Resistance Formula R = ρL/A: Effect of Length and Area
Physics · Current Electricity · NEET
The resistance of a wire is R = ρL/A, where ρ is the resistivity of the material, L is the length, and A is the cross-sectional area. Resistance goes UP when the wire is longer, and DOWN when the wire is thicker (more area). Memory hook: think of a crowded corridor — a longer corridor (more L) is harder to cross, but a wider corridor (more A) lets electrons flow more easily.
A longer, thinner wire has higher resistance; a shorter, thicker wire has lower resistance. R rises with length (direct) and falls with area (inverse), and for a round wire area depends on diameter squared.
Your doubts, answered
Does resistance depend on length or on area?
On BOTH, but in opposite ways. R = ρL/A. Resistance is directly proportional to length L (double the length -> double the resistance) and inversely proportional to area A (double the area -> half the resistance). Length and area work against each other.
Why does a thicker wire have LESS resistance?
A thicker wire has a larger cross-sectional area A. More area means more parallel paths for the free electrons to flow through, so it is easier for current to pass. Since R = ρL/A, a larger A makes R smaller. This is why thick wires are used for high-current supply lines.
Is A the radius or the area in R = ρL/A?
A is the cross-sectional AREA, not the radius or diameter. For a round wire A = πr² = πd²/4. This is a common trap: if the diameter is tripled, the area becomes 9 times bigger (3²), so R becomes 9 times smaller. Always square the radius or diameter before comparing.
Does R = ρL/A depend on the material?
Yes, through ρ (resistivity). ρ is a property of the material only — copper, aluminium, silver each have a fixed ρ at a given temperature. L and A depend on the shape of the wire. So two wires of the same size but different materials have different resistances.
If I cut a wire in half, what happens to its resistance?
Each half has the same area A but half the length. Since R is proportional to L, each half-piece has half the original resistance (R/2). The area does not change when you simply cut a wire — only the length does.
⚠️ The NEET trap ✗ If a wire's diameter is made 3 times larger, its resistance becomes 3 times smaller (R/3). ✓ Area depends on diameter SQUARED: A = πd²/4. Tripling the diameter makes the area 9 times bigger, so resistance becomes 9 times smaller (R/9). NTA loves testing whether you square the radius or diameter. 🧠 Diameter vs area — the squared factor.
Real NEET questions
2023
Wire A has resistance 81 Ω. Wire B is of the same material, has equal length, but a diameter three times that of A. The resistance of wire B is:
A · 729 Ω
B · 243 Ω
C · 81 Ω
D · 9 Ω ✓
Solution: Step 1: Use R = ρL/A with A = πd²/4, so R = ρL/(πd²/4) = 4ρL/(πd²). Step 2: Same material (same ρ) and equal length (same L), so R ∝ 1/d². Step 3: Diameter of B is 3 times that of A, so d² becomes 9 times larger, making R_B nine times smaller. Step 4: R_B = 81/9 = 9 Ω. Answer: (D).
2022
A copper wire of length 10 m and radius (10⁻²/√π) m has resistance 10 Ω. The current density for an electric field strength of 10 V/m is:
A · 10⁴ A/m²
B · 10⁶ A/m²
C · 10⁻⁵ A/m²
D · 10⁵ A/m² ✓
Solution: Step 1: Area A = πr² = π × (10⁻²/√π)² = π × (10⁻⁴/π) = 10⁻⁴ m². Step 2: Find ρ from R = ρL/A -> ρ = R·A/L = (10 × 10⁻⁴)/10 = 10⁻⁴ Ω·m. Step 3: Current density J = E/ρ = 10 / 10⁻⁴ = 10⁵ A/m². Answer: (D).
Solved Current Electricity NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Rearranging, ρ = RA/L, so the unit is ohm·metre² / metre = ohm·metre (Ω·m). Do not confuse it with the unit of resistance, which is the ohm (Ω).
How does resistance change if length is doubled and area is also doubled?
R = ρL/A. If both L and A double, the factors cancel: R_new = ρ(2L)/(2A) = ρL/A = R. The resistance stays the same.
Why is R directly proportional to length?
A longer wire means electrons travel a greater distance and collide with more ions on the way, so they lose more energy and face more opposition. Double the length gives double the collisions, hence double the resistance.
Does R = ρL/A work for any shape?
It works for a uniform conductor with a constant cross-sectional area (like a straight wire). For odd or changing shapes you must integrate, but for NEET the wire is always uniform.
