Kirkchhoff's law

1. Mar 2, 2016

nothing909

1. The problem statement, all variables and given/known data
Using kirkchhoff's, how do I work out the branch currents using three or more loops?
2. Relevant equations

3. The attempt at a solution
Usually what I do is just assume just make two loops like this:

I do clockwise = anti-clockwise for both loops and then I do simultaneous equations.

For three loops, after I've got all the equations, do I still use simultaneous equations? If so, how do I do simultaneous equations? What two do I use to calculate each branch?

2. Mar 2, 2016

Staff: Mentor

Yes you still use simultaneous equations. You will have as many equations as there are loop currents (equal to the numebr of independent loops).

There are several methods to solve simultaneous equations. That's a matter of mathematics and largely a personal choice that depends upon your educational experience in equation solving. Some people stick to manual equation manipulation and substitutions, others use matrix methods such as Cramer's rule. When the number of loops exceeds three or four, manual methods become tedious and error prone, and most will resort to computer based solvers or matrix methods.

3. Mar 2, 2016

nothing909

If I'm doing a circuit like this:

If I have the equation for loop 1, 2 and 3, how is it I use them when I do simultaneous equations?

To find I1, do I just do (loop1-loop2).

To find I1, can I just them sub I1 into loop2's equation?

Then to find I3, do I do loop1-loop2-loop3?

4. Mar 2, 2016

cnh1995

You'll have three equations with variables as i1, i2 and i3. To find each, you'll need to solve the equations simultaneously using a convenient method.

5. Mar 2, 2016

nothing909

I don't understand how to use it "in a convenient way."

Is this the correct way to do it:

To find I1, do I just do (loop1-loop2).

To find I1, can I just them sub I1 into loop2's equation?

Then to find I3, do I do loop1-loop2-loop3?

6. Mar 2, 2016

Staff: Mentor

There are several methods to solve simultaneous equations. They are called "simultaneous" because all three equations are interdependent. That is, in general the independent variables (the three currents in this case) all influence each other in some way. So you don't get to solve for one of them without considering the effects of all of them together.

Your proposed approach may be a viable method, but it depends upon what you really have in mind when you say something like "To find I1, do I just do (loop1-loop2)". What are "loop1" and "loop2" here? I think it would be beneficial at this point if you presented an example in some detail, showing the actual equations you obtain from a circuit. It doesn't matter if you reach a complete solution immediately, we can evaluate your approach and make suggestions.

7. Mar 2, 2016

nothing909

Yea, I was only using that circuit as an example I found on the internet. The one I'm doing is different. I was just trying to find a quick answer.

If you look at the attachment, that is the circuit I'm actually using. It doesn't have two voltage sources, it has only one, but it does have three loops, which I highlighted as 1, 2 and 3.

For this, would I do the same thing?

i1 = loop1-loop2
i2= i1 substituted into the loop2 equation
i3 = loop1-loop2-loop3

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8. Mar 2, 2016

Staff: Mentor

Again, you should show the actual equations that you are referring to as "loop1" through "loop3". I have to presume that your "loop2" equation will contain all three variables i1, i2, and i3 since loop2 boarders both other loops. So you won't find i1 directly by taking the difference between two equations.

So, what are your three loop equations for this circuit?

9. Mar 2, 2016

nothing909

That diagram that is shown above, is that the correct direction that I put the loops?

Usually when I do kirkchhoff's, I have two sources, one at the left and one and the right, and two loops. Usually I have two loops going different directions.

I'm trying to do clockwise = anti-clockwise, but when I'm doing the second loop, I don't know what to do with R2, is that clockwise or anti-clockwise. In one loop it's clockwise, in the other it's anti-clockwise.

Am I talking rubbish? I'm confused.

10. Mar 2, 2016

Staff: Mentor

Your diagram is fine. You are free to choose the current directions any way you wish: it won't matter, the math will take care of finding correct values for the currents. If the chosen direction happened to be "wrong", the result will be a negative value for the current, indicating that the current is actually flowing in the opposite direction to your guess.

Usually one simply chooses all the currents to have the same direction, clockwise as you have done it being the most common choice.

