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 $$\rho$$ (see figure). If the battery is connected to a resistance of value $$R$$, the maximum Joule heating in $$R$$ will take place for:

We model the torch battery as an ideal emf source $$\mathcal E$$ in series with its internal resistance, which is the resistance of the electrolyte that fills the space between the inner solid rod (radius $$a$$) and the outer cylindrical shell (inner radius $$b$$). Our first task is therefore to calculate this internal resistance, which we shall denote by $$r$$.

To find $$r$$, we remember the differential formula for the resistance of a conductor:

$$dR=\rho\,\dfrac{dl}{A},$$

where $$\rho$$ is the resistivity, $$dl$$ is the infinitesimal length of the current path, and $$A$$ is the cross-sectional area perpendicular to the current.

In the present arrangement the current flows radially outward from the inner rod to the outer shell. Choose an imaginary cylindrical surface of radius $$r$$ (with $$a<r<b$$) and thickness $$dr$$. The current crosses this surface normally, so:

• The path length for the current element is the radial thickness $$dl=dr$$.
• The area normal to the current is the curved area of the cylinder, $$A = 2\pi r\,l$$, where $$l$$ is the length of the battery.

Substituting these into the differential resistance formula, we obtain

$$dR=\rho\,\dfrac{dr}{2\pi r\,l}.$$

To get the total internal resistance, integrate from the inner radius $$a$$ to the outer radius $$b$$:

$$r = \int_{a}^{b}dR = \int_{a}^{b}\rho\,\dfrac{dr}{2\pi r\,l} = \dfrac{\rho}{2\pi l}\int_{a}^{b}\dfrac{dr}{r} = \dfrac{\rho}{2\pi l}\,\Bigl[\ln r\Bigr]_{a}^{b} = \dfrac{\rho}{2\pi l}\,\ln\!\left(\dfrac{b}{a}\right).$$

Now the circuit consists of the emf $$\mathcal E$$, the internal resistance $$r$$ and the external resistance $$R$$ in series. The Joule heating (power) in the external resistor is

$$P=\dfrac{\mathcal E^{2}R}{\left(R+r\right)^{2}}.$$

The maximum power transfer theorem states that, for a fixed internal resistance $$r$$, the power delivered to $$R$$ is maximum when

$$R=r.$$

Therefore, using the explicit form of $$r$$ we just calculated, the condition for maximum Joule heating is

$$R=\dfrac{\rho}{2\pi l}\,\ln\!\left(\dfrac{b}{a}\right).$$

This expression matches Option B.

Hence, the correct answer is Option B.