Question

A thermistor of resistance $1k\Omega$, temperature coefficient of resistance $4.5\left(\%{/}^{\circ }C\right)$ and an internal temperature rise of ${0.2}^{\circ }C/mW$ around ${27}^{\circ }C$ is included in a d.c bridge with three fixed resistors each of value of $1k\Omega$. A d.c amplifier of high input impedance is connected across the bridge output and an indicating device connected to the amplifier output the maximum voltage sensitivity of the bridge in $\left(\frac{mV}{{}^{\circ }C}\right)$ if the internal temperature rise is not to exceed ${0.1}^{\circ }C$ is

• A. 0
• B. 11.9
• C. 15.9
• D. 12.9

Answer 1

The correct option is C 15.9

power dissipation in the thermistor due to a change of ${0.1}^{\circ }C$

$P={0.1}^{\circ }C\times \frac{1}{0.2{(}^{\circ }C/mw)}=0.5\left(mW\right)$

Let $V_i$ is the maximum allowable input voltage across the bridge as each arm has resistance of $1k\Omega .$

$\text{Voltage across each arm}=\frac{V_i}{2}$

Power dissipation in the thermistor
$P={\left(\frac{V_i}{2}\right)}^2\times \frac{1}{R}$

$0.5\times {10}^{-3}={\left(\frac{V_i}{2}\right)}^2\times \frac{1}{1000}$

$V_i=1.41\left(V\right)$

Change in resistance of thermistor

$\Delta R=\epsilon \Delta \theta R$

$=\frac{4.5}{100}\times 0.1\times 1000$

$\Delta R=4.5\Omega$

Bridge output voltage when all the arms have initial resistance

$\Delta e_0=\frac{\Delta R/R}{4}{e}_i$

$=\frac{4.5/1000}{4}\times 1.41$

$\Delta e_0=1.59\left(mV\right)$

This voltage change is caused by temperature input of ${0.1}^{\circ }C$ maximum bridge sensitivity

$S=\frac{\Delta e_0}{\Delta \theta }=\frac{1.59}{0.1}=15.9\left(\frac{mV}{{}^{\circ }C}\right)$