Prerequisites
Before studying this topic, make sure you understand:
- Ohm’s Law - Foundation of resistance concept
- Drift Velocity - Microscopic understanding of current
The Hook: Why Are Extension Cords Made of Thick Copper Wire?
You’re watching Jawan (2023) on your TV. Suddenly, you plug in an electric heater using a thin, long extension cord. The lights dim! What just happened?
The culprit: RESISTANCE
The thin, long cord has HIGH resistance. When heavy current flows (from heater + TV), voltage drops across the cord, leaving less for your appliances. This is why:
- House wiring uses THICK copper wires (low resistance)
- Extension cords have limits (thin → high R → voltage drop)
- Gold connectors in premium cables (very low resistivity!)
Question: Why does copper work better than iron? And why does a hot wire have MORE resistance than a cold one? Let’s find out!
Interactive Demo
See how length, area, and material affect resistance:
The Core Concept: What is Resistance?
Resistance is the opposition to current flow in a conductor.
Think of it as “electrical friction” - just like mechanical friction opposes motion, resistance opposes current.
Ohm’s Definition
$$\boxed{R = \frac{V}{I}}$$Unit: Ohm (Ω)
In simple terms: “How much voltage is needed to push 1 ampere of current through the conductor”
Factors Affecting Resistance
Resistance of a conductor depends on:
1. Length (L)
$$R \propto L$$Longer wire → More resistance
Like a longer obstacle course - more length means more collisions for electrons!
2. Cross-sectional Area (A)
$$R \propto \frac{1}{A}$$Thicker wire → Less resistance
Like a wider highway - more lanes for electrons to flow!
3. Material (ρ - resistivity)
$$R \propto \rho$$Different materials, different resistances
Some materials let electrons flow easily (conductors), others don’t (insulators).
Combined Formula
$$\boxed{R = \rho \frac{L}{A}}$$where:
- $\rho$ = resistivity of material (Ω·m)
- $L$ = length (m)
- $A$ = cross-sectional area (m²)
Tony Stark’s arc reactor needs to deliver HUGE currents to his suit. That’s why:
- Cables are extremely SHORT (minimize L)
- Cables are very THICK (maximize A)
- Made of best conductors, likely superconducting! (minimize ρ)
Real engineering: High-power cables in real life use multiple thick copper strands for the same reason!
Resistivity (ρ): The Material Property
Resistivity is an intrinsic property of a material - doesn’t depend on shape or size.
$$\boxed{\rho = R \frac{A}{L}}$$Unit: Ohm-meter (Ω·m)
Typical Values (at 20°C)
| Material | Resistivity (Ω·m) | Type |
|---|---|---|
| Silver | $1.6 \times 10^{-8}$ | Best conductor |
| Copper | $1.7 \times 10^{-8}$ | Excellent conductor |
| Gold | $2.4 \times 10^{-8}$ | Good conductor |
| Aluminum | $2.8 \times 10^{-8}$ | Good conductor |
| Tungsten | $5.6 \times 10^{-8}$ | Used in bulbs |
| Iron | $10 \times 10^{-8}$ | Fair conductor |
| Nichrome | $100 \times 10^{-8}$ | Used in heaters |
| Glass | $10^{10} - 10^{14}$ | Insulator |
| Rubber | $10^{13} - 10^{16}$ | Insulator |
Pattern recognition:
- Conductors: $\rho \sim 10^{-8}$ Ω·m
- Semiconductors: $\rho \sim 10^{-3}$ to $10^{3}$ Ω·m
- Insulators: $\rho > 10^{8}$ Ω·m
Silver has the lowest resistivity, so why do we use copper?
Answer: COST!
