Hi, Sorry, I thought I had solved this but Ive run into a snag :/ Uploaded with ImageShack.us

Through Kirchhoffs voltage laws Ive found

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

 

Doesn't matter, Ive just redone the voltage equations going clockwise. At this point it'll have to do

Thanks anyway

 

You can use either electron current or conventional currrent, as long as you do so consistently. The difference will simply be that the sign of the voltage drops in all the resistors will be the opposite compared to using the other choice. Now, if you have a current source in the circuit, you have to be sure that you know which type of current it is using.

 

In the end I inadvertently did it right the first time.

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/DesignOffic.../DC/DC_10.html

This website says as you're walking around the loop and you come across for example a resistor like this (- +) 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 guess as long as your presumed polarity for the way the resistors are connected in the circuit is correct, and as long as youre consistent in the way you work it out it doesn't matter which way you go round

Thanks again for helping again, hope you and everyone else has a nice christmas

 

It really sounds like you are focusing on memorizing a set of steps instead of understanding the fundamentals. My recommendation is to focus on the fundamentals so that you can understand why the steps are what they are. With that under your belt, you won't have to rely on correctly recalling some memorized steps.