equivalent resist.

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

  1. bahire

    Thread Starter New Member

    Apr 28, 2009
    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.
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    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
  3. Wendy


    Mar 24, 2008
  4. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
    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. :)