Question

A strain gauge having gauge factor of 2.5 and nominal resistance $1000\Omega$ is made from a metallic wire having resistance temperature coefficient of $\alpha =0.004{/}^{\circ }Cat{25}^{\circ }C$. If the dissipation factor is given as $P_D=25mW{/}^{\circ }C$ then the maximum current that can passed through the strain gauge to keep self-heating error below or upto 1 microstrain is ________ $\mu A$

Answer 1

Since, Resistance change due to applied strain is given by

$(\Delta R{)}_{ϵ}=R\times G\times ϵ$

and Resistance change due to temperature change is given by

$(\Delta R{)}_T=R\times \alpha \times \Delta T$

Now $\Delta T$ is the temperature rise due to the electrical power dissipation which is given as,

$P_D=i^{2}RWatt$

$Since25mW\to 1^{\circ }Criseintemperature$

$1mW\to \frac{1}{25\times {10}^{-3}}{}^{\circ }Criseintemperature$

$i^{2}R\to \frac{i^{2}R}{25\times {10}^{-3}}{}^{\circ }C\text{rise in temperature}$

$There,\Delta T=\frac{i^{2}R}{25\times {10}^{-3}}{}^{\circ }C$

$So,Fromequation\left(ii\right)(\Delta R{)}_T=R\times \alpha \times \frac{i^{2}R}{25\times {10}^{-3}}$

Now, according to condition given in problem

$(\Delta R{)}_T\le (\Delta R{)}_{ϵ=1micron}$

Therefore,

$R\times \alpha \times \frac{i^{2}R}{25\times {10}^{-3}}\le R\times G\times ϵ{|}_{ϵ=1micron}$

$i\le \sqrt{\frac{25\times {10}^{-3}\times G\times ϵ}{\alpha R}}$

$i\le \sqrt{\frac{25\times {10}^{-3}\times 2.5\times {10}^{-6}}{0.004\times 1000}}$

$i\le 125\mu A$