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I am trying to find how to break apart a Y-Y connection to find the equivalent resistance. I understand how to find the equivalent resistance for 2 Y circuits in series, but not parallel. I understand how to transform from delta to Y and back, but am unable to find a simplified circuit for this design:

o--------o--------------------o
|         \        o        /
|          \      /\       /
|           R1   /  \     R6
|            \  R3   R4  /
|             \ /     \ /
|              o       o
PS             |       |
|              R2      R5
|              |       |
o--------------o-------o

Where "PS" is the power source and "o" is the nodes and "Rx" is the different resistors. Please Help.

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    \$\begingroup\$ That drawing makes it look harder than it is. R3 and R4 are in series, so can be replaced by a single resistor that connects the two nodes R1-R2 and R5-R6. So the topology is an H, not two Y's. \$\endgroup\$
    – Pete Becker
    Commented Jul 7, 2015 at 1:07

2 Answers 2

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Thanks to Pete Becker and Josh Jobin, the answer did appear. The H topology is a rewrite of the unbalanced wheatstone bridge, which can be rewritten as 2 wye's in series or:

o-------------o
|            / \
|          R1    R6
|          /      \
|         o        o
|          \      /
|           R8  R9
|            \  /
PS            \/
|             o
|             |
|            R10
|             |
o-------------o

Which is easily solvable. Thank you for your help.

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R1 and R3 and in parallel, lets say this can be simplified to Ra. Now R4 and R6 are in parallel, call them Rb. Now Ra and R2 are in series, call that Rc. And Rb and R5 are in series, call that Rd. Now we have Rc and Rd in parallel and we can find the total resistance. So: Req = {[(R1*R3)/(R1+R3)] + R2}*{[(R4*R6)/(R4+R6)] + R5}/({[(R1*R3)/(R1+R3)] + R2} + {[(R4*R6)/(R4+R6)] + R5})

Edit: Nice picture -

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    \$\begingroup\$ R1 and R3 aren't in parallel. That node where R3 and R4 join isn't connected to anything else. \$\endgroup\$
    – Pete Becker
    Commented Jul 7, 2015 at 1:08
  • \$\begingroup\$ My apologies! I thought that was a small typo. \$\endgroup\$
    – Josh Jobin
    Commented Jul 7, 2015 at 1:43
  • \$\begingroup\$ R3 and R4 are in series. They then create a delta shape with R2 and R5. You would then convert that delta to wye. After that, you would have R1 in series with a wye resistor, and R6 in series with another wye resistor. Then it would be simple to simplify. Again, my apologies for misreading that. \$\endgroup\$
    – Josh Jobin
    Commented Jul 7, 2015 at 1:45
  • \$\begingroup\$ Thank you Pete, Josh. I posted the answer using both of your inputs. \$\endgroup\$
    – Mark Walsh
    Commented Jul 7, 2015 at 3:17

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