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When attempting to find the Thévenin-equivalent circuit to the left of terminals a and b in the image below, should one consider the 4.7 kΩ resistor? Why or why not?

How to find Thévenin-equivalent resistance?

When taking the 4.7 kΩ resistor into account, the process goes as follows:

$$ \begin{aligned} 3.3\parallel\big[(1\parallel1.5)+(4.7\parallel1.5)\big] &= 3.3\parallel\left[\frac{1\times1.5}{1 + 1.5}+\frac{4.7\times1.5}{4.7+1.5}\right]&&\\ &= 3.3\parallel1.737098&&\\ &= \frac{3.3\times1.737098}{3.3+1.737098}&&\\ &= 1.138041&&\\ &= 1.14\text{ kΩ} \end{aligned} $$

When omitting the 4.7 kΩ resistor:

$$ \begin{aligned} 3.3\parallel\big[(1\parallel1.5)+1.5\big] &= 3.3\parallel\left[\frac{1\times1.5}{1+1.5}+1.5\right]&&\\ &= 3.3\parallel2.1&&\\ &= \frac{3.3\times2.1}{3.3+2.1}&&\\ &= 1.283333&&\\ &= 1.28\text{ kΩ} \end{aligned} $$

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    \$\begingroup\$ You need to ask yourself whether the 4.7k resistor is to the left of the terminals a and b. If it is, then yes, if not, then no. BTW, I don't get the answer you get. \$\endgroup\$
    – Neil_UK
    Commented Oct 26, 2023 at 14:23
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    \$\begingroup\$ As Neil_UK said, your results either way aren't correct. To give you hint, the 3.3K isn't in parallel with anything. Try redrawing the schematic after you've removed \$R_L\$ and shorted the voltage source. \$\endgroup\$
    – GodJihyo
    Commented Oct 26, 2023 at 14:50
  • \$\begingroup\$ @GodJihyo Here is my simplified schematic: mediafire.com/convkey/21f9/moa3fcu1h2mhczsbg.jpg?size_id=4 I'm still confused as to why the 3.3 kΩ resistor is not in parallel with the rest. To my eye, both its terminals share nodes with the terminals of the structure below it. Thanks. \$\endgroup\$
    – Someone.
    Commented Oct 26, 2023 at 16:52
  • \$\begingroup\$ @Someone. You have to take note of where the resistance is being measured across, points \$a\$ and \$b\$. You shouldn't use curved lines in a schematic, they make it difficult to read. Redraw it with straight lines, and add points \$a\$ and \$b\$, then simplify by combining resistors that are in parallel, and combining resistors that are in series, keep combining until you're left with one resistor across points \$a\$ and \$b\$.. Hopefully that will make it clearer. \$\endgroup\$
    – GodJihyo
    Commented Oct 26, 2023 at 17:01
  • \$\begingroup\$ @GodJihyo Thank you for the follow-up. I followed your suggestion by breaking it up into steps and replacing resistors until I got a single one. Is the answer something near 880.4 Ω? My second last step had three parallel resistors, 3.9k, 1.5k and 4.7k. If I got it wrong, then I can show you my steps on paper. Thank you again. \$\endgroup\$
    – Someone.
    Commented Oct 26, 2023 at 20:28

1 Answer 1

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Well of course it should be considered; it is part of the circuit for which you are finding the Thevenin equivalent. Also, mentioned in the comments, your calculations aren't correct. The 3.3k resistor is connected in delta, not parallel. To find the Thevenin resistance you could do a delta-wye conversion and analyze that equivalent network, or from calculating the open-circuit voltage and short-circuit current

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    \$\begingroup\$ You don't need to do a delta-wye, it can be figured out with just parallel and series resistance formulas. \$\endgroup\$
    – GodJihyo
    Commented Oct 26, 2023 at 15:22

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