I'm trying to simplify the following circuit. In the picture, you can see 4 resistors and a voltage source. There are also two terminals at which I measure the output voltage. I suppose that I can replace these output terminals with a voltmeter since I'm measuring the voltage there.

The question is, how can I join these resistors together? Which ones are in parallel and which ones are in series and why? I get mostly confused by the output terminals. If I remove them, then everything is clear.


Picture of the circuit

  • \$\begingroup\$ Theveninos: quanta of the Thevenin field? \$\endgroup\$ – Alfred Centauri Jun 10 '14 at 2:36

To find the Thevenin equivalent resistance (Rth) we need to...

  • Replace voltage sources with short circuits
  • Replace current sources with open circuits
  • Calculate the resistance as it appears from the terminal.

For your case, short circuit the voltage source. This will eliminate R4 as current will take the path of least resistance and now that you have a s/c across R4, current will flow through that.

So we redraw the schematic with this new information...


simulate this circuit – Schematic created using CircuitLab

Now looking at the circuit, the observed resistance between point A and B can be calculated...

Rth = (R1 + R2) || R3

That is, Rth is R3 in parallel with R1 and R2 in series. You should remember how to calculate resistors in parallel.

Now to determine what the equivalent voltage source would be, lets look at the circuit in a different light...


simulate this circuit

All we've done here is move some components around to make it easier to understand. As you'll see, the voltage drop across R4 is the same as the voltage drop across R1,2,3 so we can ignore it as it doesn't help us determine Vth.

Using the voltage divider rule, we can figure out Vth as the voltage drop across R3. In this case...

Vth = V * (R3 / (R1 + R2 + R3))

That is, R3 will have a voltage drop that is proportional to the resistance is adds to the branch compared to the total resistance of the branch. Conveniently, if all values for R1, R2 and R3 are the same...

Vth = V / 3

This is because 1/3 of the voltage drop will occur across R3.

Finally, redraw the circuit to have a voltage source of Vth and a series resistor of Rth.

  1. R4 has no effect on the circuit because the voltage across it is absolutely fixed by the perfect voltage source connected in parallel with it. So you can simply remove R4.

  2. Now R2, V1, R1 are in series and you can get a Thevenin equivalent for the R2, V1, R1 combination by combining R2 and R1 as in-series resistors. Call the new resistor replacing R1 and R2 as R1'.

  3. Now R1' and R3 are in parallel.

  • \$\begingroup\$ Oh, ok, I didn't know that two resistors that are separated by a voltage source count as resistors in series. Now, does it matter whether the new R1' is put before (on top of) the voltage source or after (on the bottom)? \$\endgroup\$ – leopik Jun 9 '14 at 22:15
  • \$\begingroup\$ It doesn't matter. Vout = V1 - I * R1' either way. But if you put it on top you'll see what you did is make the Thevenin equivalent for the V1, R1, R2 subcircuit. \$\endgroup\$ – The Photon Jun 9 '14 at 22:19
  • \$\begingroup\$ That counts as series because it obeys the definition of "series", i.e. the same current flows through all these bipoles. \$\endgroup\$ – Vladimir Cravero Jun 9 '14 at 22:41

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