Q.An electric toaster uses nichrome for its heating element. When a negligibly small current passes through it, its resistance at room temperature (27.0 ∘C) is found to be 75.3 Ω. When the toaster is connected to a 230 V supply, the current settles, after a few seconds, to a steady value of 2.68 A. What is the steady temperature of the nichrome element? The temperature coefficient of resistance of nichrome averaged over the temperature range involved, is 1.70×10−4 ∘C−1.
Concept understanding — Temperature Dependence of Resistance
Temperature Dependence of Resistance
Imagine you're trying to walk through a crowded market. When the market is cool and calm, people move slowly and you can weave through easily. Now imagine the same market on a hot, chaotic day — everyone is jostling, moving faster, bumping into each other. Getting from one end to the other becomes much harder.
That's exactly what happens inside a metal wire when you heat it up.
The Intuition
In a metal, electric current is carried by free electrons drifting through a fixed lattice of positive ions. At room temperature, these ions are vibrating slightly around their positions. When you heat the metal, the ions vibrate more vigorously — they shake faster and with larger amplitude.
Think of the vibrating ions as a row of swinging doors. At low temperature, the doors barely move, so electrons slip through easily. At high temperature, the doors swing wildly, and electrons get knocked off course constantly. Each collision with a vibrating ion scatters the electron, making it harder for the current to flow.
The result: resistance increases as temperature increases — for most conductors.
The Precise Statement
For a metallic conductor over a moderate temperature range (not too close to absolute zero), the resistance changes linearly with temperature:
R(T)=R0[1+α(T−T0)]
Where:
- R(T) is the resistance at temperature T
- R0 is the resistance at a reference temperature T0 (often 0∘C or 20∘C)
- α is the temperature coefficient of resistance (units: per °C or per K)
R=R0(1+αΔT)
The coefficient α tells you how sensitive the material is to temperature changes. For copper, α≈0.0039/∘C — meaning for every 1°C rise, resistance increases by about 0.39%.
What About Other Materials?
Not everything behaves like metals.
Semiconductors (like silicon, germanium) do the opposite: their resistance decreases sharply as temperature rises. Why? Because heating frees more electrons from their bonds, creating many more charge carriers. Even though the lattice vibrates more, the huge increase in available carriers overwhelms that effect, so resistance drops.
Insulators also show decreasing resistance with temperature, but the effect is much smaller than in semiconductors.
Alloys like constantan (copper-nickel) have a very small α — their resistance barely changes with temperature. This is useful for making precision resistors that stay stable.
Superconductors are a special case: below a critical temperature, resistance drops to exactly zero.
A common mistake is to think that all materials have higher resistance when hot. That's only true for pure metals. Semiconductors and insulators behave in the opposite way.
Why This Matters in Exams
You'll often be asked to:
- Calculate the new resistance after a temperature change using R=R0(1+αΔT)
- Find α from experimental data
- Explain why resistance changes — always mention increased lattice vibrations for metals, and increased carrier concentration for semiconductors
The key is to remember: for metals, heat makes ions shake more → more collisions → higher resistance. For semiconductors, heat breaks bonds → more free electrons → lower resistance.
That's the whole story in a nutshell. The formula is just a way to quantify what your intuition already tells you.
How resistance changes with temperature for conductors, semiconductors and alloys is covered in the NCERT Class 12 Physics chapter on current electricity, and comparing metals with semiconductors on this point is a common CBSE board and JEE Main question. Searches for "temperature coefficient of resistance formula class 12 physics" will find this lattice-vibration-versus-carrier-concentration explanation is the standard NCERT reasoning.
Why this formula?
Temperature Dependence of Resistance — Why the Formula Holds
Let’s build this from the ground up. The key formula you’ll see in exams is:
RT=R0(1+αT)
But why does resistance change with temperature? It’s not magic — it’s about what happens inside the wire.
1. What determines resistance?
Resistance R of a conductor depends on three things:
- Length L (longer → more resistance)
- Cross-sectional area A (thicker → less resistance)
- Resistivity ρ — a material property
The formula is:
R=ρAL
When temperature changes, L and A change very slightly (thermal expansion), but the big effect is on ρ.
2. Why does resistivity change with temperature?
Resistivity ρ depends on how easily electrons can move through the material.
- In metals: Atoms vibrate more as temperature rises. These vibrations scatter electrons, making it harder for them to flow. So ρ increases.
