Electronics - Kirchhoffs' Law - Anyone Help?

[Walleee]

I may be no help at all, but isn't it just that what goes in must come out?

Have you a picture of the circiut your doing? would make things mucho easier.

[David]

Got the circuit diagram? Looks like it might need some network analysis...although it depends how complex it is.

[Bionic Balls]

yeah man....its gonna need a picture!

kirchoff law just means you sum all the individual bits basically...

[basher]

You havnt actually told us what you need to do. Do you have to draw a circuit with the things you have said? work out a value using others?

[PaRtZ]

OK I've just done some uni coursework on this so here goes:

Kirchoff has 2 main rules for working out circuits: The junction rule and the loop rule

Junction rule

At any junction the some of the current I is equal to 0. This is because there is electricity flowing into the junction (considered positive) and electricity leaving the junction (considered negative). Obviously as electricity doesn't just disappear, the positives cancel out the negatives

Loop rule

In a loop abcda (Of a closed loop), the sum of the voltage V is equal to 0. This is because of the fact the closed loop doesn't have any electricity input other than EMF sources so resistance will cancel out the voltage given.

Now the whole point of these two laws is to find out voltage in a parallel circuit maybe?

First step: roughly work out the direction of current. Draw little arrows on the diagram, dont worry if they're not write, it will just give you a negative answer if your wrong

Second step: Reduce the number of unknowns. If you draw the diagram in loops, then label each loop with a different I. (like I₁, I ₂) just to make things easier. Maybe they'll all have the same current, but you wont know till you work it out.

If it helps, start from the largest power source and work around the circuit labelling each line with the right I label. Then when you get to a split (remembering junction rule the current for both paths WON'T be the same, because the sum of the paths adds up the current of the wire you came from (might need to re-read that a few times)

Third step: Now using the junction rule, you know that the positive current is equal to the negative current so you can re-write one in the form of the other:

If you have current A being split to current B and current C, then A = B - C. Change current A labels to: Current B - Current C (or I₂ - I₃ etc)

Fourth step: Now you have to work out the current in one of the loops. In parallel arrangement, you know that PD is the same across the circuit so now we introduce the loop rule

Fifth step: If you have two unknowns, I₂ and I₃ then you need 2 equations to input into a simultaneous equation. This is done by looking at 2 different loops. Label the top left, bottom left, top right and bottom right corners of the circuit a,b,c,d and the crossing in the middle, top = e, bottom = f. From top left it should go a,e,b,c,f,d. It doesn't matter, but thats what I'll use for this

Sixth step: Start at a and we'll first look at loop a,b,c,d,a So from a, follow the circuit around until encounter something. Don't forget we're looking at VOLTAGE here so any resistance needs an I_x value accompanying it (5Ω resistor in the I₁ loop = 5I₁) To save my hands, Ill assume you know about direction of travel and if a voltage is going to be negative or postive, if not, quote me and highlight this bit and ask.

Seventh step: So hopefully you have say an xI₂ ± yI₃ = zV. This is from the loop abcda but we cant sove I₂ and I₃ from this, so we need to do another loop to get another equation, so do the same step for loop a,e,f,d,a

Eighth step: 2 equations: Use youre favourite method to solve using simultaneous equations

Ninth step: You should now have values for I₂, I₃ and substituing back into the junction rule at the start, a value for I₁ too. Now you know that V = IR so you have a value of I for each part of the circuit, use another loop method to get from a to c. IMPORTANT (my downfall everytime!) Make sure you MULTIPLY THE RESISTANCE BY THE CORRECT I_x VALUE AND NOT JUST USE THE RESISTANCE VALUE. REMEMBER V = IR NOT JUST V = R!

Tenth step: rearrange your answer and you should end up with a voltage. You can check if this is right by going round the circuit in another direction/route and so long as you go from one side to the other, you should end up with the same value.

Ill be honest, its been a while since I managed to get that to work correctly, but I pretty much just copied my lecturer's notes onto here. If anyone see's a mistake Ill be glad to know what it is so I know what Im doing wrong

Sorry its long lol