R Network- finding total resistance.

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SUMMARY

The discussion focuses on calculating the total resistance in a resistor network with equal resistances of R = R' = 1.97 Ω. The key findings indicate that the net resistance between points a and c can be simplified using symmetry and voltage divider principles, leading to a calculation of 1/R + 1/2R + 1/2R = 1/Rac. The participant initially struggled with a complex branch in the circuit but was guided to eliminate certain resistors, simplifying the analysis significantly.

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  • Understanding of series and parallel resistor combinations
  • Familiarity with voltage dividers in electrical circuits
  • Basic knowledge of circuit symmetry principles
  • Ability to interpret circuit diagrams
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Homework Statement


Assuming R = R' = 1.97 Ω in the network shown above, determine the net resistance

a) between the points a and c
b) between points a and b.
HELP: Use symmetry.
HELP: Imagine applying a voltage between a and c and think about what current will flow in each resistor.

https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/Knox/phys130a/spring/homework/09/02/P26_36.jpg


Homework Equations


Series: Rtotal=R1+R2
Parallel: Rtotal=(R1-1+R2-1)-1


The Attempt at a Solution



Well, I redrew the series in my notebook as parallel and series, but in both attempts at a re-draw, I ended up with a weird branch that wasn't really either series or parallel, because of how it connected. I tried solving it anyway (and have come up with 1.313 for the resistance between a and c) but nothing's been right. I'm just not sure what to do with the weird branch. :/
 
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Observe that with all the resistors equal to each other that you can drop out the R' resistors that cross between the || branches. At each leg of your || you have a voltage divider with no net voltage across the R'. This should make your calculation a bit easier.

For instance between AC:

1/R + 1/2R + 1/2R = 1/Rac
 
Oh wow, that makes things a lot simpler. I wasn't aware that I could drop out that resistor. Thank you!
 

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