- In semiconductors: More electrons get enough energy to jump into the conduction band. So ρ decreases.
For most metals (and many conductors), the change is linear over a moderate temperature range.
3. Deriving the linear formula
Let ρ0 be resistivity at a reference temperature T0 (often 0∘C or 20∘C).
For a small change ΔT=T−T0, the change in resistivity is proportional to ΔT and to ρ0:
Δρ∝ρ0ΔT
Introduce the temperature coefficient of resistivity α:
Δρ=αρ0ΔT
So the new resistivity is:
ρ=ρ0+Δρ=ρ0(1+αΔT)
Now, since R=ρAL, and L and A change negligibly (for small ΔT), we get:
R=ρAL=ρ0(1+αΔT)AL=R0(1+αΔT)
That’s the formula:
RT=R0(1+αΔT)
Where:
- RT = resistance at temperature T
- R0 = resistance at reference temperature T0
- α = temperature coefficient of resistance (unit: ∘C−1 or K−1)
- ΔT=T−T0
4. Important exam notes
- α is positive for metals (resistance increases with temperature).
- α is negative for semiconductors (resistance decreases).
- The formula is linear approximation — valid only for moderate temperature ranges (not near melting point or absolute zero).
- For very precise work, a quadratic term is sometimes added: R=R0(1+αT+βT2).
5. Quick intuition check
Think of a light bulb filament (tungsten):
- When cold, resistance is low → large current flows.
- As it heats up, resistance rises → current stabilises.
- That’s why bulbs often blow when first switched on (cold resistance is much lower).
Bottom line: The formula comes from the fact that resistivity changes linearly with temperature for most conductors, and the geometric changes (L, A) are negligible. The coefficient α captures how strongly the material’s atomic vibrations impede electron flow.
Concept: Temperature Dependence of Resistance — resistance changes linearly with temperature: R=R0(1+αΔT).
- Hot resistance from Ohm's law: R=IV=2.68230≈85.82 Ω.
- R0R−1=75.385.82−75.3≈0.1397.
- ΔT=1.70×10−40.1397≈821.8 ∘C, so T=27.0+821.8≈848.8 ∘C.
The steady temperature of the nichrome element is ≈849 ∘C.
The hot resistance is R=IV=2.68230≈85.82 Ω. Using R=R0[1+α(T−T0)] with full precision throughout gives a steady element temperature of about 849 ∘C.
A metal's resistance rises with temperature. At room temperature the nichrome element has a known (cold) resistance R0; once it settles into steady glowing operation, it draws a fixed current from the 230 V supply, which fixes its (hot) resistance R. Comparing the two through the linear temperature law gives the operating temperature.
Given: T0=27.0 ∘C, R0=75.3 Ω, V=230 V, I=2.68 A, α=1.70×10−4 ∘C−1 (already an average over the temperature range involved, so the linear law can be used directly).
Step 1 — Hot resistance (Ohm's law):
R=IV=2.68230≈85.82 Ω.
Step 2 — Temperature dependence of resistance:
R=R0[1+α(T−T0)]⇒R0R−1=α(T−T0).
Step 3 — Substitute (keeping full precision):
R0R−1=75.385.82−75.3=75.310.52≈0.1397.
Step 4 — Solve for T:
T−T0=1.70×10−40.1397≈821.8 ∘C,T=27.0+821.8≈848.8 ∘C≈849 ∘C.
Rounding the hot resistance to 85.8 Ω before subtracting (instead of carrying the full 85.82 Ω) shifts the answer down to about 847 ∘C — a 2 ∘C difference purely from intermediate rounding. Carrying full precision through every step, as above, gives the more accurate ≈849 ∘C.
The steady temperature of the nichrome heating element is approximately 849 ∘C (about 8.5×102 ∘C).
