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You can only use the simple techniques you use to combine parallel and series resistors. Nothing else.

The difficult thing about this network of resistor is that: every node besides a and b has 3 or more branches connected to it, so none of them are in parallel.

Each resistor has 1 Ohm resistance.

enter image description here

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  • \$\begingroup\$ You should number the resistors - how can anyone comment on a specific component without a reference. Your statement about no resistors being in parallel is not quite true - I can see two sets of resistors that can be reduced. \$\endgroup\$ – Kevin White Sep 13 '17 at 1:57
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This can be solved with a Star-Mesh transform. The centre 6 resistors C-H can be replaced by a 4-resistor star, which can be all 1/4 Ohm, if I have it right.

See [this question] (Under what conditions is the star-mesh transform invertible?) for a more thorough discussion.

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  • \$\begingroup\$ He is asking to use parallel and series only, nothing else! \$\endgroup\$ – Hazem Sep 13 '17 at 5:17
  • \$\begingroup\$ well actually yes star delta transformation is accepted! \$\endgroup\$ – most venerable sir Sep 14 '17 at 21:44
  • \$\begingroup\$ What do next after replacing C and H to the outside can you draw a picture of the step or tell me which other triangle to transform? \$\endgroup\$ – most venerable sir Sep 14 '17 at 22:05
  • \$\begingroup\$ Well I admit the star-delta transformation isn't helping me reduce the general case. It might be time to break out nodal analysis... In the specific case of all 1 Ohm, you can remove C and H entirely, there's no potential across them. Consider them short circuited. Then DEFG are all in parallel, AB are in parallel and IJ are in parallel. \$\endgroup\$ – tomnexus Sep 15 '17 at 18:59
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Each resistor has 1 Ohm resistance

This makes it easier (hopefully). Also, by observation, the circuit has symmetry and this makes solving it quite simple.

If all resistors have the same value then the current from node a (top) to node b (bottom) is shared equally between resistor A and B and given that they have the same value there is zero voltage across resistor C. If there is zero voltage across C then it can be shorted out without any expectation of a change in impedance.

The same can be said for resistor H and this leads to a simplified circuit thus: -

schematic

simulate this circuit – Schematic created using CircuitLab

Hence, total resistance from a to b is 1.25 ohm

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