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I am really confused by the concept of ignoring resistors while doing Thévenin theorem, especially for this problem.

Some online videos, like this one, and other examples that I watched seem to suggest otherwise.

Could someone please explain why the 16 Ω resistor is ignored here:

enter image description here

enter image description here

The example link above seems to contradict the solution given by the book.

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  • \$\begingroup\$ Academic questions need an attempt at a solution, please post a source for the images. \$\endgroup\$
    – Voltage Spike
    Dec 17, 2023 at 3:45
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    \$\begingroup\$ I vote to reopen. The question is not to privide a solution but to understand whythe given solution works \$\endgroup\$
    – RussellH
    Dec 17, 2023 at 4:45
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    \$\begingroup\$ I posted an example of other circuits with similar set up where the resistor close to the shorted voltage is not excluded. I am confused as to why 16ohm is not considered in the Rth calculation and is "bypassed" as the book claims. \$\endgroup\$ Dec 17, 2023 at 4:47
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    \$\begingroup\$ Yes I am not asking for a solution. I am asking why the solution works \$\endgroup\$ Dec 17, 2023 at 4:48
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    \$\begingroup\$ @VoltageSpike The answer to the course work question is already given at 6. OP's actual question is to explain why that is the case. \$\endgroup\$
    – Graham Nye
    Dec 17, 2023 at 17:59

2 Answers 2

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As summarised in the paragraph 6 you quoted, and documented elsewhere, the Thévenin-equivalent resistance \$R_{Th}\$ is calculated by replacing ideal voltage sources with a short circuit (and, although not needed in your circuit, replacing ideal current sources with an open circuit).

The 60 V source in your circuit is in parallel with the 16 Ω resistor. When you replace the 60 V source with a short circuit you are putting a short circuit (0 Ω, by definition for an ideal short circuit) across the 16 Ω resistor. A 0 Ω resistor and a 16 Ω resistor in parallel have a combined resistance of 0 Ω, so the 16 Ω resistor disappears from the calculation.

Now to your Savvy Engineer question:

Savvy Engineer question
Source

The 6 Ω resistor here is in series with the 12 V source. When we replace the 12 V source with a short circuit we have a 0 Ω resistor in series with a 6 Ω resistor. Their combined resistance is 6 Ω. So as this is a series combination this time the value of the 6 Ω resistor does matter. There is no contradiction as your question deals with a parallel circuit whilst the Savvy Engineer question deals with a series circuit (for the parts of the circuit we are discussing).

You might find it useful to revise series and parallel circuits, and the formulas for combining resistances, described on Wikipedia (as well as lots of other places online and in books, if you don't get on with the Wikipedia description).

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    \$\begingroup\$ Thanks! This is really helpful, I appreciate your insight \$\endgroup\$ Dec 19, 2023 at 1:09
  • \$\begingroup\$ @DuaTheEngineer I'm glad you found it useful. On this site, rather than leaving messages of thanks, you are requested to vote up and, if appropriate, accept an answer that addresses your question, as described here. \$\endgroup\$
    – Graham Nye
    Dec 19, 2023 at 2:11
  • \$\begingroup\$ @DuaTheEngineer You still need to accept whichever answer best answers your question. If neither answer completely answers your question add a comment explaining what else you need to know to whichever question is closest to what you need. Accepting answers is how this site works. If you fail to accept answers people will be less likely to provide you with them, such as your reactive power question. \$\endgroup\$
    – Graham Nye
    Dec 22, 2023 at 22:56
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Could someone please explain why the 16ohm resistor is ignored here

Yes. See the steps shown below:

schematic

simulate this circuit – Schematic created using CircuitLab

  1. You may choose one and only one node to call \$0\:\text{V}\$. I've chosen \$V_\text{B}\$. This does NOT in any way change the circuit. It's just getting rid of a wire on the schematic and using the ground label, instead. Nothing else implied.

  2. There's no need to show the voltage supply itself on the schematic. You can simply label the node with that voltage and get rid of the voltage supply icon. It's just an extra distraction, so why not get rid of it to clean things up a bit more?

  3. Breaking up the node into two, with the exact same voltage doesn't actually change the circuit if both are connected to the same voltage supply. I think you must agree with that. No change. It's just another way of writing the same thing.

    But take this just one step further. Imagine creating two separate voltage supplies, each with the exact same voltage and reference point (ground.) Certainly would not do that due to cost and space and weight, etc., of course. But technically could do that. The circuit would still behave the same. So there's no harm in mentally taking this step to simplify the schematic, further. And it won't change the behavior of node A.

So you can see that these are two different circuits, each having no impact on the other. Which means you can throw away the one that includes \$R_1\$ and focus on the other one:

schematic

simulate this circuit

It's just a resistor voltage divider, now.

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    \$\begingroup\$ Wow! this is great thanks for the detailed explanation , I really appreciate it! \$\endgroup\$ Dec 19, 2023 at 1:09

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