The figure below shows a rod of resistive material. Its resistance per unit length increases in the positive direction of the x axis. At any position x along the rod, the resistance dR of a narrow section of width dx is given by dR=5x dx , where dR is in ohms and x is in meters. I want to slice off a length of the rod between x=0 and some position x=L and then connect that length to a battery with a potential difference V=5V . I want the current in the lenght to transfer energy to thermal energy at the rate of 200 W . I am going to show how to find the position x=L such that these conditions are met.

Since I am dealing with a resistor, I can relate the given rate of energy transfer to thermal energy, P , to the Resistance, R , using the equation P=\frac{V^2}{R} , which is the resistive dissipation of the resistive battery. Essentially electric potential energy is converted to internal thermal energy via collisions between charge carriers and atoms. After plugging in the given voltage for the battery and the rate of thermal energy transfer, and then solving for R , I obtain R=1.25\times10^{-1}\Omega , which is the resistance of the battery. Next, I can integrate the given differential, dR=5x dx , which will equal R and set it equal to R=1.25\times10^{-1}\Omega . I am doing this because the rate of thermal energy transfer depends upon the Resistance, which in turn depends on the length of the rod. After applying the Fundamental Theorem of Calculus to the integral and solving for R , I obtain L=2.236\times10^{-1}m