchhoff's Law problems, most of my answers are invalidated only for using standard convention one instead of the flipped one.

Then you were consistently making some other mistake also.

 

[Ibix]

This could happen because in my high school, when doing Kirchhoff's Law problems, most of my answers are invalidated only for using standard convention one instead of the flipped one.

If you are having trouble in school, the problem may be that your working is "wrong" because the sign convention is opposite, rather than because your answers are wrong. Just use their convention - getting used to odd sign conventions is an important part of the "school of hard knocks" bit of physics education.

However, I note that the equations you present in your OP appear to have a sign error. Are you sure you aren't just making mistakes?

 

[bagasme]

If you are having trouble in school, the problem may be that your working is "wrong" because the sign convention is opposite, rather than because your answers are wrong. Just use their convention - getting used to odd sign conventions is an important part of the "school of hard knocks" bit of physics education.

However, I note that the equations you present in your OP appear to have a sign error. Are you sure you aren't just making mistakes?

But @Dale said that either conventions works as long as consistent, and always double-check my solution and compare it to other one.

Personally, I'd rather follow conventions on my textbook (I use book by Giancoli) than on book supplied by the high school.

Sorry for the typo/sign error from second equation from "flipped" convention one. What I mean is $-6 + 3I_1 + 3I_3 = 0$.

 

[Ibix]

But @Dale said that either conventions works as long as consistent, and always double-check my solution and compare it to other one.

I agree with Dale. I'm just suggesting that you may be being marked down for not using your school's chosen convention, right answer or not. In that case switching conventions is the easy solution - and a useful skill, since this is very far from the only sign convention issue in physics.

So I'd suggest talking to whoever marks your work and asking for feedback around a question of this type. See if you can find out if you are being marked down because you aren't using their convention (some marking schemes are very restrictive, for good reasons and bad). Then you can decide if you want to flip convention, persuade them to allow different conventions, or if there's some other issue.

 

[vanhees71]

Kirchhoff's Law is nothing else than the integral form of Farday's Law. If all parts of the circuit are at rest, it reads
$$\int_{\partial A} \mathrm{d} \vec{r} \cdot \vec{E}=-\frac{1}{c} \mathrm{d}_t \int_A \mathrm{d}^2 \vec{f} \cdot \vec{B}=-\frac{1}{c} \mathrm{d}_t \Phi_{\vec{B},A},$$
where one takes $\partial A$ as a closed path usually going along the wires and circuit elements (resistors, capacitances, inductances, batteries,...) and $A$ is an area with the integration path $\partial A$ its boundary. The signs have to be carefully determined according to the right-hand rule, i.e., given a direction of the line integral along $\partial A$ the direction of the surface-normal vectors $\mathrm{d}^2 \vec{f}$ are determined by the right-hand rule (thumb of the right hand along $\partial A$, fingers along $\mathrm{d}^2 \vec{f}$. It's of course completely arbitrary in which direction you let $\partial A$ be oriented. Important is only the correct relative orientation of $\mathrm{d} \vec{r}$ along $\partial A$ and $\mathrm{d}^2 \vec{f}$ along $A$, which is determined by the right-hand rule as described.