An RTD is assumed to be represented in the range 0 - C to 100-C by a linear model asRT - R01 - 0-004 Twhere T is the temperature in -C- R0 - 100 - with a variation of -2 - The true model of the RTD isRT- R01 - 0-004 T - 6 - 10-7 T2 The worst-case error magnitude in the resistance value that will be introduced- if the linear model is used over a range of 0 to 100-C is

True value at 100°C, R_100^∘ C = R_0(1 + 0.004 T+ 6 × 10^ 7 T^2) = 100(1+0.004 × 100+6 × nbsp; 10^7 × 100^2) = 140. 6Ω Value of R_0 = 100 ± 2Ω = 98Ω,102 Ω ...

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Standard XII

Physics

Series Combination of Cells

An RTD is ass...

Question

An RTD is assumed to be represented in the range $0^{\circ} C$ to 100°C by a linear model as
$R_{T} = R_{0} (1 + 0.004 T)$
where T is the temperature in $° C$ . $R_{0} = 100 Ω$ with a variation of $\pm 2 Ω$ The true model of the RTD is
$R_{T} = R_{0} (1 + 0.004 T + 6 \times 10^{- 7} T^{2})$ The worst-case error magnitude in the resistance value that will be introduced, if the linear model is used over a range of 0 to 100°C is

2 Ω

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2.8 Ω

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3.0 Ω

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3.4 Ω

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Solution

The correct option is D $3.4 Ω$
$True value at 100°C,$

$R_{100^{\circ} C} = R_{0} (1 + 0.004 T + 6 \times 10^{- 7} T^{2})$

$= 100 (1 + 0.004 \times 100 + 6 \times 10^{7} \times 100^{2})$

$= 140. 6 Ω$

$Value of R_{0} = 100 \pm 2 Ω = 98 Ω, 102 Ω$

$R_{l a} = R_{0} (1 + 0.004 T)$

$= 98 (1 + 0.004 \times 100) = 137.2 Ω$

$= R_{l b} = R_{0} (1 + 0.004 T)$

$= 102 (1 + 0.004 \times 100) = 142.8$

$∴ Worst error = (140.6 - 137.2) Ω = 3.4 Ω$

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The correct option is D 3.4ΩTrue value at 100°C, R100∘C=R0(1+0.004T+6×10−7T2) =100(1+0.004×100+6×107×1002) =140.6Ω Value of R0=100±2Ω=98Ω,102Ω Rla=R0(1+0.004T) =98(1+0.004×100)=137.2Ω =Rlb=R0(1+0.004T) =102(1+0.004×100)=142.8 ∴ Worst error =(140.6−137.2)Ω=3.4Ω