Model a torch battery of length l to be made up of a thin cylindrical bar of radius ‘a’ and a concentric thin cylindrical shell of radius ‘b’ filled in between with an electrolyte of resistivity ρ (see figure). If the battery is connected to a resistance of value R, the maximum Joule heating in R will take place for :

Solution

Physics sir · 3 steps

Current flows outward through the electrolyte, so build the internal resistance from thin cylindrical shells.

1. Path of the current. The current goes radially from the central bar at radius a to the outer shell at radius b. Take a thin shell of electrolyte of radius r and thickness dr. Its length along the current is dr, and its area is 2π r l.

   rᵢₙₜ = ∫ₐᵇ (ρ dr)/(2π r l) = ρ/(2π l)ln(b/a)

2. Maximum power transfer. The power in R is P = (E² R)/(R + rᵢₙₜ)². Setting dP/dR = 0 gives R = rᵢₙₜ, so the Joule heating in R is greatest when R equals the internal resistance.
3. So. Maximum heating occurs for R = ρ/(2π l)ln(b/a), which is option (c).

The idea to keep. When the cross-section changes along the current path, add up thin slices with dR = ρ dx / A. And for maximum power in the load, match the load to the internal resistance.

Answer from the NTA final answer key. Solution by PrepWiser.