As an example, assume the cable came in two lengths: length A has a total resistance of 15 Ω, and length B has a total resistance of 30 Ω. At 20°C, the RTD has a resistance of 107.79 Ω. The voltage (VAB) with the influence of the cables in Fig. 23-5 can be calculated as shown.

Question

As an example, assume the cable came in two lengths: length A has a total resistance of 15 Ω, and length B has a total resistance of 30 Ω. At 20°C, the RTD has a resistance of 107.79 Ω. The voltage [latex](V_{AB})[/latex] with the influence of the cables in Fig. 23-5 can be calculated as shown.

Accepted Answer

Temperature = 20°C, Length A at 15 Ω	Temperature = 100°C, Length B at 30 Ω
[latex]V_{AB} = V_A − V_B[/latex]	[latex]V_{AB} = V_A − V_B[/latex]
[latex]V_{AB}=V_1\left(\frac{RTD+R_{Cable}}{R_1+RTD+R_{Cable}}\right) -V_1\left(\frac{R_3}{R_2+R_3} \right) [/latex]	[latex]V_{AB}=V_1\left(\frac{RTD+R_{Cable}}{R_1+RTD+R_{Cable}}\right) -V_1\left(\frac{R_3}{R_2+R_3} \right) [/latex]
[latex]R_2 = R_3 ∴[/latex]	[latex]R_2 = R_3 ∴[/latex]
[latex]V_{AB}=V_1\left(\frac{RTD+R_{Cable}}{R_1+RTD+R_{Cable}}\right) -V_1\left(\frac{1}{2} \right) [/latex]	[latex]V_{AB}=V_1\left(\frac{RTD+R_{Cable}}{R_1+RTD+R_{Cable}}\right) -V_1\left(\frac{1}{2} \right) [/latex]
[latex]V_{AB}=10 \ V\left(\frac{107.79 \ Ω+15 \ \Omega }{100 \ Ω + 107.79 \ Ω+15 \ \Omega} \right)-10 \ V \left(\frac{1}{2} \right) [/latex]	[latex]V_{AB}=10 \ V\left(\frac{107.79 \ Ω+30 \ \Omega }{100 \ Ω + 107.79 \ Ω+30 \ \Omega} \right)-10 \ V \left(\frac{1}{2} \right) [/latex]
[latex]V_{AB} = 511.47[/latex] mV	[latex]V_{AB} = 794.61 [/latex] mV

Temperature = 20°C, Length A at 15 Ω	Temperature = 100°C, Length B at 30 Ω
$V_{AB} = V_A − V_B$	$V_{AB} = V_A − V_B$
$V_{AB}=V_1\left(\frac{RTD+R_{Cable}}{R_1+RTD+R_{Cable}}\right) -V_1\left(\frac{R_3}{R_2+R_3} \right)$	$V_{AB}=V_1\left(\frac{RTD+R_{Cable}}{R_1+RTD+R_{Cable}}\right) -V_1\left(\frac{R_3}{R_2+R_3} \right)$
$R_2 = R_3 ∴$	$R_2 = R_3 ∴$
$V_{AB}=V_1\left(\frac{RTD+R_{Cable}}{R_1+RTD+R_{Cable}}\right) -V_1\left(\frac{1}{2} \right)$	$V_{AB}=V_1\left(\frac{RTD+R_{Cable}}{R_1+RTD+R_{Cable}}\right) -V_1\left(\frac{1}{2} \right)$
$V_{AB}=10 \ V\left(\frac{107.79 \ Ω+15 \ \Omega }{100 \ Ω + 107.79 \ Ω+15 \ \Omega} \right)-10 \ V \left(\frac{1}{2} \right)$	$V_{AB}=10 \ V\left(\frac{107.79 \ Ω+30 \ \Omega }{100 \ Ω + 107.79 \ Ω+30 \ \Omega} \right)-10 \ V \left(\frac{1}{2} \right)$
$V_{AB} = 511.47$ mV	$V_{AB} = 794.61$ mV

Question 23.2: As an example, assume the cable came in two lengths: length ...

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