Method: Using the Temperature Dependence of Resistance Formula
This is a direct application of the linear approximation for resistance variation with temperature, given by:
RT=R0[1+α(T−T0)]
Where:
- RT = resistance at final temperature T
- R0 = resistance at reference temperature T0
- α = temperature coefficient of resistance (assumed constant over the range)
Step 1: Find the steady-state operating resistance
When the toaster is connected to 230 V and draws 2.68 A, use Ohm’s law:
Rhot=IV=2.68230
Rhot=85.82 Ω(approximately)
Step 2: Identify the known quantities
- R0=75.3 Ω (at T0=27.0 ∘C)
- RT=85.82 Ω (at unknown T)
- α=1.70×10−4 ∘C−1
Step 3: Rearrange the formula to solve for T
From:
RT=R0[1+α(T−T0)]
Divide both sides by R0:
R0RT=1+α(T−T0)
Subtract 1:
R0RT−1=α(T−T0)
Divide by α:
T−T0=α1(R0RT−1)
Finally:
T=T0+α1(R0RT−1)
Step 4: Substitute and calculate
First compute the resistance ratio:
R0RT=75.385.82=1.1397
Then:
R0RT−1=0.1397
Now:
T−27.0=1.70×10−40.1397
T−27.0=821.8 ∘C
Step 5: Final answer
T=27.0+821.8=848.8 ∘C
Key exam tip: Always check that the temperature rise is reasonable — nichrome heating elements typically operate in the range 800 –1100 ∘C, so this answer is physically plausible.
Here are the most common mistakes students make on this problem, along with how to avoid each.
1. Using the Wrong Formula for Temperature Dependence
The Mistake:
Students often use the approximate linear formula RT=R0(1+αT) but forget that T must be the change in temperature (ΔT), not the final temperature itself. They plug in 27∘C directly into the formula incorrectly.
How to Avoid:
Always write the correct relation:
RT=R0[1+α(T−T0)]
Where:
- RT = resistance at final temperature T
- R0 = resistance at initial temperature T0
- α = temperature coefficient of resistance
Key point: The term (T−T0) is the temperature difference, not the absolute temperature.
2. Confusing R0 with the Operating Resistance
The Mistake:
Some students take R0=75.3 Ω as the resistance at the final steady temperature. They then try to solve for T using R0 as the unknown.
How to Avoid:
Clearly label:
- At room temperature (27∘C): R0=75.3 Ω
- At steady operating temperature (T): RT must be calculated from Ohm’s law using the given voltage and current.
3. Incorrectly Calculating RT from the Circuit
The Mistake:
Using R=V/I but with the wrong values — for example, using the initial current (which is negligibly small) instead of the steady current.
How to Avoid:
Use the steady-state values only:
RT=IV=2.68 A230 V
Calculate carefully:
RT≈85.82 Ω
4. Sign Errors in the Temperature Difference
The Mistake:
When rearranging the formula, students sometimes write T−T0=αR0RT−R0 but then forget that RT>R0 (since resistance increases with temperature for nichrome). They end up with a negative temperature difference.
How to Avoid:
Check physically: nichrome’s α is positive, so if the toaster gets hot, RT>R0. Therefore T−T0 must be positive. If your calculation gives a negative value, you have swapped RT and R0 or made a sign error.
5. Forgetting to Add T0 at the End
The Mistake:
After finding ΔT=T−T0, students stop and give ΔT as the final answer instead of the actual steady temperature T.
How to Avoid:
Always write the final step explicitly:
T=T0+ΔT
Here:
ΔT=αR0RT−R0=(1.70×10−4)×75.385.82−75.3
Calculate ΔT≈822∘C, then:
T=27∘C+822∘C=849∘C
Final answer: 849 ∘C (approximately).
Quick Checklist to Avoid All Mistakes
| Step | What to do |
|---|---|
| 1 | Identify R0 (at T0) and RT (at unknown T) |
| 2 | Compute RT=V/I using steady current |
| 3 | Use RT=R0[1+α(T−T0)] |
| 4 | Solve for T−T0, then add T0 |
| 5 | Check sign: ΔT>0 for heating |
By following this structured approach, you will avoid the common pitfalls and get the correct answer every time.
- KEAM 2026Set eng-2026-04184 marksMCQQ.The temperature coefficient of resistance of a coil of wire of resistance 4Ω at 30∘C and 6Ω at 70∘C (in per ∘C) is (A) 0.04 (B) 0.06 (C) 0.004 (D) 0.006 (E) 0.02
›Reveal solutionSolution
Using Rt=R0(1+αt) at the two temperatures and eliminating R0 gives α=0.02∘C−1.
With Rt=R0(1+αt):
R1=R0(1+αt1),R2=R0(1+αt2)
Eliminating R0:
α=R1t2−R2t1R2−R1
Substituting R1=4Ω at t1=30∘C, R2=6Ω at t2=70∘C:
α=4×70−6×306−4=280−1802=1002=0.02∘C−1
✓Final answerThe correct option is (E).
