GATE 2025 IN Q.58: solution and answer

GATE 2025 · INQ.58 · 2 marks · NAT

The resistance of a thermistor is measured to be 2.25 kΩ at 30 °C and 1.17 kΩ at 60 °C. Its material constant β is ___________ K (rounded off to two decimal places).

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Answer and solution

ANSWER

The material constant is $β \approx 2201.43 K$ .

Official answer: 2160 to 2210Source: Official answer key, GATE 2025 IN (IIT Roorkee)

Use the NTC thermistor law with absolute temperatures: $R_{T} = R_{0} e^{β (\frac{1}{T} - \frac{1}{T _{0}})}$ .
Convert: $T_{1} = 30 + 273.15 = 303.15 K$ and $T_{2} = 60 + 273.15 = 333.15 K$ , with $R_{1} = 2.25 k Ω$ at $T_{1}$ and $R_{2} = 1.17 k Ω$ at $T_{2}$ .
Write the law for each temperature and divide: $\frac{R _{1}}{R _{2}} = e^{β (\frac{1}{T _{1}} - \frac{1}{T _{2}})}$ . The reference resistance cancels, so only the ratio of the two readings is needed.
Take natural logs: $β = \frac{ln ( R _{1} / R _{2} )}{\frac{1}{T _{1}} - \frac{1}{T _{2}}}$ .
Numerator: $ln (\frac{2.25}{1.17}) = ln 1.9231 = 0.65393$ .
Denominator: $\frac{1}{303.15} - \frac{1}{333.15} = \frac{30}{303.15 \times 333.15} = 2.9705 \times 1 0^{- 4} K^{- 1}$ .
$β = \frac{0.65393}{2.9705 \times 1 0 ^{- 4}} \approx 2201.43 K$

KEY IDEA

For an NTC thermistor, $R_{1} / R_{2} = e^{β (1/ T_{1} - 1/ T_{2})}$ with temperatures in kelvin, so $β = ln (R_{1} / R_{2}) / (1/ T_{1} - 1/ T_{2})$ . Two resistance readings are enough, and the reference resistance cancels.

WATCH OUT FOR

Using Celsius temperatures in the exponent instead of kelvin, or inverting the ratio so the sign of $β$ comes out wrong (the higher resistance belongs to the lower temperature). Using 273 instead of 273.15 shifts the answer by a few kelvin, which is why the key is a range.

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GATE 2025 IN Q.58: solution and answer

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