Hi I'm struggling a bit with obtaining the total resistance of this circuit, I know it's an easy one I'm just a bit confused on how the resistors are merged using parallel and series operations, here's an image:

Offending Circuit

The overall goal is to obtain Thevenin's equivalent circuit so when obtaining \$R_{th}\$ the circuit is open between nodes A and B, and the voltage source is shorted out.

I've tried

  • First
    • Rp = R1//R3
    • Rs = Rp + R2
    • Rt = Rs//R4 = 2.29
  • Second
    • Rs1 = R1 + R2 = 9
    • Rs2 = R3 + R4 = 8
    • Rt = Rs1//Rs2 = 4.23

Any help is really appreciated.


If you're looking for Rth, you do exactly as you said (remove RL such that there is an open-circuit between A and B, then shut V2 off and measure equivalent resistance from A to B).

First, remove the resistor and short V2. If you label nodes at this point, it will help you to redraw the circuit in a more typical fashion:


Convince yourself that that is true, and from there you should be golden :)

  • \$\begingroup\$ thx for replying, According, to your diagram, \$ R_{th} \$ should be 1 ohms? \$\endgroup\$ – Tristian Jun 4 '12 at 6:54
  • 3
    \$\begingroup\$ +1 for the arrow using the ground symbol :) @Triztian it is \$\endgroup\$ – clabacchio Jun 4 '12 at 7:04
  • \$\begingroup\$ I was hoping someone else would appreciate that xP \$\endgroup\$ – Shamtam Jun 4 '12 at 10:59

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