The thermal conduction rate ' P_ cond ' of a slab is given by P_...

The rate of heat flow (thermal conduction rate) is given by $P_{cond} = \dfrac{kA(T_1 - T_2)}{x}$.

This can be rewritten as $P_{cond} = \dfrac{T_1 - T_2}{\dfrac{x}{kA}}$.

Comparing this with the electrical analogy $I = \dfrac{\Delta V}{R}$, where $I$ is the current, $\Delta V$ is the potential difference, and $R$ is the electrical resistance.

Here, $P_{cond}$ is analogous to current $I$, and the temperature difference $(T_1 - T_2)$ is analogous to potential difference $\Delta V$.

Thus, the thermal resistance $R_T$ is analogous to electrical resistance $R$.

Therefore, $R_T = \dfrac{x}{kA}$.

Answer: $\dfrac{x}{kA}$