- Silver: Expensive, used only in critical applications (high-frequency circuits, aerospace)
- Copper: 100x cheaper, almost as good (1.7 vs 1.6 × 10⁻⁸)
- Aluminum: Even cheaper, lighter (used in power transmission lines despite higher ρ)
- Gold: Doesn’t corrode (used in connectors, not for bulk wiring)
Conductivity (σ): The Inverse
Conductivity measures how well a material conducts electricity.
$$\boxed{\sigma = \frac{1}{\rho}}$$Unit: Siemens per meter (S/m) or (Ω·m)⁻¹
In Ohm’s law:
$$J = \sigma E$$- High σ (low ρ) → Good conductor
- Low σ (high ρ) → Poor conductor (insulator)
Temperature Dependence of Resistance
For Metals (Positive Temperature Coefficient)
$$\boxed{R_T = R_0(1 + \alpha \Delta T)}$$ $$R_T = R_0[1 + \alpha(T - T_0)]$$where:
- $R_T$ = resistance at temperature T
- $R_0$ = resistance at reference temperature $T_0$ (usually 0°C or 20°C)
- $\alpha$ = temperature coefficient of resistance (per °C or per K)
- $\Delta T = T - T_0$
For metals: $\alpha > 0$ (resistance INCREASES with temperature)
Why Does This Happen?
Microscopic explanation:
Higher temperature → Atoms vibrate more → More collisions for electrons → Higher resistance
Think of it like:
- Cold wire: Atoms are calm, electrons drift smoothly
- Hot wire: Atoms shake violently, electrons keep bumping into them!
Temperature Coefficient (α)
| Material | α (per °C) | Behavior |
|---|---|---|
| Silver | $3.8 \times 10^{-3}$ | R increases with T |
| Copper | $3.9 \times 10^{-3}$ | R increases with T |
| Aluminum | $3.9 \times 10^{-3}$ | R increases with T |
| Tungsten | $4.5 \times 10^{-3}$ | R increases with T |
| Nichrome | $0.4 \times 10^{-3}$ | Nearly constant |
| Manganin | $0.002 \times 10^{-3}$ | Nearly constant |
| Semiconductors | Negative | R decreases with T |
| Electrolytes | Negative | R decreases with T |
Question: Why does a bulb often burn out when you switch it ON, not when it’s already glowing?
Answer:
- When OFF (cold): $R_{cold}$ is LOW
- When ON (hot, ~2500°C): $R_{hot}$ is HIGH (can be 10-15 times higher!)
- At switch-on: Huge surge current ($I = V/R_{cold}$) flows
- This surge can break the filament!
This is why: The I-V curve of a bulb is non-linear and bends upward!
For Semiconductors (Negative Temperature Coefficient)
$$R_T = R_0 e^{-\beta \Delta T}$$For small ΔT:
$$R_T \approx R_0(1 - \alpha \Delta T)$$where $\alpha < 0$ (negative)
Semiconductors: Higher temperature → More free electrons → LOWER resistance
Used in:
- Thermistors (temperature sensors)
- Temperature compensation circuits
Special Cases and Applications
1. Wire Stretching Problems (JEE Favorite!)
When a wire is stretched, volume remains constant.
Original: Length $L$, Area $A$, Resistance $R$
After stretching to length $nL$:
Volume constant: $AL = A'L'$
If $L' = nL$, then $A' = \frac{A}{n}$
New resistance:
$$R' = \rho \frac{L'}{A'} = \rho \frac{nL}{A/n} = \rho \frac{n^2 L}{A} = n^2 R$$ $$\boxed{R' = n^2 R}$$Key formula: If length becomes $n$ times, resistance becomes $n^2$ times!
2. Combining Same Wire into n Parts
Wire cut into n equal parts:
Each part: $R_{part} = \frac{R}{n}$
If all n parts connected in parallel:
$$\frac{1}{R_{eq}} = n \times \frac{1}{R/n} = \frac{n^2}{R}$$ $$\boxed{R_{eq} = \frac{R}{n^2}}$$3. Cylindrical Shell
For a hollow cylinder (tube):
$$R = \rho \frac{L}{\pi(r_2^2 - r_1^2)}$$where $r_2$ = outer radius, $r_1$ = inner radius
Conductance (G): The Inverse of Resistance
$$\boxed{G = \frac{1}{R} = \frac{I}{V}}$$Unit: Siemens (S) or mho (℧)
From $R = \rho \frac{L}{A}$:
$$G = \frac{1}{R} = \frac{\sigma A}{L}$$Uses:
- Parallel circuit calculations become easier (add conductances directly!)
