# Kirchhoff Laws: Conventional Or Electron Flow

1. Dec 20, 2012

### wellcoughed

Hi,

Ive trying to work through Kirchhoffs voltage and current laws.

Ive so far calculated:

Current in Loop one (I1): 3 Amps going anticlockwise
Current in Loop Two (I2): 4 amps going anticlockwise
Current in Loop Three (I3): 8 amps going anticlockwise

R1: 5 amps going up = (Current in loop three - current in loop one)
-5amps going down = (current in loop one - current in loop three)

R2: 3 amps going up
-3 amps going down

R3: 4 amps going up = (Current in loop three - current in loop two)
-4 amps going down = (current in loop two - current in loop three)

R4: 4 amps going up
-4 amps going down

R5 -1 amps left to right = (current in loop one - current in loop two)
1 amps right to left = (current in loop two - current in loop one)

All these currents are going anticlockwise but the problem they've given me has 8 amps written on it going clockwise.

So when I apply Kirchhoffs Current Laws (The sum of currents going into a node plus the currents going out of a node = 0) starting with the 8amps clockwise they've given me I find the opposite to what I've found when I solved the voltage laws.

Specifically using the current laws starting with 8amps clockwise I now have:

R1: 5 amps going down the resistor
R2: 3 Amps going down the Resistor
R3: 4 Amps going down the resistor
R4: 4 Amps going down the Resistor
R5: 1 Amps going left to right

Am I right in saying that By using Kirchoffs voltage laws Ive found the electron flow. The direction the electrons flow from the negative of the battery to the positive.

But if I use kirchoffs current laws and start with the "8 Amps clockwise" the lecturer has given me in the problem Im now actually finding the direction of the conventional flow?

Which is why it's directly opposite to the electron flow I've found when I used Kirchhoffs voltage laws?

many thanks if you could clear this up :)

Last edited: Dec 20, 2012
2. Dec 20, 2012

### phinds

"current", which is the term universally used when discussing electrical circuits, flows (by definition) from high voltage to low voltage, and in your circuit, the 8 amps is properly labeled (as positive).

3. Dec 20, 2012

### wellcoughed

I dont understand what you're trying to say or how it helps my questions sorry :/

4. Dec 20, 2012

### SammyS

Staff Emeritus

Usually circuit analysis is done using "conventional current", current flowing out of the positive terminal of the voltage source, returning to the negative terminal -- that is as if the charge carriers were positive, even though the actual charge carriers are electrons, which are negatively charged and flow in a direction opposite to the direction of flow of conventional current.

The results obtained using conventional current are equivalent to results that would be obtained with electron current.

5. Dec 20, 2012

### wellcoughed

ach I still dont understand :/

I've just resorted to redoing the voltage equations going clockwise instead of anticlockwise.

And that gives me the right currents going down.

Thanks anyway :)

Last edited: Dec 20, 2012
6. Dec 20, 2012

### SammyS

Staff Emeritus
I was in the process of giving you a more complete answer when I saw that the image you posted has been taken down.

What's up with that?

7. Dec 20, 2012

### SammyS

Staff Emeritus
I take it that the above results are what you get for electron current.

When I first looked at your figure, i was confused because although you said the current for Loop 3 is 8 amps anticlockwise, the figure had current in Loop 3 going clockwise.

You certainly can use electron current rather than conventional current to do circuit analysis. You just have to be consistent. Of course, when you report your results back to your teacher, you had better convert all the electron current results back to conventional current by reversing directions.

8. Dec 20, 2012

### wellcoughed

Thanks for taking the time to help Sammy :)

I took the pic down because of that inaccuracy you mentioned. All the loops that ive derived my figures from in my original post go anticlockwise.

However I think Ive inadvertently done it right.

When I originally did it I followed the advice from this website on how to do it: (its about a quarter of the way down)

http://www.eng.cam.ac.uk/DesignOffice/mdp/electric_web/DC/DC_10.html [Broken]

This website says as you're walking around the loop and you come across for example a resistor like this (- +) then you add (10ohms x Current.) and when you come across something thats (+ -) you subtract. So you basically bunny hop the first symbol and act on the second.

Using this method and going anticlockwise for loop 3 I had:
-54v + 2ohms(I3-I1)ohms + 11ohms(I3-I2) = 0

But ive looked on youtube and this teacher went clockwise but when he came across an emf (- +) he subtracted (10ohms x Current.) and when he come across a resistor like this `(+ -) he added (10ohms x Current.)

Using this method on the same circuit but going clockwise now I had:
-54v + 2ohms(I3-I1)ohms + 11ohms(I3-I2) = 0

Bunny hopping the first symbol and going anticlockwise was the same as going clockwise but using the first symbol you come.

I havent gone through it all because its getting late and i need shleep but I'm hoping the Kirchhoff voltage calcs ive done going anticlockwise can be used for the clockwise motion as well and that I'll be alright.

Thanks again for helping, hope you have a nice christmas :)

Last edited by a moderator: May 6, 2017
9. Dec 20, 2012

### SammyS

Staff Emeritus
That in an interesting link. The authors definitely use electron current. Except for the fact of using electron current, which is opposite the direction of conventional current, all of their results are the same as you get with conventional current. Even their equations from KVL are the same as if you use conventional current. They label the (+ -) on batteries the same as for conventional current. The (+ -) on resistors is opposite, but so is the current's direction so it turns out the same.

Going around your Loop 3 in a clockwise direction should give
54V - 2Ω(I3-I1) - 11Ω(I3-I2) = 0​
for conventional current with all the currents flowing clockwise. .

Last edited by a moderator: May 6, 2017