In this problem I'm trying to find the Thevenin resistance in the circuit below:


simulate this circuit – Schematic created using CircuitLab

The first thing we do is remove all sources and the load resistor. So now we have the following circuit:


simulate this circuit

The part I'm stuck on is computing the total resistance between the load terminals a and b. If I remove R4, I can figure it out; R2, R7, and R8 all terminate at the ground node. We have R2 and R7 in parallel, which is in series with R5, which is in parallel with R8. Thinking of it this way, but then adding R4 into the mix, doesn't R4 begin and end at the same node?

  • \$\begingroup\$ Yes it does. Since it's shorted out, does that suggest something? \$\endgroup\$ – Spehro Pefhany Sep 18 '14 at 0:29
  • 1
    \$\begingroup\$ @SpehroPefhany, so that means the current would prefer to flow through the wire than the resistor, so it's like the resistor isn't even there. Right? \$\endgroup\$ – Lefty Sep 18 '14 at 0:30
  • \$\begingroup\$ Exactly correct. \$\endgroup\$ – Spehro Pefhany Sep 18 '14 at 0:43
  • \$\begingroup\$ I think of it this way: a resistor in parallel with a short (a 0 ohm resistor) has a resistance of R // 0 = (R * 0) / (R + 0) = 0 ohms. That means you can ignore R4. \$\endgroup\$ – Greg d'Eon Sep 18 '14 at 0:44
  • \$\begingroup\$ Why shorted resistor is ignorable? \$\endgroup\$ – Amit Hasan Sep 18 '14 at 1:53

You can use T to Pi transformation which would make things a little more practical. Reference to : Pi and T networks .

Hope it helped.


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