- $G_{eq} = G_1 + G_2 + G_3$ (in parallel)
Memory Tricks & Patterns
Mnemonic for Resistance Formula
“Really Long And Pointless” → $R = \rho \frac{L}{A}$
Or: “Rho Loves Africa” → $R = \rho L / A$
Temperature Coefficient Memory
“Metals Get MAD when heated” → $\alpha$ is positive, R increases
“Semiconductors Stay COOL” → $\alpha$ is negative, R decreases
Wire Stretching
“Stretch to N, Resistance to N-squared”
$L \to nL$, then $R \to n^2R$
JEE Pattern Recognition
- If ρ is given: Problem involves different materials
- If wire is stretched: Use $R' = n^2 R$
- If temperature mentioned: Use $R_T = R_0(1 + \alpha \Delta T)$
- If wire is cut and rearranged: Use $R = \rho L/A$ for each part
When to Use This
Given geometry (L, A) and material (ρ):
- Use $R = \rho L/A$
Given R at one temperature, find at another:
- Use $R_T = R_0(1 + \alpha \Delta T)$
Wire is stretched/compressed:
- Use volume constant: $AL = A'L'$
- Then $R' = n^2 R$ if stretched to $nL$
Comparing resistances of different wires:
- Use $\frac{R_1}{R_2} = \frac{\rho_1}{\rho_2} \times \frac{L_1}{L_2} \times \frac{A_2}{A_1}$
For alloys (Manganin, Nichrome, Constantan):
- Nearly constant R with temperature
- Used in standard resistors
Common Mistakes to Avoid
Wrong: “Resistivity depends on length and area”
Correct:
- Resistivity (ρ): Material property, independent of dimensions
- Resistance (R): Depends on both material (ρ) AND dimensions (L, A)
JEE Trap: “Find resistivity of a 2m copper wire” - resistivity is SAME for all copper, regardless of length!
Wrong: “Wire stretched to 2L, so R becomes 2R”
Correct: Must account for BOTH length increase AND area decrease!
- $L' = 2L$ → Factor of 2
- $A' = A/2$ → Factor of 2 (in denominator)
- Combined: $R' = 2 \times 2 \times R = 4R$
Formula: $R' = n^2 R$ where $n$ is stretching factor
Wrong: “For all materials, $R_T = R_0(1 + \alpha \Delta T)$”
Correct:
- Metals: Use $+\alpha$ (α positive)
- Semiconductors: Use $-\alpha$ (α negative) OR different formula!
Common error: Using metal formula for semiconductors!
Common mistake: Mixing cm and m, mm² and m²
Always convert to SI:
- Length → meters (m)
- Area → m²
- Resistivity → Ω·m
Example: Area = 2 mm² = $2 \times 10^{-6}$ m² (NOT $2 \times 10^{-3}$!)
Practice Problems
Level 1: Foundation (NCERT/Basic)
A copper wire has length 10 m and cross-sectional area 2 mm². If resistivity of copper is $1.7 \times 10^{-8}$ Ω·m, find its resistance.
Solution:
Given:
- $L = 10$ m
- $A = 2 \text{ mm}^2 = 2 \times 10^{-6}$ m²
- $\rho = 1.7 \times 10^{-8}$ Ω·m
Answer: $R = 0.085$ Ω = 85 mΩ
Insight: Even 10 m of copper wire has very small resistance!
A wire has resistance 20 Ω at 0°C. Find its resistance at 100°C if temperature coefficient is $4 \times 10^{-3}$ per °C.
Solution:
$$R_T = R_0(1 + \alpha \Delta T)$$ $$R_{100} = 20[1 + 4 \times 10^{-3} \times (100 - 0)]$$ $$R_{100} = 20[1 + 0.4] = 20 \times 1.4 = 28 \text{ Ω}$$Answer: $R_{100} = 28$ Ω
Concept: Resistance increased by 40% for 100°C rise!
Two wires of same material have equal lengths but radii in ratio 2:1. What is the ratio of their resistances?
Solution:
$$R = \rho \frac{L}{A} = \rho \frac{L}{\pi r^2}$$Same material (ρ), same length (L):
$$\frac{R_1}{R_2} = \frac{r_2^2}{r_1^2}$$Given $\frac{r_1}{r_2} = \frac{2}{1}$:
$$\frac{R_1}{R_2} = \frac{1^2}{2^2} = \frac{1}{4}$$Answer: $R_1 : R_2 = 1:4$
Key point: Resistance inversely proportional to square of radius!
