Q: The resistance of iron wire is 10 Ω and α = 5 × 10^{-3}/°C. If a current of 30 A is flowing in it at 20 °C, keeping the potential difference across its length constant, if the temperature is increased to 120 °C, what is the current flowing through that wire ?

Sol: $\large \alpha = \frac{R_{120}-R_{20}}{R_{20}(120-20)}$

$\large 5 \times 10^{-3} = \frac{R_{120}-10}{10 \times 100 }$

R_{120} = 15 Ω ; But V = I R

Here V is constant. Hence,

$\large \frac{I_{120}}{I_{20}} = \frac{R_{20}}{R_{120}} $

$\large \frac{I_{120}}{30} = \frac{10}{15} $

I_{120} = 20 A