R2 has two loop currents (also called "mesh" currents) flowing through it. Your equations must take into account both of them. Once you've solved for the loop currents, then the actual current flowing through R2 will be given by their superposition: $I_{R2} = i_1 - i_2$ given the choice of current directions that you've drawn.

11. Mar 2, 2016

nothing909

The equation I got for loop 1 is:

(1470+j0)i1 + (470+j0)i2 = 10+0j

The equation I got for loop 2 is:

(470+j0)i1 + (1470-j677.26) + (0-j677.26)i3 = 0

Is that correct?

12. Mar 2, 2016

Staff: Mentor

No, the signs for your i2 and i3 terms are both incorrect. You need to respect the chosen loop current directions when determining whether they contribute a potential drop or a potential rise when doing your "KVL walk" around a given loop.

For example, when you follow the current i1 around loop 1 and come to R2 you are proceeding in the same direction as the current so the resulting potential change is a drop. But note that loop current i2 is also flowing through that resistor but in the opposite direction. Thus your "KVL walk" is against that current, so its contribution to the potential change there is a rise. So since you're summing potential drops you would write for the R2 term: $470 ( i_1 - i_2)$. That gives a negative sign to the i2 term in your first equation.

13. Mar 2, 2016

nothing909

I corrected my answers, is this now correct?:

Loop 1:

(1470+j0)i1 -(470+j0)i2 = 10+j0

Loop 2:

(470+j0)i1 + (530-j677.26)i2 - (0-j677.26)i3 = 0

14. Mar 2, 2016

Staff: Mentor

Your Loop 1 equation looks fine.

There are problems with your Loop 2 equation. How did the first term end up being positive? If you do a KVL walk around loop 2 following the clockwise direction of i2 then current i1 is flowing against the direction of that walk as you pass through R2. That means i1 provides a potential rise for that walk. I also don't see how you found a real component of 530 for i2; the total real impedance in that loop is 1000 + 470, or 1470 Ohms.

It may help you to mark up your circuit diagram with the expected directions of potential changes for the loop currents before you start writing your equations. Then you can compare the terms in your equations with what you expect to find:

15. Mar 2, 2016

nothing909

Is this now the correct value for loop 2:

-(470+j0)i1 + (1470-j677.26)i2 - (0-j677.26)i3 = 0

16. Mar 2, 2016

Staff: Mentor

Yes, that looks fine.

17. Mar 2, 2016

nothing909

Can you just explain briefly how I do simultaneous equations with the two loop equations I found?

The only reason I'm confused is because there's three unknowns in one and two in the other.

18. Mar 2, 2016

Staff: Mentor

That is why I've repeatedly been saying that you need three equations to solve three unknowns. All of the variables affect each other. You can't solve for one variable without involving three equations. You need three independent equations to give you three independent bits of information about three independent variables.

I think you need to investigate solving simultaneous equations. This is a general math issue, not an engineering or physics issue. Why don't you do a Google search on "simultaneous equation methods" and review some of the offerings there. Then perhaps address any questions that arise to the Precalculus Mathematics Homework forum. I doubt that the helpers here in this forum will be willing to teach a basic course in equation solving; we're just not set up to teach subjects from scratch.

I'm surprised that you haven't covered simultaneous equations in your education prior to dealing with AC circuit problems.

19. Mar 2, 2016

nothing909

Well you told me both my equations were correct, but obviously not if I need three unknowns to work it out.

I needed you to briefly tell me how to do it with three unknowns. I've used simultaneous equations for a long time, but until I started doing engineering, I've never had to work out three unknowns. It might be an easy thing to do, and I know it is, but I've never learned it. I'm pretty sure I could have picked it up quickly if you literally wrote about 2 sentences on how to work it out.

20. Mar 2, 2016

Staff: Mentor

That's why you need the equation for the third loop. Three loops, three currents, three variables, three equations.
There are several different methods, as I've mentioned. If your chosen method is to use substitution, then "solve" one equation for one of the variables and replace all occurrences of that variable in the other two equations with that expression. Then "solve" one of the remaining two equations for one of the remaining two variables and substitute that expression into the third equation. You'll end up with a single equation in one variable. Solve for that variable. This will be the actual solution for that variable. Then use that value for back-substitution into previous expressions to solve for the values of the other two variables.

You should look for examples on the net. Google something like: "simultaneous equations examples 3 variables".