Question

A platinum RTD having resistance temperature coefficient of $0.004{/}^{\circ }C$ is connected in one arm of the bridge circuit as shown in the figure. The output of the bridge is given to an instrumentation amplifier so that the temperature range from ${25}^{\circ }Cto{75}^{\circ }C$ is converted to 0 to 10 V output. If the nominal resistance of RTD (at $T={25}^{\circ }C$) is $150\Omega$ then the resistance value $R_G$ of the instrumentation amplifier is ________ $k\Omega$.

• A. 9
• B. 7
• C. 5
• D. 3

Answer 1

The correct option is B 7

Since output of instrumentation amplifier is given by,

$V_0=[1+\frac{2R_G}{R}](V_1-V_2)...\left(i\right)$

$\text{Now RTD resistance at}{25}^{\circ }C=150\Omega \left[given\right]$

$At{75}^{\circ }C,R_T=150[1+0.004\times 50]=180\Omega$

$\text{Therefore at}{25}^{\circ }C\text{,}{V}_1=V_2\Rightarrow V_0=0$

At ${75}^{\circ }C,V_0=10V\left[Given\right]$

$V_2=\frac{150}{150+120}\times 15=\frac{25}{3}V$

$V_1=\frac{180}{180+120}\times 15=9V$

So,

$10=\left(\frac{2}{3}\right)(1+2R_G)$

$R_G=7k\Omega$