Measurements & Instrumentation · Problem type

Thermistor by the β equation

Use this for “A 10 kΩ NTC thermistor (at 25 °C, β = 3950 K) is at 50 °C. What is its resistance?”

Chart

R(T) = 10,000e3,950·(1/(T + 273.15) − 1/(25 + 273.15))010203040506005,00010,00015,00020,00025,000Temperature T (°C)Resistance R (Ω)(50, 3,588)R₀ = 10,000T = 25
R(50) ≈ 3,588
Level R₀ = 10,000 meets the curve at T = 25
This is an example. Type your own values to start a new problem.
R₀
Reference resistance
example · Ω
T₀
Reference temperature
example · °C
β
Beta constant
example · K
T
Temperature
example · °C
R
Resistance
calculated · Ω

Formulas

R
=
R₀
×
e^(
β
×
(
1
T
+ 273.15
−
1
T₀
+ 273.15
))
3,588.1826
=
10,000
×
e^(3,950
×
(
1
50 + 273.15
−
1
25 + 273.15
))

Assumptions

•
Temperatures go into the exponent in kelvin, T + 273.15.
•
β is fitted over the range used, so the curve is only as good as that fit.
•
An NTC thermistor: its resistance falls as it warms.

Step-by-step

WE KNOW
T = 50 °C, R₀ = 10,000 Ω, T₀ = 25 °C, β = 3,950 K
FIND
resistance (R)
Step 1 · Find resistance (R)
Use R = R₀ × e^(β × (1 ÷ (T + 273.15) − 1 ÷ (T₀ + 273.15)))
Turn
both
temperatures
into
kelvin;
warmer
than
T₀
makes
the
bracket
negative,
so
R
falls
below
R₀
.
R
=
10,000
×
e^(3,950
×
(
1
50 + 273.15
−
1
25 + 273.15
))
R
=
10,000
×
e
3,950 × (0.0030945381 − 0.0033540164)
R
=
10,000
×
e
3,950 × (−0.00025948)
R
=
10,000
×
e
−1.0249
R
=
10,000
×
0.35883
R
=
3,588.183
Ω
Check
Put the numbers back in. Both sides should match.
3,588.183
=
10,000
×
e^(3,950
×
(
1
50 + 273.15
−
1
25 + 273.15
))
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.

Related lessons

© 2026 SME. All rights reserved.