Question

The resistance of an iron wire and a copper wire at 20°C are 3.9 Ω and 4.1 Ω, respectively. At what temperature will the resistance be equal? Temperature coefficient of resistivity for iron is 5.0 × 10^–3 K^–1 and for copper, it is 4.0 × 10^–3 K^–1. Neglect any thermal expansion.

Answer 1

Given:
Resistance R_Fe of the iron wire at 20°C = 3.9 Ω
Resistance R_Cu of the copper wire at 20°C = 4.1 Ω
Temperature coefficient α_Fe for iron = 5.0 × 10^–3 K^–1
Temperature coefficient α_Cu for copper = 4.0 × 10^–3 K^–1
Let
The temperature at which the resistance be equal = T
Resistance of iron wire at T °C = R_Fe'
Resistance of copper wire at T °C = R_Cu'
We know:
$R=R_0(1+\alpha ∆T)$
Here, ΔT = T - 20
$$\begin{gathered} \Rightarrow R_{\mathrm{Fe}}\text{'}=R_{\mathrm{Fe}}\left[1+{\alpha }_{\mathrm{Fe}}\left(T-20\right)\right] \\ {R}_{\mathrm{Cu}}\text{'}=R_{\mathrm{Cu}}\left[1+{\alpha }_{\mathrm{Cu}}\left(T-20\right)\right] \\ {R}_{\mathrm{Fe}}\mathit{\text{'}}\mathit{=}{R}_{\mathrm{Cu}}\text{'} \\ \mathit{\Rightarrow }{R}_{\mathrm{Fe}}\left[1+{\alpha }_{\mathrm{Fe}}\left(T-20\right)\right]=R_{\mathrm{Cu}}\left[1+{\alpha }_{\mathrm{Cu}}\left(T-20\right)\right] \\ \Rightarrow 3.9\left[1+5\times {10}^{-3}\left(T-20\right)\right]=4.1\left[1+4\times {10}^{-3}\left(T-20\right)\right] \\ \Rightarrow 3.9+3.9\times 5\times {10}^{-3}\left(T-20\right)=4.1+4.1\times 4\times {10}^{-3}\left(T-20\right) \\ \Rightarrow 4.1\times 4\times {10}^{-3}\left(T-20\right)-3.9\times 5\times {10}^{-3}\left(T-20\right)=-4.1+3.9 \\ \Rightarrow 16.4\left(T-20\right)-19.5\left(T-20\right)=-0.2\times {10}^3 \\ \Rightarrow \left(T-20\right)\left(-3.1\right)=-0.2\times {10}^3 \\ \Rightarrow T-20=64.5 \\ \Rightarrow T=84.5°\mathrm{C} \end{gathered}$$