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The question asks to find the equivalent resistance of such a network of resistors.

schematic

simulate this circuit – Schematic created using CircuitLab

I have managed to simplify it down to the following:

schematic

simulate this circuit

Is there a way to find the equivalent resistance without the use of a delta-transform (as I do not believe those are allowed on this assignment). From a logical point of view since the left and right nodes of the 0.5 ohm resistor are at the same voltage, no current would flow through it and hence it would offer no resistance. Would the equivalent resistance then be the parallel connection of the 1 and 1.5 ohm resistor? or am I completely wrong.

Thank you in advance for your responses!

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    \$\begingroup\$ It's correct: R3//short = short. But you shouldn't have wasted time in calculating the parallel of R6 and R7 - they are both shorted. \$\endgroup\$ – Sredni Vashtar Jun 18 '15 at 0:00
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    \$\begingroup\$ so the correct answer is 0.6 ohms = 3/5 ohms? And thank you for your help :) \$\endgroup\$ – Filip Gajowniczek Jun 18 '15 at 0:07
  • \$\begingroup\$ @FilipGajowniczek Yes, \$0.6\Omega\$. \$\endgroup\$ – Null Jun 18 '15 at 2:21
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Observe that in the original circuit the resistors R6 and R7 are short as they have a wire connecting their ends. You can just replace them with a short. Replacing it with a short you are left with R4 and R5 in parallel. It should be fairly easy to solve from here. It is

$$ R3 || (R2 + (R4 || R5)) = 0.6 \Omega $$.

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