Question

In the circuit below, the op-amps are ideal and the sensor connected to one arm of resistive bridge is the platinum RTD whose nominal resistance $(atT=0^{\circ }C)$ is $100\Omega$ and Resistance temperature coefficient of ${0.004}^{\circ }C.$ If the output voltage of the circuit at some unknown tempreature $T^{\circ }C$ is 5 V, then the value of T is

• A. 100
• B. 25
• C. 12.5
• D. 125

Answer 1

The correct option is D 125

Redraw the circuit,



In the given circuit, the output of the bridge is given to the differential amplifier whose output is given by

$V_o=\frac{R_f}{R_1}[V_1-V_2]$

$Where,\frac{R_f}{R_1}=\frac{100k\Omega }{10k\Omega }=10$

$\therefore V_0=10[V_1-V_2]$

$5=10[V_1-V_2]$

$0.5=V_1-V_2...\left(1\right)$

Now,

$V_1=5\times \frac{R_T}{R_T+100}[\because \text{Ideal opamp takes zero input current}]$

$V_2=5\times \frac{100}{100+100}=2.5V$

From equation (i),

$0.5=5\left[\frac{R_T}{R_T+100}\right]-2.5$

$\frac{3}{5}=\frac{R_T}{R_T+100}$

$R_T=150\Omega$

For platinum RTD, $R_T$ is given by

$R_T=R_0[1+\alpha T]$

$150=100[1+(0.004\times T\left)\right]$

$T={125}^{\circ }C$