Level 2: JEE Main
A wire of resistance 100 Ω is stretched to twice its original length. What is its new resistance?
Solution:
Method 1: Direct formula
Wire stretched to $n = 2$ times:
$$R' = n^2 R = 2^2 \times 100 = 400 \text{ Ω}$$Method 2: From first principles
Original: $L$, $A$, $R = 100$ Ω
Stretched: $L' = 2L$
Volume constant: $AL = A'L'$
$$A' = \frac{AL}{L'} = \frac{AL}{2L} = \frac{A}{2}$$ $$R' = \rho \frac{L'}{A'} = \rho \frac{2L}{A/2} = 4 \rho \frac{L}{A} = 4R = 400 \text{ Ω}$$Answer: $R' = 400$ Ω
JEE Shortcut: Remember $R' = n^2 R$ - saves 30 seconds in exam!
A wire of resistance R is cut into 5 equal parts. These parts are connected in parallel. What is the equivalent resistance?
Solution:
Each part: $R_{part} = \frac{R}{5}$
5 parts in parallel:
$$\frac{1}{R_{eq}} = \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5}$$ $$\frac{1}{R_{eq}} = 5 \times \frac{5}{R} = \frac{25}{R}$$ $$R_{eq} = \frac{R}{25}$$Quick formula: Cut into $n$ parts, connected in parallel → $R_{eq} = \frac{R}{n^2}$
Answer: $R_{eq} = \frac{R}{25}$
Resistance of a metal wire at 0°C is 100 Ω. At what temperature will it be 120 Ω? (α = $4 \times 10^{-3}$ per °C)
Solution:
$$R_T = R_0(1 + \alpha \Delta T)$$ $$120 = 100[1 + 4 \times 10^{-3} \times (T - 0)]$$ $$1.2 = 1 + 4 \times 10^{-3} \times T$$ $$0.2 = 4 \times 10^{-3} \times T$$ $$T = \frac{0.2}{4 \times 10^{-3}} = \frac{0.2}{0.004} = 50°\text{C}$$Answer: $T = 50°$C
Check: 20% increase needs 50°C rise (since α = 0.004 = 0.4% per °C)
Level 3: JEE Advanced
A wire of resistance R is stretched uniformly so that its radius becomes half. What is the ratio of new resistance to original resistance?
Solution:
If radius becomes half: $r' = \frac{r}{2}$
Area: $A' = \pi (r')^2 = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4} = \frac{A}{4}$
Volume constant: $AL = A'L'$
$$L' = \frac{AL}{A'} = \frac{AL}{A/4} = 4L$$New resistance:
$$R' = \rho \frac{L'}{A'} = \rho \frac{4L}{A/4} = 16 \rho \frac{L}{A} = 16R$$Answer: $\frac{R'}{R} = 16$
Pattern: If radius → r/2, then R → 16R
- Area reduces by 4
- Length increases by 4
- Combined effect: 4 × 4 = 16
Alternative: $n = L'/L = 4$, so $R' = n^2 R = 16R$
Two wires A and B of same material have their lengths in ratio 1:2 and radii in ratio 2:3. They are connected in series. Find the ratio of potential drops across them.
Solution:
$$R = \rho \frac{L}{\pi r^2}$$ $$\frac{R_A}{R_B} = \frac{L_A}{L_B} \times \frac{r_B^2}{r_A^2} = \frac{1}{2} \times \frac{3^2}{2^2} = \frac{1}{2} \times \frac{9}{4} = \frac{9}{8}$$In series, same current flows through both.
Potential drop: $V = IR$
$$\frac{V_A}{V_B} = \frac{R_A}{R_B} = \frac{9}{8}$$Answer: $V_A : V_B = 9:8$
Key concept: In series, voltage divides in ratio of resistances.
A metal rod of length L and cross-section A has resistivity varying with distance x from one end as $\rho(x) = \rho_0(1 + \frac{x}{L})$. Find the resistance of the rod.
Solution:
For non-uniform resistivity, integrate!
