Adaal
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How would I be able to find the total resistance of the setup below?
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Adaal
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lordaxil
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First of all, you need to define the question unambiguously: I assume you mean find the total resistance of network between opposite ends (horizonally).

There are two basic strategies for this kind of question. The first is to use Kirchhoff's circuit laws (sum of all currents in/out of a point is zero, sum of all voltages around a closed loop is zero) together with Ohm's law (V=IR) to form simultaneous equations in all (five) resistances and solve. This will always work, provided you identify the right number of equations to match the number of unknowns, but is tedious to solve by hand for large networks.

A quicker way, but one which requires more thought, is to try to decompose the resistor network into combinations of resistors in parallel and series, then use the well-known formulae for combining those resistances. This has to be done step-by-step until you end up with either a single resistor or a combination which is easy to solve. However, there are cases where this approach fails, and the example you have drawn is one of them. You cannot break the network down into combinations of resistors in parallel/series as it stands, and there are five unknowns, which is tedious to solve with Kirchhoff.

The solution is to use something called the star-delta transformation, which repacles a "star" of 3 resistors around a point with a "delta" (triangle) of 3 resistors in a closed loop. There is a worked example (Example 2) for the network topology you have drawn at the web page below (they call it a delta-wye transformation, but it's the same thing). See if you can follow the steps.

https://spinningnumbers.org/a/delta-...-networks.html
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Adaal
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(Original post by lordaxil)
First of all, you need to define the question unambiguously: I assume you mean find the total resistance of network between opposite ends (horizonally).

There are two basic strategies for this kind of question. The first is to use Kirchhoff's circuit laws (sum of all currents in/out of a point is zero, sum of all voltages around a closed loop is zero) together with Ohm's law (V=IR) to form simultaneous equations in all (five) resistances and solve. This will always work, provided you identify the right number of equations to match the number of unknowns, but is tedious to solve by hand for large networks.

A quicker way, but one which requires more thought, is to try to decompose the resistor network into combinations of resistors in parallel and series, then use the well-known formulae for combining those resistances. This has to be done step-by-step until you end up with either a single resistor or a combination which is easy to solve. However, there are cases where this approach fails, and the example you have drawn is one of them. You cannot break the network down into combinations of resistors in parallel/series as it stands, and there are five unknowns, which is tedious to solve with Kirchhoff.

The solution is to use something called the star-delta transformation, which repacles a "star" of 3 resistors around a point with a "delta" (triangle) of 3 resistors in a closed loop. There is a worked example (Example 2) for the network topology you have drawn at the web page below (they call it a delta-wye transformation, but it's the same thing). See if you can follow the steps.

https://spinningnumbers.org/a/delta-...-networks.html
Sorry for the ambiguity, I meant what you've inferred it to mean. I've tried the second method, but obviously realised it wouldn't work in this case. Thanks for sharing the link!
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lordaxil
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(Original post by Adaal)
Sorry for the ambiguity, I meant what you've inferred it to mean. I've tried the second method, but obviously realised it wouldn't work in this case. Thanks for sharing the link!
No problem - try the star-delta transformation if you are stuck using second method.

I wrote it out for you in attachment to check your answer.
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Adaal
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(Original post by lordaxil)
No problem - try the star-delta transformation if you are stuck using second method.

I wrote it out for you in attachment to check your answer.
Got it right! Tysm!
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