• Support PF! Buy your school textbooks, materials and every day products Here!

Kichoff Simultaneous Equations

  • Thread starter jase951
  • Start date
  • #1
9
0

Homework Statement



I what to know how I form an equation for the middle loop, as there is no EMF in that part. Or, how do I go about forming the equations? Also, am I right to say that I should have 3 equations?

Homework Equations



http://www.theground.co.uk/CCF18012009_00000.jpg [Broken]

Thanks!

Jase
 
Last edited by a moderator:

Answers and Replies

  • #2
Doc Al
Mentor
44,945
1,206
In any loop, the sum of the voltage drops (IR) and the EMFs must be zero.

One way is to assign each branch its own current. You'll need as many equations as branches.
 
  • #3
9
0
Thanks!

So the middle section, we'd have 0V = (Ix20) + (Ix10) + (Ix8) + (Ix20)?
 
  • #4
Doc Al
Mentor
44,945
1,206
So the middle section, we'd have 0V = (Ix20) + (Ix10) + (Ix8) + (Ix20)?
Yes, but each segment of that loop will have a different current. And take care with signs when you add up the voltage drops.
 
  • #5
9
0
Thanks for the reply.

I'm trying it again. I've updated the image.

I have managed to have do this with 2 loops, but the the 3 loops. Using Kichoff's second law, I'll have 3 equations I understand.

What I am confused about is where I need to lable the current changes to enable me to use Kichoff's first law.

Thanks!
 
  • #6
Doc Al
Mentor
44,945
1,206
I would label each segment of the circuit with its own current. I see six separate segements, thus I'd have six currents. Then I'd use 3 loops and 3 nodes to produce my equations. (It's not as hard as it sounds.)
 
  • #7
14
0
Another thing I found helpful when I took this class is, once you follow Doc Al's advice, to set up all the equations in a matrix. It helps make the algebra easier! :smile:
 
  • #8
9
0
I would assume that the 6 currents would be that of where the resitors are. where would the 6th be?

For the first loop on the left I have:
200V = 10.I1 (thus I1 = 10A).

2nd loop I have:
0V = 8.I2 + 10.I1 + 20.I3 + 20.I5

3rd:
-125V = 30.I4 + 20.I5

For the nodes, and K's 1st Law:

I3 + I4 = I5

And I don't understand how to get the other 2 when I have 2 currents coming out of the node: I5 = I4 + I2 ???

I've got to understand this, doing a Product Design degree. Thanks.
 
  • #9
Doc Al
Mentor
44,945
1,206
I would assume that the 6 currents would be that of where the resitors are. where would the 6th be?
The 6th current will be through E1.

For the first loop on the left I have:
200V = 10.I1 (thus I1 = 10A).
Good.

2nd loop I have:
0V = 8.I2 + 10.I1 + 20.I3 + 20.I5
Good.

3rd:
-125V = 30.I4 + 20.I5
Good.

For the nodes, and K's 1st Law:

I3 + I4 = I5
Good.

And I don't understand how to get the other 2 when I have 2 currents coming out of the node: I5 = I4 + I2 ???
That works.

Get one more node equation (which will involve the 6th current).
 
  • #10
9
0
Thanks!

So I6 + I2 = I1.

I should be able to do the algebra from these.

Could you tell me why you would have I6 there, because of the EMF? As I thought that that current through the EMF would be the same as I1 - or is it because the bottom node is split?

Thanks again!
 
  • #11
Doc Al
Mentor
44,945
1,206
So I6 + I2 = I1.
That works.
Could you tell me why you would have I6 there, because of the EMF? As I thought that that current through the EMF would be the same as I1 - or is it because the bottom node is split?
Because the current splits. (From the above equation, the only way that I6 = I1 would be if I2 = 0.)

Whenever you have a node, assume that each segment intersecting that node has its own current.
 
  • #12
9
0
I'm understanding it now. Many Thanks again! :)
 

Related Threads on Kichoff Simultaneous Equations

  • Last Post
Replies
6
Views
945
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
3
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
8
Views
7K
Replies
2
Views
1K
Replies
3
Views
4K
Top