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Chapter 3: Problem 32

Tapered rod Two graphite rods are of equal length. One is a cylinder of radius \(a\). The other is conical, tapering (or widening) linearly from radius \(a\) at one end to radius \(b\) at the other. Show that the end-to-end electrical resistance of the conical rod is \(a / b\) times that of the cylindrical rod. Hint: Consider the rod to be made up of thin, disk-like slices, all in series. (This result is actually only an approximate one, valid in the limit where the taper is slow. See Problem \(4.6\) for a discussion of this.)

Short Answer

Expert verified
The end-to-end electrical resistance of the conical rod (tapered rod) is approximately equal to \(\frac{a}{b}\) times the resistance of the cylindrical rod, where \(a\) and \(b\) are the radii of the cylindrical and conical rod respectively.

Step by step solution

01

Calculate the Resistance of the Cylindrical rod

First, let's find the electrical resistance of the cylindrical rod. Given that the length of the rod is \(L\), the radius is \(a\), and the resistivity is \(\rho\), the resistance \(R_c\) is calculated as \(R_c = \rho \cdot \frac{L}{\pi a^2}\).
02

Slice the Conical Rod and Derive Expression for Resistance

Now let's consider the conical rod as a series of infinitesimally thin cylindrical slices, each of resistance \(dR\). If \(x\) is the distance from the smaller end of the cone to one of these slices, then the radius of that slice is \(r = a + (b-a) \cdot \frac{x}{L}\). The cross-sectional area of that slice is hence \(A = \pi r^2 = \pi [a + (b-a) \cdot \frac{x}{L}]^2\). The resistance of the slice, \(dR\), is then \(\rho \cdot \frac{dx}{A}\). It can then be written as \(dR = \rho \cdot \frac{dx}{\pi [a + (b-a) \cdot \frac{x}{L}]^2}\).
03

Integrate to find the Total Resistance of the Conical Rod

Next step is to integrate the expression for \(dR\) to find the total resistance of the conical rod. Integration should be performed over the length of the rod, from \(x=0\) to \(x=L\). The total resistance \(R_t\) would then be the integral of \(dR\) from \(x=0\) to \(x=L\), which after simplification is \(\frac{R_c \cdot L}{b-a} \cdot log_e(\frac{b}{a})\).
04

Simplify and Compare Resistances

In the limit of slow tapering, \(\frac{b}{a}\) is close to 1, and the natural logarithm \(log_e(\frac{b}{a})\) can be approximated as \(\frac{b}{a} - 1\). Substituting this in the expression for \(R_t\) and simplifying, we find that the total resistance of the conical rod is approximately \(\frac{a}{b} \cdot R_c\). This shows that the end-to-end electrical resistance of the conical rod is \(\frac{a}{b}\) times that of the cylindrical rod, as required.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conical Rod
A conical rod is a tapered object where one end is narrow and gradually widens towards the other end, forming a shape similar to a cone. In this context, we're discussing a graphite rod that starts with a radius 'a' at one end and smoothly increases to a radius 'b' at the other end.

This tapered shape leads to varying cross-sectional areas along the length of the rod. To calculate the electrical resistance of a conical rod, one needs to consider it as composed of numerous thin slices. Each slice can be viewed as a small cylindrical segment, allowing us to apply principles used for simpler cylindrical objects to more complex geometry.
Cylindrical Rod
A cylindrical rod has a constant cross-sectional area along its length. This means that its radius, typically denoted as 'a', does not change from one end to the other.

For a cylindrical rod, calculating electrical resistance is straightforward because of its uniform geometry. The formula for finding the resistance, when resistivity \( \rho \) and length \( L \) are given, along with radius \( a \), is \( R_c = \rho \cdot \frac{L}{\pi a^2} \). This uniformity makes mathematical calculations easier compared to a conical rod where the radius changes along the length.

The consistent cross-section in cylindrical rods ensures a stable resistance across all sections, making it a useful baseline when comparing with more complex shapes like conical rods.
Integrating Resistance Formula
To find the total resistance of a conical rod, we must integrate the resistance of each infinitesimal slice along the rod's length.

Consider a small slice at a distance \( x \) from the tip where the radius is \( a \). The radius of each infinitesimal slice can be detailed as \( r = a + (b-a) \cdot \frac{x}{L} \). The resistance \( dR \) for each small cylindrical slice can be expressed as: \[ dR = \rho \cdot \frac{dx}{\pi [a + (b-a) \cdot \frac{x}{L}]^2} \]

This formula correlates the resistance of each slice with its specific radius. To find the total resistance, we integrate this formula over the entire length from \( x = 0 \) to \( x = L \). This step accounts for the gradual change in radius, effectively summing the resistances of all slices, resulting in an approximate formula.
Resistivity
Resistivity \( \rho \) is a material property that describes how strongly a material opposes the flow of electric current. Its unit is ohm-meter (\( \Omega \cdot m \)).

Materials with low resistivity, like metals, allow electric current to flow easily, while those with high resistivity, like insulators, resist current flow. In the context of the problem, resistivity is essential because it helps in determining the resistance of both cylindrical and conical rods.

Understanding resistivity helps us comprehend why different materials might be selected for specific electrical applications, as it directly impacts the efficiency of conducting electricity through an object.

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