# equivalent resist.

Discussion in 'General Electronics Chat' started by bahire, Apr 28, 2009.

1. ### bahire Thread Starter New Member

Apr 28, 2009
1
0
Hi,
i tried hard to analyze the equivalent resistance of the following cubed-structured configuration of resistors. but i couldn't figure it out. so can anyone help me hint how to reduce the circuit to its simplest equivalent ckt. i tried to simplify it using the wye-delta conversion but i fail in the middle as i coildn't visualize the three dimensional arrangment of the resistors.

all the resistors are of the same value, say R and a current is going in Node A and out of node B.

File size:
52 KB
Views:
21
2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
One way to approach the problem is to imagine you "inject" a current of (say) 1Amp into node A and the same current comes out of the other node B.

Because the network is perfectly symmetrical you would imagine the current divides equally among the 3 branches at the first node A. Again, by symmetry each of those branch currents will re-combine equally at the exit node B.

It remains to determine how the current divides at the intermediate nodes {think symmetry}. Once you know the current in each branch you can find the voltage drop across those branches which add up to the the voltage VA-to-B.

RA-to-B is then found by (VA-to-B)/Iinject

Mar 24, 2008
21,839
3,051
4. ### KL7AJ AAC Fanatic!

Nov 4, 2008
2,208
427
One way you can simplify this a great deal is by the principle of "virtual shorts." By inspection, insert a current into the input terminals and determine which nodes of the cube will have equal voltages. Connect those points togeter with a wire, shorting out the resistors in question. Now solve the remaining currents.

eric