Q: The temperature coefficient of resistance of platinum is α = 3.92 × 10^{-3} K^{-1} at 0 °C. Find the temperature at which the increase in the resistance of platinum wire is 10 % of its value at 0 °C.

Sol: $\large R_2 = R_1 + R_1 \times \frac{10}{100} = \frac{110 R_1}{100}= 1.1 R_1 $

$\large \Delta t = \frac{R_2 – R_1}{R_1 \alpha } $

$\large \Delta t = \frac{1.1 R_1 – R_1}{R_1 \alpha } = \frac{0.1}{\alpha} $

$\large \Delta t = \frac{0.1}{3.92 \times 10^{-3}} $

Δt = 25.51 °C ; t_{2} = 25.51 + 20 = 45.51 °C.