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Consider a temperature measurement scheme shown in the adjoining figure. It uses an RTD whose resistance at 0cC is 100 W and temperature coefficient of resistance ^ h a is 0 00392 . /Cc Q. 18 The differential gain of the instrumentation amplifier to achieve a voltage sensitivity of 10 mV C /c , should be approximately (A) 13.41 (B) 26.02 (C) 57.53 (D) 90.1?
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Consider a temperature measurement scheme shown in the adjoining figur...
Solution:

Given parameters:

- RTD resistance at 0°C, R0 = 100 Ω
- Temperature coefficient of resistance, α = 0.00392 /°C
- Voltage sensitivity required, Vsens = 10 mV/°C

We need to determine the differential gain of the instrumentation amplifier to achieve the required voltage sensitivity.

Calculation:

The resistance of the RTD at any temperature T can be computed using the following formula:

R(T) = R0 [1 + α (T - 0)]

The change in resistance of the RTD for a temperature change of ΔT is given by:

ΔR = R(T2) - R(T1) = R0 [1 + α (T2 - 0)] - R0 [1 + α (T1 - 0)]

ΔR = R0 α (T2 - T1)

The voltage across the RTD for a change in resistance of ΔR is given by:

ΔV = ΔR × I

where I is the excitation current flowing through the RTD.

Assuming I = 1 mA, we have:

ΔV = 0.001 × R0 α (T2 - T1)

To achieve a voltage sensitivity of 10 mV/°C, we need:

ΔV/ΔT = 10 mV/°C

0.001 × R0 α = 10 mV/°C

α = 0.00392 /°C

Substituting the values, we get:

R0 = 100 Ω

0.001 × 100 × 0.00392 = 0.000392 V/°C

ΔV/ΔT = 10 mV/°C

Differential gain of the instrumentation amplifier, A = ΔV/ΔT × Rf/Rg

where Rf and Rg are the feedback and gain resistances of the instrumentation amplifier, respectively.

Therefore,

A = (10 × 10^-3 V/°C) × (Rf/Rg)

Substituting Rf = 10 kΩ and Rg = 1 kΩ, we get:

A = (10 × 10^-3 V/°C) × (10 kΩ/1 kΩ) = 100

Therefore, the differential gain of the instrumentation amplifier to achieve a voltage sensitivity of 10 mV/°C should be approximately 100.

Answer: (D) 90.1
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Consider a temperature measurement scheme shown in the adjoining figure. It uses an RTD whose resistance at 0cC is 100 W and temperature coefficient of resistance ^ h a is 0 00392 . /Cc Q. 18 The differential gain of the instrumentation amplifier to achieve a voltage sensitivity of 10 mV C /c , should be approximately (A) 13.41 (B) 26.02 (C) 57.53 (D) 90.1?
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Consider a temperature measurement scheme shown in the adjoining figure. It uses an RTD whose resistance at 0cC is 100 W and temperature coefficient of resistance ^ h a is 0 00392 . /Cc Q. 18 The differential gain of the instrumentation amplifier to achieve a voltage sensitivity of 10 mV C /c , should be approximately (A) 13.41 (B) 26.02 (C) 57.53 (D) 90.1? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about Consider a temperature measurement scheme shown in the adjoining figure. It uses an RTD whose resistance at 0cC is 100 W and temperature coefficient of resistance ^ h a is 0 00392 . /Cc Q. 18 The differential gain of the instrumentation amplifier to achieve a voltage sensitivity of 10 mV C /c , should be approximately (A) 13.41 (B) 26.02 (C) 57.53 (D) 90.1? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Consider a temperature measurement scheme shown in the adjoining figure. It uses an RTD whose resistance at 0cC is 100 W and temperature coefficient of resistance ^ h a is 0 00392 . /Cc Q. 18 The differential gain of the instrumentation amplifier to achieve a voltage sensitivity of 10 mV C /c , should be approximately (A) 13.41 (B) 26.02 (C) 57.53 (D) 90.1?.
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