Resistance of a material is not constant — it changes with temperature. This property plays a crucial role in designing electrical circuits and components. In this blog, we’ll understand how resistance changes with temperature, what the temperature coefficient of resistance is, and how to solve numerical problems based on this concept.
Table of Contents
- Effect of Temperature on Resistance
- Temperature Coefficient of Resistance (α)
- Resistance-Temperature Formulas
- Solved Numericals
- FAQs
What is Temperature Coefficient of Resistance (α)?
The Temperature Coefficient of Resistance (α) is a numerical value that tells us how much the resistance of a material changes with a change in temperature. In simple words, it indicates how sensitive a material’s resistance is to temperature changes.
Definition:
The temperature coefficient of resistance (α) is defined as the increase in resistance per ohm of original resistance per °C rise in temperature.
Formula:
If a conductor has:
- Resistance = R0 at 0°C
- Resistance = Rt at t°C
α0 = (Rt − R0) / (R0 × t)
This equation tells us how much the resistance increases for each degree Celsius rise in temperature, compared to the original resistance.
Alternate Formula to Find Resistance at Any Temperature:
Rt = R0 (1 + α0 × t)
Where:
- Rt = Resistance at temperature t°C
- R0 = Resistance at 0°C
- α0 = Temperature Coefficient at 0°C
Types of Temperature Coefficients:
- Positive Temperature Coefficient (PTC): Resistance increases with temperature.
Example: Metals like copper, aluminum. - Negative Temperature Coefficient (NTC): Resistance decreases with temperature.
Example: Semiconductors like silicon, thermistors. - Zero Temperature Coefficient: Resistance remains almost constant over a wide range.
Example: Alloys like Manganin, Constantan.
Why is α Important?
Knowing the temperature coefficient is important for:
- Designing precision resistors and measuring instruments
- Understanding behavior of coils, transformers, and motors under heating
- Predicting failures or efficiency drops due to overheating
Typical Values of Temperature Coefficient (α) at 0°C:
| Material | α at 0°C (/°C) |
|---|---|
| Copper | 0.00426 |
| Aluminum | 0.00429 |
| Iron | 0.0065 |
| Manganin | ~0.00002 |
| Silicon (Semiconductor) | Negative |
Quick Summary:
- Temperature coefficient α tells how resistance changes with temperature.
- Positive α → resistance increases with heat (metals).
- Negative α → resistance decreases with heat (semiconductors).
- Used to design safe and efficient electrical systems.
Resistance of materials changes with temperature based on the nature of the material:
- Pure Metals (e.g., copper, aluminum): Resistance increases with temperature. These have a positive temperature coefficient.
- Semiconductors and Insulators (e.g., silicon, rubber): Resistance decreases with temperature. These have a negative temperature coefficient.
- Alloys (e.g., Constantan, Manganin): Resistance changes very slightly and irregularly, often considered stable over a range of temperatures.
Resistance-Temperature Formulas
- Rt = R0(1 + α0t)
- α1 = α0 / (1 + α0t1)
- If resistance changes from R1 to R2 with temperature t1 to t2:
R2 = R1(1 + α1(t2 − t1))
Solved Numericals on Temperature and Resistance
Example 1:
Question: The shunt winding of a motor has a resistance of 35.1 Ω at 20°C. Find its resistance at 32.6°C. Given: α0 = 0.00427/°C
Solution:
Rt = R0(1 + α0t) = 35.1 × (1 + 0.00427 × (32.6 − 20))
= 35.1 × (1 + 0.00427 × 12.6) = 35.1 × 1.0538 ≈ 39.6 Ω
Example 2:
Question: The resistance of a coil increases from 40 Ω at 10°C to 48.25 Ω at 60°C. Find the temperature coefficient of the material.
Solution:
α0 = (R2 − R1) / (R1 × (t2 − t1)) = (48.25 − 40) / (40 × (60 − 10))
= 8.25 / 2000 = 0.004125/°C ≈ 0.0043/°C
Example 3:
Question: A copper coil has resistance 4 Ω at 22°C. After operation, its current is 42 A at 210 V. Find its average temperature.
Solution:
R = V/I = 210/42 = 5 Ω
Now use:
5 = 4 × (1 + 0.00427 × t)
1.25 = 1 + 0.00427 × t → 0.25 = 0.00427 × t → t ≈ 58.5°C
Final temp = 22 + 58.5 = 80.5°C (rounded: 86.1°C using exact data)
High Order Thinking Questions
1. A nichrome heater is operated at 1500°C. What is the percentage increase in its resistance over that at room temperature (20°C)?
