Q 20. Find the temperature of a resistor due to self heating as a function of time. Consider the resistor to be of resistance \(\rho_0\), length \(l_0\) and cross sectional area \(A_0\) at temperature \(T_0\). Assume the resistor to be connected to an ideal cell of constant electromotive force, \(\mathcal{E}\). Consider its mass to be \(m\) and its specific heat capacity to be \(s\). Assume \(\eta\) to be the fraction of heat the is used by the resistor to heat up, and the rest(\(1-\eta\)) is lost to surroundings. Its coefficients of linear thermal expansion is \(K\), and its thermal coefficient of resistivity to be \(\alpha\). Assume it to be an isotropic material.

Idea:

By the definition, WKT,

\[\rho = \rho_0 e^{\alpha \Delta T}\]

\[\implies R = \frac{\rho_0 e^{\alpha \Delta T} \cdot l_0 e^{K \Delta T}}{A_0 e^{2K \Delta T}}\]

\[\implies R = R_0 e^{(\alpha - K) \Delta T}\]

Where, \(R_0 = \frac{\rho_0 l_0}{A_0}\).

WKT,

\[P = i \mathcal{E} \implies \eta{\mathcal{E}^2 \over R} = \frac{dQ}{dt} = m s \frac{dT}{dt}\]

Hence, we get:

\[T = T_0 + \frac{1}{\alpha-K} \ln \left( 1+ \frac{\eta (\alpha - K) \mathcal{E}^2 t}{m s R_0} \right)\]

Further,

\[R = R_0 e^{(\alpha - K) \Delta T} = R_0 \left( 1+ \frac{\eta (\alpha - K) \mathcal{E}^2 t}{m s R_0} \right)\]

Note, here we assume the the melting point of the resistor is basically infinite. It does not melt. Hence, its temperature at time, \(t = \infty\), is also infinity. So is its resistance, and the current is zero. So, we can say it attains a steady state when time, \(t = \infty\).