- KEAM 2026Set eng-2026-04204 marksMCQQ.A copper wire of temperature coefficient of resistance 4×10−3K−1 has resistance 10 Ω at 20∘ C. Its resistance at 80∘ C is (A) 1.22 Ω (B) 12.2 Ω (C) 24.4 Ω (D) 38.8 Ω (E) 2.44 Ω
›Reveal solutionSolution
Linear resistance–temperature law: R=R0(1+αΔT) gives 12.4 Ω.
Compute. ΔT=80−20=60∘C.
R80=10[1+(4×10−3)(60)]=10(1+0.24)=10×1.24=12.4 Ω.
✓Final answerThe correct option is (B).
- KEAM 2026Set eng-2026-04224 marksMCQQ.With increase in temperature, the resistivity of (A) both Cu and Ge increases (B) both nichrome and Ge decreases (C) nichrome decreases and that of Ge increases (D) both Cu and nichrome increases (E) Cu, nichrome and Ge increases
›Reveal solutionSolution
Metals and alloys have positive temperature coefficients; semiconductors (Ge) have negative.
With rising temperature: Cu (a metal) and nichrome (an alloy) both show increasing resistivity, while germanium (a semiconductor) shows decreasing resistivity.
Hence the correct statement is that both Cu and nichrome increase.
✓Final answerThe correct option is (D).
- KEAM 2025Set eng-2025-04264 marksMCQQ.The resistance of a wire at 30ºC and 40ºC are respectively 5 Ω and 6 Ω. The temperature coefficient of resistance of the material of the wire (in per degree Celcius) is (A) 0.04 (B) 0.05 (C) 0.02 (D) 0.03 (E) 0.01
›Reveal solutionSolution
The temperature coefficient of resistance is 0.05/∘C.
Using R=R0(1+αt) with R30=5Ω and R40=6Ω:
R30R40=1+30α1+40α=56.
Cross-multiplying:
5(1+40α)=6(1+30α) ⇒ 5+200α=6+180α ⇒ 20α=1,
α=0.05 per ∘C.
✓Final answerThe correct option is (B).
- KEAM 2025Set eng-2025-04264 marksMCQQ.A wire of 25 Ω resistance is cut into n pieces of equal length. If these pieces of wires are connected in parallel, their equivalent resistance is 1 Ω, then the value of n is (A) 3 (B) 6 (C) 8 (D) 5 (E) 4
›Reveal solutionSolution
Cutting the wire into n equal pieces makes each 25/n Ω; connecting n of them in parallel gives 25/n2=1 Ω, so n=5.
Resistance of each piece:
Rpiece=n25 Ω
Parallel of n equal resistors:
Req=nRpiece=n25/n=n225
Set equal to 1 Ω:
n225=1⟹n2=25⟹n=5
✓Final answerThe correct option is (D).
- KEAM 2025Set eng-2025-04274 marksMCQQ.The current carrying rail of a subway track is made of steel and has a cross-sectional area of about 20 cm2. The resistance of 2 km of the track is (in ohm) as a multiple of the specific resistance of steel, ρ is: (A) 102ρ (B) 103ρ (C) 104ρ (D) 105ρ (E) 106ρ
›Reveal solutionSolution
R=ρL/A with L=2000 m and A=20 cm2=2×10−3 m2 gives R=106ρ.
Resistance of a uniform conductor is
R=ρAL.
Convert units: L=2 km=2000 m and A=20 cm2=20×10−4 m2=2×10−3 m2. Then
R=ρ2×10−32000=ρ×106=106ρ.
✓Final answerThe correct option is (E).
- KEAM 2025Set eng-2025-04294 marksMCQQ.If both the length and area of cross-section of a linear conductor are halved, its resistance would (A) be doubled (B) remain unchanged (C) be halved (D) be tripled (E) be quadrupled
›Reveal solutionSolution
Resistance R=ρL/A. Halving the length and the area gives R′=ρ(L/2)/(A/2)=ρL/A=R — unchanged.
The resistance of a conductor is R=AρL. If both L and A are halved, R′=(A/2)ρ(L/2)=AρL=R. The two factors of 21 cancel, so the resistance remains unchanged.
✓Final answerThe correct option is (B).
