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I'm learning from Paul Scherz's Practical Electronics for Inventors (2nd ed.) and I've become confused at what was I thought I was comfortable with: namely parallel and series resistors. The offending example is:

Worked example of Thévenin's Theorem

Looking at the uppermost diagram and working out the simple Thévenin equivalent circuit for the circled loop, I don't understand why the total Thévenin resistance is the sum of individual parallel resistors and not series resistors.

My reasoning tells me that since the other half of the circuit has been shorted, then an equivalent current flows through both resistors in this loop i.e. implying a series arrangement. Going further, even if the circuit were complete, these two resistors would still be in series, so I'm confused as to why the Thévenin resistance has been calculated as if this is a parallel arrangement.

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The Thevenin equivalent resistance is one that looks to the circuit you want to replace. To find this value, all independent sources must be passivated.

Passivate ideal voltage source means replacing it by a short circuit. Passivate ideal current source means replacing it by an open circuit.

If the 5V supply is passivated in this specific case, the two resistors are connected in parallel, and here the value of the Thevenin equivalent resistance.

schematic

simulate this circuit – Schematic created using CircuitLab

ERRATA: In the schematic, "V2" must be \$V_{th}\$, Thevenin's equivalent voltage source.

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    \$\begingroup\$ that should be "all independent sources passivated". Also, it is necessarily the case that said equivalent resistance is the ratio of the open-circuit voltage to short-circuit current, \$R_{th} = \frac{V_{OC}}{I_{SC}}\$ so one doesn't necessarily need to zero the independent sources to find the equivalent resistance. \$\endgroup\$ – Alfred Centauri Sep 16 '14 at 13:43
  • \$\begingroup\$ @AlfredCentauri Yes!. You right! Tanks for your comment... \$\endgroup\$ – Martin Petrei Sep 16 '14 at 13:45
  • \$\begingroup\$ Why is it that passivating the independent source makes the resistors parallel from their original series configuration? I'm not finding this incredibly obvious. Is it because they are now at the same voltage i.e zero? \$\endgroup\$ – TheJerseyChemist Sep 16 '14 at 14:18
  • \$\begingroup\$ @TheJsyChemist Think that: Passivating the voltage source is the same as remove it of the circuit and instead set a conductor. With a wire instead of the source, the two resistors are connected in parallel. \$\endgroup\$ – Martin Petrei Sep 16 '14 at 14:22
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first you write norton of DC power supply and then two resistor are parallel then write thevnin. enter image description here

enter image description here

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    \$\begingroup\$ Although it applies in this case, this response may be confused about the method to determine the value of the Thevenin equivalent resistance. The equivalent resistance (both Norton to Thevenin) is one that "sees" the load circuit with all sources passivated. \$\endgroup\$ – Martin Petrei Sep 16 '14 at 13:37

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