Question

What is fractional change in the resistance of a filament of $\alpha =10\times {10}^{-4}{/}^{\circ }C$, when temperature changes from ${25}^{\circ }$ C to ${40}^{\circ }$ C ?

Answer 1

Using $R_t=R_o(1+\alpha \Delta t)$

$⟹$ Fractional change in resistance $\frac{\Delta R}{R}=\alpha \Delta T$

$\therefore$ $\frac{\Delta R}{R}=10\times {10}^{-4}\times (40-25)=0.015$