- KEAM 2024Set eng-2024-06064 marksMCQQ.The resistance of a wire at 0 ∘C is 4 Ω. If the temperature coefficient of resistance of the material of the wire is 5×10−3/∘C, then the resistance of a wire at 50 ∘C is (A) 20 Ω (B) 10 Ω (C) 6 Ω (D) 8 Ω (E) 5 Ω
›Reveal solutionSolution
Resistance rises linearly with temperature: R=R0(1+αT).
Using R=R0(1+αT) with R0=4 Ω, α=5×10−3/∘C and T=50∘C:
R=4(1+5×10−3×50)=4(1+0.25)=4×1.25=5 Ω.
✓Final answerThe correct option is (E).
- KEAM 2024Set eng-2024-06074 marksMCQQ.Masses of three copper wires are in the ratio 1 : 3 : 5 and their lengths are in the ratio 5 : 3 : 1. Then the ratio of their electric resistances is (A) 125 : 15 : 1 (B) 5 : 3 : 1 (C) 1 : 25 : 125 (D) 1 : 3 : 5 (E) 5 : 21 : 25
›Reveal solutionSolution
Resistance of a wire of fixed material is R∝L2/m, giving 25:3:0.2=125:15:1.
Resistance R=AρL. The cross-sectional area is A=Lvolume=Lm/d, where d is the (common) density. Hence R=m/(dL)ρL=mρdL2, i.e. R∝mL2.
With masses 1:3:5 and lengths 5:3:1:
mL2=125:39:51=25:3:0.2.
Multiplying by 5: 125:15:1.
✓Final answerThe correct option is (A).
- KEAM 2024Set pha-2024-06104 marksMCQQ.Resistivity of a conductor increases with (A) increase in its length (B) decrease in its length (C) increase in its area of cross-section (D) decrease in its area of cross-section (E) increase in its temperature
›Reveal solutionSolution
Resistivity is a material property, independent of geometry.
Resistivity ρ depends on the material and temperature, not on length or cross-section (those affect resistance R=ρl/A, not ρ). For a metallic conductor, ρ increases with temperature due to increased lattice vibrations.
✓Final answerThe correct option is (E). Resistivity increases with increase in temperature.
- KEAM 2022Set eng-2022-P1-A14 marksMCQQ.The INCORRECT statement is (A) Resistivity of copper increases with increase of temperature (B) Resistivity of germanium decreases with the increase of temperature (C) Resistivity of semiconductors is higher than that of the conductors (D) Resistivity of nichrome shows a weak dependence with temperature (E) Resistivity of insulators is independent of temperature
›Reveal solutionSolution
The false statement is that insulators' resistivity is independent of temperature; it actually varies (decreasing) with temperature.
Concept and Intuition
Resistivity's temperature behaviour differs by material type: metals (copper, nichrome) increase with temperature; semiconductors (germanium) decrease with temperature; insulators are not temperature-independent — like semiconductors, their resistivity falls with rising temperature as more carriers become available.
Step-by-Step Solution
- (A) Copper is a metal: resistivity increases with temperature — correct.
- (B) Germanium is a semiconductor: resistivity decreases with temperature — correct.
- (C) Semiconductors have higher resistivity than conductors — correct.
- (D) Nichrome (alloy) has a weak temperature dependence — correct.
- (E) Insulators being temperature-independent is false; their resistivity changes with temperature.
Common Mistakes
- Assuming insulators simply do not conduct at all; their resistivity is very high but still temperature-dependent.
✓Final answerThe correct option is (E) — Resistivity of insulators is independent of temperature.
ANSWER: E
- KEAM 2022Set eng-2022-P1-A14 marksMCQQ.Material that is widely used to make wire bound standard resistors is (A) manganin (B) iron (C) copper (D) tungsten (E) germanium
›Reveal solutionSolution
Standard wire-wound resistors are made of manganin because of its very low temperature coefficient of resistance.
Concept and Intuition
A standard resistor must keep its resistance essentially constant with temperature and time. Manganin (Cu–Mn–Ni alloy) has an extremely small temperature coefficient of resistivity, making its resistance stable — ideal for precision standard resistors. Copper, iron and tungsten have significant temperature coefficients; germanium is a semiconductor.
Step-by-Step Solution
- Requirement: near-zero temperature coefficient for a stable standard.
- Manganin (and constantan) satisfy this; the others do not.
- Hence manganin is the standard choice.
Common Mistakes
- Choosing copper because it is a good conductor; low resistivity is not the criterion — temperature stability is.
✓Final answerThe correct option is (A) — manganin.
ANSWER: A
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