An Important Deduction

This concept shows how the resistance R of a conductor depends on its length l and cross-sectional area A, using the logic of series and parallel resistor combinations. It builds the foundation for the resistivity formula used throughout Current Electricity.

This derivation is a recurring board-exam question and forms the base for resistivity, temperature dependence, and Wheatstone bridge topics.

Consider a conductor of length l, imagined as l identical unit-length pieces joined end to end (in series).

• Each unit-length piece has resistance R′.
• For resistors in series, total resistance adds up:
  R = R′ + R′ + … (l times) = lR′

Key Result 1: R ∝ l — Resistance increases with the length of the conductor.

Consider a conductor of cross-sectional area A, imagined as A identical unit-area conductors placed side by side (in parallel).

• Each unit-area conductor has resistance R′.
• For resistors in parallel, reciprocals add up:
  \[\frac {1}{R}=\frac {1}{R′}+\frac {1}{R′}\] + …(A times) = \[\frac {A}{R′}\]
  R = \[frac {R′}{A}\]

Key Result 2: R ∝ \[\frac {1}{A}\]​ — Resistance decreases as cross-sectional area increases.

This is the final deduction: resistance is directly proportional to length and inversely proportional to cross-sectional area.

Given three resistors: 1 Ω, 2 Ω, and 3 Ω. Find combinations producing each target resistance.

Think of resistance like a crowd moving through a corridor:

• A longer corridor (greater length) means more distance to push through → more resistance to flow.
• A wider corridor (greater area) means more space for people to pass simultaneously → less resistance to flow.

This mirrors why long, thin wires heat up more (higher resistance) than short, thick wires.