Derive a formula for equivalent resistance

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SUMMARY

The discussion focuses on deriving the formula for equivalent resistance in circuits with resistors arranged in series and parallel configurations. The established formulas are: for series, the equivalent resistance is calculated as Reqv = R1 + R2 + ... + Rn, and for parallel, it is Reqv = 1/{(1/R1) + (1/R2) + (1/R3) + ... + (1/Rn)}. The conversation explores the complexity of cascading multiple resistors and suggests a method for combining resistors in series and parallel to find the overall equivalent resistance. The proposed approach involves recursively applying these formulas to groups of resistors.

PREREQUISITES
  • Understanding of basic electrical concepts, specifically Ohm's Law.
  • Familiarity with series and parallel resistor configurations.
  • Knowledge of algebraic manipulation for solving equations.
  • Basic circuit analysis skills.
NEXT STEPS
  • Study the derivation of equivalent resistance in complex circuits.
  • Learn about Kirchhoff's laws for circuit analysis.
  • Explore the concept of Thevenin's and Norton's theorems.
  • Investigate practical applications of equivalent resistance in real-world circuits.
USEFUL FOR

Electrical engineering students, circuit designers, and hobbyists interested in understanding and calculating equivalent resistance in various circuit configurations.

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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