Derive a formula for equivalent resistance

AI Thread Summary
The discussion focuses on deriving a formula for equivalent resistance in cascaded sections, specifically in series and parallel configurations. For resistors in series, the equivalent resistance is the sum of individual resistances: Reqv = R1 + R2 + ... + Rn. For parallel resistors, the formula is given by Reqv = 1/{(1/R1) + (1/R2) + ... + (1/Rn)}. The conversation explores the approach to calculate the equivalent resistance of multiple stages, suggesting that combining resistors in series and parallel can lead to a recursive formula. The analysis indicates that combining stages involves summing series resistances and applying the parallel formula to groups of resistors.
boo_lufc
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Is it possible to derive a formula for the equivalent resistance of n such sections in cascade?
Show your analysis.

Series: Reqv = R1 + R2 +...Rn

Parallel: Reqv = 1/{(1/R1) + (1/R2) + (1/R3) +...(1/Rn)}
 

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boo_lufc said:
Is it possible to derive a formula for the equivalent resistance of n such sections in cascade?
Show your analysis.

Series: Reqv = R1 + R2 +...Rn

Parallel: Reqv = 1/{(1/R1) + (1/R2) + (1/R3) +...(1/Rn)}

...and how would you go about attempting to solve this? What is the resistance of one stage? And the resistance of two stages in series? And...
 
I was thinking Rn is the last resistor so the last branch of any similar circuit would be
Rn + Rn-1 in series
Then this would be in parallel with Rn-2 + Rn-3 and so on
i.e Reqv = 1/ {(1/Rn + Rn-1) + (1/(Rn-2 + Rn-3) + (Rn-4 + Rn-5)}
Is this along the right lines at
 
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