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RTD Resistance-Temperature Relation

Models how a resistance temperature detector's resistance changes linearly with temperature.

InstrumentationTemperature Sensing

RTD Resistance-Temperature Calculator RT = R₀(1 + α ΔT)

RT = R₀ · (1 + α · ΔT)
RT = resistance at temperature T (Ω)  ·  R₀ = resistance at reference (Ω)  ·  α = temperature coefficient (1/°C)  ·  ΔT = temperature change (°C)
⟹ Solve RT, R₀, α, ΔT
Ω
Ω
1/°C
°C
Please fix the errors above.
Solve for:
Presets:
RT
RT: R₀: α: ΔT:
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RT Gauge
Low (< 100 Ω) Medium (100–200 Ω) High (> 200 Ω)
RT = R₀(1 + αΔT)  ·  Valid for linear RTD approximation. Typical α for platinum: 0.00385 /°C.

Interpretation

RTD resistance‑temperature relation: R_T = R₀(1 + αΔT) is an approximately linear model for resistance temperature detectors.
It is accurate over small temperature ranges.
Example: R₀=100Ω, α=0.00385/°C, ΔT=50°C → R_T = 100(1 + 0.00385×50) = 119.25Ω.

R_T = R₀(1 + αΔT)
RTD Resistance-Temperature Relation

Variables

SymbolQuantityUnit
R_TResistance at temperature TΩ
R₀Resistance at reference temperatureΩ
αTemperature coefficient of resistance1/°C
ΔTChange in temperature°C

What it means

Resistance Temperature Detectors (RTDs) have a resistance that increases almost linearly with temperature. The relation is R_T = R₀(1 + αΔT), where R₀ is the resistance at 0°C, α is the temperature coefficient of resistance (TCR), and ΔT is the temperature change. For platinum RTDs, α ≈ 0.00385 per °C. This linear approximation is valid over a limited range. RTDs are used for precise temperature measurement in industrial and laboratory settings. They are more accurate than thermocouples but are slower and more expensive. Example: A Pt100 RTD has R₀ = 100Ω at 0°C. At 50°C, ΔT = 50°C, so R_T = 100(1 + 0.00385*50) = 100(1+0.1925) = 119.25Ω. This resistance is measured to infer the temperature.

Worked example

RTD Resistance–Temperature – Practical Example

Real‑World
Scenario: A Pt100 RTD has R₀ = 100 Ω at 0 °C, α = 0.00385 /°C. What is the resistance at 50 °C?
ParameterValue
R₀100 Ω
α0.00385 /°C
ΔT50 °C
FormulaRT = R₀(1 + αΔT)
1αΔT = 0.00385 × 50 = 0.1925
2RT = 100 × (1 + 0.1925) = 119.25 Ω
Final Design R50°C = 119.25 Ω ✓ RTD resistance
Why: RTDs have a nearly linear positive temperature coefficient – used for precise temperature measurement.

Common mistakes

Watch unit consistency and the assumptions behind the formula; misapplying it outside its valid conditions is the most frequent error.

Applications

RTD resistance‑temperature relation R_T = R₀(1 + αΔT) is a linear approximation for resistance temperature detectors, used for accurate temperature measurement. Engineers use RTDs in precision industrial applications, where stability and repeatability are required. By measuring the resistance, temperature is inferred. This formula is the basis for RTD calibration and compensation.

  • Precision temperature measurement in process control
  • HVAC and building automation
  • Laboratory instrumentation and calibration
  • Food processing and pharmaceuticals
  • Educational understanding of RTD sensors

Frequently Asked Questions

Q01What is the RTD resistance‑temperature relation?
A01

For platinum RTDs, the resistance is approximated as R_T = R₀(1 + αΔT) for a limited temperature range, where α is the temperature coefficient of resistance (typically 0.00385 /°C for Pt100).

Q02What is the Callendar‑Van Dusen equation?
A02

It is a more accurate polynomial for RTDs: R_T = R₀[1 + A·T + B·T² + C·(T−100)·T³] for T<0°C. It provides higher accuracy over a wide range.

Q03What are typical α values for platinum RTDs?
A03

α = 0.00385 /°C for IEC 60751 (European curve). α = 0.00392 /°C for the US curve (older standard).

Q04What are common mistakes when using the RTD relation?
A04

Common errors: 1) using the wrong α, 2) applying the linear approximation outside its valid range, 3) forgetting the self‑heating effect, 4) using the wrong R₀ (100 Ω for Pt100), 5) not accounting for lead resistance.

Q05What are practical applications?
A05

Precision temperature measurement in industrial, laboratory, and medical applications.

Q06How do you measure the RTD resistance?
A06

Using a Wheatstone bridge, a current source (4‑wire method), or a resistance meter. The 4‑wire method eliminates lead resistance.

Q07What is the temperature coefficient for a Pt100 RTD?
A07

α = 0.00385 /°C, so for every 1°C change, resistance changes by 0.385 Ω.

Q08How does the RTD compare to a thermocouple?
A08

RTDs are more accurate and stable but have a narrower range and slower response. Thermocouples have a wider range and faster response but lower accuracy.