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I'm having a little trouble finding the Thevenin resistance equivalent of this circuit. Without the 7.5k resistor, I expect it to be just 15k||15k. Does the 7.5k resistor here act in parallel with the lower 7.5k resistor (to become a 5k resistor) or is it more complicated than that?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ The 7.5K would be in parallel with the lower 15K resistor. However we cannot be fully certain if it is truly connected since you did not dot the connections. I even see a small gap at the top of the 7.5K. \$\endgroup\$ – Michael Karas Apr 9 '17 at 0:48
  • \$\begingroup\$ Cover up the right side resistor and focus on the Thevenin for the rest. \$V_{TH}=15\:\textrm{V}\cdot\frac{15\:\textrm{k}\Omega}{15\:\textrm{k} \Omega+15\:\textrm{k} \Omega}=7.5\:\textrm{V}\$ and \$R_{TH}=\frac{15\:\textrm{k}\Omega\cdot 15\:\textrm{k}\Omega}{15\:\textrm{k} \Omega+15\:\textrm{k} \Omega}=7.5\:\textrm{k} \Omega\$. That's a voltage and a series resistance. This sources the right side. Does the rest look pretty easy to complete, now? \$\endgroup\$ – jonk Apr 9 '17 at 2:11
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You might find it easier to forget you ever heard the word "Thevenin." You're trying to find a single voltage source and single resistor that would have the same open circuit output voltage and short circuit output current as the circuit in the problem (definition of Thevenin and Norton equivalents). There's no special rule that suspends Ohm's law - you surely do know how to solve the problem. The 7.5k and 15k resistors are in parallel, so of course you can combine them, getting 5k. That leaves you with a voltage divider whose output is the Thevenin voltage. Then, to get the Thevenin resistance, first calculate how much current flows when the output of the problem circuit is short-circuited, and determine what size of resistor would give you the same short-circuit current from the Thevenin source.

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Yes to 5k. No it is not more complicated than that

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