Consider small element at distance x of length dx:
$$dR = \rho(x) \frac{dx}{A} = \frac{\rho_0}{A}\left(1 + \frac{x}{L}\right)dx$$Total resistance:
$$R = \int_0^L dR = \int_0^L \frac{\rho_0}{A}\left(1 + \frac{x}{L}\right)dx$$ $$R = \frac{\rho_0}{A} \int_0^L \left(1 + \frac{x}{L}\right)dx$$ $$R = \frac{\rho_0}{A} \left[x + \frac{x^2}{2L}\right]_0^L$$ $$R = \frac{\rho_0}{A} \left[L + \frac{L^2}{2L}\right] = \frac{\rho_0}{A}\left[L + \frac{L}{2}\right]$$ $$R = \frac{\rho_0 L}{A} \times \frac{3}{2}$$Answer:
$$\boxed{R = \frac{3\rho_0 L}{2A}}$$Advanced concept: For non-uniform materials, use calculus!
Sanity check: If ρ were constant at ρ₀, we’d get $R = \frac{\rho_0 L}{A}$. Our answer is 1.5 times this, which makes sense since ρ increases along the rod!
Quick Revision Box
| Situation | Formula/Approach |
|---|---|
| Basic resistance formula | $R = \rho \frac{L}{A}$ |
| Temperature dependence (metal) | $R_T = R_0(1 + \alpha \Delta T)$ |
| Wire stretched to $nL$ | $R_{new} = n^2 R$ |
| Wire cut into $n$ parts, parallel | $R_{eq} = \frac{R}{n^2}$ |
| Conductivity | $\sigma = \frac{1}{\rho}$ |
| Conductance | $G = \frac{1}{R}$ |
| Radius halved | $R$ becomes 16 times |
| Comparing two wires | $\frac{R_1}{R_2} = \frac{\rho_1}{\rho_2} \cdot \frac{L_1}{L_2} \cdot \frac{A_2}{A_1}$ |
JEE Strategy: High-Yield Points
Wire stretching problems (EVERY YEAR!)
- Formula: $R' = n^2 R$ where wire stretched to $n$ times length
- Practice 10-15 variations before exam
Temperature dependence
- $R_T = R_0(1 + \alpha \Delta T)$ for metals
- Don’t forget: α positive for metals, negative for semiconductors
Comparing resistances
- Two wires with different L, A, ρ
- Use ratio method to avoid calculation errors
Material properties
- Memorize: Copper ρ ≈ $1.7 \times 10^{-8}$ Ω·m
- Know order: Silver < Copper < Gold < Aluminum
Tricky scenarios
- Wire cut and rearranged
- Non-uniform cross-section or resistivity
- Hollow cylinders
Time-saving tricks:
- For stretched wire, directly use $R' = n^2 R$
- For cut wire in parallel, use $R_{eq} = R/n^2$
- Always check if volume constant mentioned (stretching problems)
Related Topics
Within Current Electricity
- Ohm’s Law - Foundation for understanding resistance
- Drift Velocity - Microscopic explanation of resistivity
- Electrical Power - Energy dissipation in resistors
- Kirchhoff’s Laws - Analyzing resistor networks
Connected Chapters
- Thermal Physics - Temperature effects on resistance
- Atomic Structure - Why different materials have different ρ
- Semiconductors - Negative temperature coefficient explained
Math Connections
- Proportionality - $R \propto L$, $R \propto 1/A$
- Integration - For non-uniform resistivity problems
- Logarithms - For exponential temperature dependence
Teacher’s Summary
Resistance formula $R = \rho L/A$ combines material property (ρ) with geometry (L, A) - longer and thinner means higher R!
Resistivity (ρ) is intrinsic to material, resistance (R) depends on dimensions - don’t confuse the two!
Wire stretching: $R' = n^2 R$ when length becomes $nL$ - most common JEE problem type in this topic!
Temperature dependence: Metals have positive α (R increases), semiconductors have negative α (R decreases) - explains filament bulb behavior!
Know your materials: Silver < Copper < Gold < Aluminum in conductivity. Nichrome and Manganin have nearly constant R with temperature.
“Resistance is like traffic - longer roads (L ↑), narrower lanes (A ↓), or difficult terrain (ρ ↑) all slow down the flow. Master this, and circuit problems become easy!”