A torch battery of length l is to be made up of a thin cylindrical bar of radius a and a concentric

Q: Why do we have to use integration for this problem?

Because the current flows radially outwards. As it moves from the inner cylinder to the outer cylinder, the cross-sectional area it passes through ($2\pi r l$) continuously increases. Since the area is not constant, we must sum up the resistances of infinitesimally thin shells using integration.

Q: Why is the area $2\pi r l$ and not $\pi r^2$?

The area in the resistance formula $R = \rho \frac{l}{A}$ must be perpendicular to the direction of current flow. Here, current flows radially outward, so it crosses the curved surface of the cylinder, which has an area of $2\pi r l$. The flat circular ends ($\pi r^2$) are parallel to the current flow, not perpendicular.

Q: What exactly is the Maximum Power Transfer Theorem?

It's a principle stating that to obtain maximum external power from a source with a finite internal resistance, the resistance of the load must equal the resistance of the source as viewed from its output terminals.

Q: Why are the elemental shells considered to be in series?

Because the exact same total current $I$ must flow through each successive concentric shell to get from the inner electrode to the outer electrode. When the same current flows through multiple components, they are in series.

Q: What if the current flowed along the length of the cylinder instead of radially?

If the current flowed longitudinally (from one flat end to the other), the cross-sectional area would be constant at $A = \pi(b^2 - a^2)$, and the length would be $l$. In that case, no integration would be needed, and the resistance would simply be $R = \rho \frac{l}{\pi(b^2 - a^2)}$.

Question

A torch battery of length $l$ is to be made up of a thin cylindrical bar of radius $a$ and a concentric thin cylindrical shell of radius $b$ is filled in between with an electrolyte of resistivity $\rho$ (see figure). If the battery is connected to a resistance $R$, the maximum joule's heating in $R$ will takes place for

Options:

(A) $R = \frac{\rho}{2\pi l}\left(\frac{b}{a}\right)$

(B) $R = \frac{\rho}{2\pi l}\ln\left(\frac{b}{a}\right)$

(C) $R = \frac{\rho}{\pi l}\ln\left(\frac{b}{a}\right)$

(D) $R = \frac{2\rho}{\pi l}\ln\left(\frac{b}{a}\right)$

Sigma Percentile · Accepted Answer

### Understanding the Battery Geometry

Imagine this battery as two concentric cylinders. The current flows radially outwards from the inner bar of radius $a$ to the outer shell of radius $b$, passing through the electrolyte. This radial flow is the key to understanding how the internal resistance builds up. Unlike a standard wire where current flows along the length, here the current spreads outwards, meaning the cross-sectional area it passes through is constantly increasing.

### The Maximum Power Transfer Theorem

The question asks for the maximum Joule's heating in the external resistance $R$. According to the **Maximum Power Transfer Theorem**, maximum power is delivered to an external load when the load resistance exactly matches the internal resistance of the source. Therefore, our primary goal is to calculate the internal resistance $R_{	ext{internal}}$ of this cylindrical battery. Once we find it, we simply set $R = R_{	ext{internal}}$.

### Calculating the Internal Resistance

To calculate the total internal resistance, we cannot use the simple formula $R = \frac{ho l}{A}$ directly because the area $A$ is not constant. Instead, we take a thin elemental cylindrical shell of radius $r$ and thickness $dr$ inside the electrolyte.

The resistance of any conductor is given by its resistivity $ho$ multiplied by its length divided by its cross-sectional area. For our elemental shell, the length along the direction of current flow is simply its thickness $dr$. The cross-sectional area it presents to the current is its curved surface area, which is $2\pi r l$.

So, the small resistance $dR$ of this elemental shell is:

$$dR = \frac{ho \cdot dr}{2\pi r l}$$

### The Integration Phase

Since all such elemental shells from the inner radius $a$ to the outer radius $b$ are arranged in series (the current must pass through each one sequentially), we integrate $dR$ from $a$ to $b$ to find the total internal resistance.

$$R_{	ext{internal}} = \int_{a}^{b} \frac{ho \cdot dr}{2\pi r l}$$

We can pull the constants $ho$, $2\pi$, and $l$ out of the integral. We are left with the integral of $\frac{1}{r}$ with respect to $r$, from limits $a$ to $b$.

$$R_{	ext{internal}} = \frac{ho}{2\pi l} \int_{a}^{b} \frac{1}{r} dr$$

The integral of $\frac{1}{r}$ is the natural logarithm of $r$. Applying the limits, we get:

$$R_{	ext{internal}} = \frac{ho}{2\pi l} [\ln r]_{a}^{b}$$

$$R_{	ext{internal}} = \frac{ho}{2\pi l} (\ln b - \ln a)$$

### Final Conclusion

Using the property of logarithms, $\ln b - \ln a$ becomes $\ln\left(\frac{b}{a}ight)$. So, we have our final expression for the internal resistance:

$$R_{	ext{internal}} = \frac{ho}{2\pi l} \ln\left(\frac{b}{a}ight)$$

For maximum heating, the external resistance $R$ must equal this exact value. Therefore, the correct option is **(b)**.

Select Answer:

Visualized Solution

\text{Battery Structure}

\text{Maximum Power Theorem}

\text{Elemental Shell}

\text{Resistance of Elemental Shell}

\text{Differential Resistance}

\text{Total Internal Resistance}

\text{Integration}

\text{Applying Limits}

\text{Final Answer}

\text{The Way Forward}

The Sigma Insight: Cells, EMF, and Internal Resistance

Understanding the Battery Geometry

The Maximum Power Transfer Theorem

Calculating the Internal Resistance

The Integration Phase

Final Conclusion

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