---
title: "NTC thermistor characteristics and resistance"
subject: "Instrumentation and Measurement"
chapter: 5
key: "semester_4/instrumentation_and_measurement:ntc-thermistor-resistance-numerical"
source: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/ntc-thermistor-resistance-numerical
site: BCT Engineering
---

# NTC thermistor characteristics and resistance

Instrumentation and Measurement · Chapter 5: Sensors

- Asked in: year not given, 4+4 marks
- Video: [Thermistors - NTC & PTC - Thermal Resistors - Temperature Sensors & Resettable Fuses](https://www.youtube.com/watch?v=nXmkkyw8v5A) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/ntc-thermistor-resistance-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_and_measurement/chapter-5

## Question

(a) Explain the working of thermistors. Differentiate between NTC and PTC types and state the advantages and limitations of thermistors.

(b) An NTC thermistor has a resistance of 10 k$\Omega$ at 25 $^\circ$C and a material constant $\beta = 3950$ K. Find its resistance at 50 $^\circ$C and 100 $^\circ$C and its temperature coefficient of resistance at 50 $^\circ$C.

## Short answer

NTC thermistors are semiconductor resistors with negative temperature coefficient; PTC types have sharp positive rise. High sensitivity, small size, nonlinear. Resistance: R = R₀ exp[β(1/T - 1/T₀)]; at 50°C R = 3.59 kΩ, at 100°C R = 697.5 Ω. Temperature coefficient α = -β/T² = -0.0378 K⁻¹ (−3.78%/°C) at 50°C.

## Answer

#### (a) Thermistors
A thermistor (thermal resistor) is a semiconductor (sintered metal oxides of Mn, Ni, Co, Cu) whose resistance changes strongly with temperature. Its sensing element is a bead, disc or rod.

For an NTC thermistor, the resistance falls as temperature rises:

$$
R = R_0\exp\left[\beta\left(\frac{1}{T} - \frac{1}{T_0}\right)\right]
$$

where $T$ and $T_0$ are in kelvin. The temperature coefficient is $\alpha = \dfrac{1}{R}\dfrac{dR}{dT} = -\dfrac{\beta}{T^2}$.

| Point | NTC | PTC |
|---|---|---|
| Coefficient | Negative | Positive (sharp rise above a switching temperature) |
| Material | Metal oxides | Doped barium titanate |
| Use | Temperature sensing, compensation | Overcurrent protection, self-regulating heaters |

**Advantages:** high sensitivity (about 4%/K), small size and fast response, high resistance so lead resistance is negligible.

**Limitations:** strongly nonlinear, small range (up to about 300 $^\circ$C), interchangeability poor, self-heating error.

#### (b) Numerical
Given $R_0 = 10$ k$\Omega$ at $T_0 = 298.15$ K, $\beta = 3950$ K.

At 50 $^\circ$C ($T = 323.15$ K):

$$
R_{50} = 10{,}000\exp\left[3950\left(\frac{1}{323.15} - \frac{1}{298.15}\right)\right] = 10{,}000\,e^{-1.0249} = 3588\ \Omega
$$

At 100 $^\circ$C ($T = 373.15$ K):

$$
R_{100} = 10{,}000\exp\left[3950\left(\frac{1}{373.15} - \frac{1}{298.15}\right)\right] = 10{,}000\,e^{-2.6628} = 697.5\ \Omega
$$

Temperature coefficient at 50 $^\circ$C:

$$
\alpha = -\frac{3950}{323.15^2} = -0.0378\ \text{K}^{-1} = -3.78\%/\text{K}
$$

**Answer: $R_{50} = 3.59$ k$\Omega$, $R_{100} = 697.5\ \Omega$, $\alpha_{50} = -3.78\%/^\circ$C.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For an NTC thermistor as temperature rises, resistance
   A. Rises
   B. Falls
   C. Stays constant
   D. Rises sharply at a switching point
   Answer: B. NTC has a negative temperature coefficient. (Key point: NTC behaviour)

2. Which formula gives NTC resistance?
   A. R = R0 exp[β(1/T - 1/T0)] with T in kelvin
   B. R = R0 exp[β(T - T0)] with T in °C
   C. R = R0 (1 + βT)
   D. R = R0 β/T
   Answer: A. Temperatures must be in kelvin. (Key point: Resistance formula)

3. Resistance at 50 °C is about
   A. 3.59 kΩ
   B. 6.9 kΩ
   C. 697 Ω
   D. 10 kΩ
   Answer: A. 10000·e^-1.0249 = 3588 Ω. (Key point: Resistance at 50 °C)

4. The temperature coefficient α = -β/T^2 at 50 °C is
   A. -0.0378 per K (-3.78%/K)
   B. -0.395 per K
   C. +0.0378 per K
   D. -0.00378 per K
   Answer: A. 3950/323.15^2 = 0.0378. (Key point: Temperature coefficient)

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