Given:
T1 = 20°C, T2 = 1500°C
α = 0.00016 /°C
Formula:
% Increase = α × (T2 - T1) × 100
Calculation:
ΔT = 1500 - 20 = 1480°C
% Increase = 0.00016 × 1480 × 100 = 23.68%
Answer: 23.6%
2. Two wires A and B are connected in series at 0°C. Resistance of B is 3.5 times that of A. The resistance temperature coefficient of A is 0.4% and that of the combination is 0.1%. Find the resistance temperature coefficient of B.
Let:
RA = R, RB = 3.5R
αA = 0.004, αcomb = 0.001
αB = ?
Formula:
αeq = (RAαA + RBαB) / (RA + RB)
Substitute:
0.001 = (R × 0.004 + 3.5R × Î±B) / (R + 3.5R)
0.001 = (0.004 + 3.5αB) / 4.5
Solve:
0.0045 = 0.004 + 3.5αB
3.5αB = 0.0005
αB = 0.000143 or 0.0143%
3. A DC shunt motor takes a field current of 1.6 A at 400 V. If the temperature rise is 40°C, what extra resistance is needed to get the same field current when starting at 20°C? (α = 0.0043/°C)
Given:
V = 400 V, I = 1.6 A, α = 0.0043, Trise = 40°C
Step 1: Calculate hot resistance:
Rhot = V / I = 400 / 1.6 = 250 Ω
Step 2: Find cold resistance:
Rcold = R / (1 + α × Î”T) = 250 / (1 + 0.0043 × 40) = 250 / 1.172 = 213.31 Ω
Step 3: Extra resistance needed =
Rextra = Rhot - Rcold = 250 - 213.31 = 36.69 Ω
4. A potential difference of 250 V is applied to a copper field coil at 15°C. The current is 5 A. What will be the mean temperature of the coil when the current drops to 3.91 A? (Voltage same)
Given:
V = 250 V, I1 = 5 A, I2 = 3.91 A, T1 = 15°C, α = 0.00428/°C
Step 1: Find initial resistance:
R1 = V / I1 = 250 / 5 = 50 Ω
R2 = 250 / 3.91 = 63.94 Ω
Step 2: Use formula:
R2 = R1[1 + α × (T - T1)]
63.94 = 50 × [1 + 0.00428 × (T - 15)]
Solve:
1.2788 = 1 + 0.00428(T - 15)
0.2788 = 0.00428(T - 15)
T - 15 = 65.15
T = 80.15°C ≈ 85°C
5. An insulating material has insulation resistance of 100% at 0°C. For every 5°C rise, resistance drops by 10%. At what temperature is resistance halved?
Let: Resistance halves = 50%
Every 5°C → 90% of previous
Let n = number of 5°C steps:
(0.9)n = 0.5
Take log:
log(0.9)n = log(0.5)
n log(0.9) = log(0.5)
n = log(0.5)/log(0.9) ≈ -0.3010 / -0.0458 ≈ 6.57
Temperature = 6.57 × 5 = 32.85°C ≈ 33°C
6. A carbon electrode has a resistance of 0.125 Ω at 20°C. The temperature coefficient of carbon is -0.0005/°C. What is the resistance at 85°C?
Given:
R0 = 0.125 Ω, T = 85°C, T0 = 20°C, α = -0.0005
Formula:
R = R0[1 + α(T - T0)]
Calculation:
R = 0.125 × [1 + (-0.0005 × 65)]
= 0.125 × (1 - 0.0325) = 0.125 × 0.9675 = 0.1209 Ω
Frequently Asked Questions (FAQs)
Q1. Why does resistance increase with temperature in metals?
In metals, atoms vibrate more with increased temperature, obstructing the flow of electrons and increasing resistance.
Q2. What is a negative temperature coefficient of resistance?
When resistance decreases with an increase in temperature, the material has a negative temperature coefficient — commonly seen in semiconductors.
Q3. Do all conductors behave the same with temperature?
No. Metals, alloys, and semiconductors behave differently. Metals generally show a linear increase in resistance, while semiconductors show a decrease.
Conclusion
Understanding the effect of temperature on resistance is essential for every electrical engineer or science student. It helps in selecting proper materials for specific applications like resistors, coils, heating elements, and semiconductors.
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