Total Resistance in this Circuit

  • Thread starter AROD
  • Start date
  • #1
18
0

Homework Statement



I have a network of resistors in a box-formation. There is 1 resistor of equal resistance in between every corner of the box, so 12 in total.

Homework Equations



In series R = r1 + r2 + ... ri
In parallel, 1/R = 1/r1 + 1/r2 + ... 1/ri

The Attempt at a Solution



I can't really draw a 2D analogous network due to the connections. I know kirchhof's rules must be applied somehow... that no current will flow between two points of equal potential difference, but not sure where that would be.
 

Attachments

Answers and Replies

  • #2
1,482
3
Just apply kirchhof law, that the sum of currents at node is zero. So first introduce some current in the circuit by applying a voltage source of 1 volt to make it easy. Then analyze every node to see how the currents are distributed in the branches.
 
  • #3
18
0
first node brances to 3, each of those branch twice, 6 branches then converging on 3 nodes, which tnen go back to 1 node.

since all resistors are equal, can the 3 node to 3 node connection be just expressed as 3 parallel with 2 each? my reaction is to make each little group of 2 parallel, giving a total of 5/6 for the entire thing, but i dont know.

this is also then accounting for all of the resistors contributing. it looks to me like charge would flow through everything.
 

Related Threads on Total Resistance in this Circuit

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
4
Views
2K
Replies
1
Views
1K
Replies
3
Views
1K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
12K
Replies
9
Views
4K
Top