Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/398

This method in the case of wires whose section varies slowly with the length gives a result very near the truth, but it is really only a lower limit, for the true resistance is always greater than this, except in the case where the section is perfectly uniform.

307.] To find the higher limit of the resistance, let us suppose a surface drawn in the conductor to be rendered impermeable to electricity. The effect of this must be to increase the resistance of the conductor unless the surface is one of the natural surfaces of flow. By means of two systems of surfaces we can form a set of tubes which will completely regulate the flow, and the effect, if there is any, of this system of impermeable surfaces must be to increase the resistance above its natural value.

The resistance of each of the tubes may be calculated by the method already given for a fine wire, and the resistance of the whole conductor is the reciprocal of the sum of the reciprocals of the resistances of all the tubes. The resistance thus found is greater than the natural resistance, except when the tubes follow the natural lines of flow.

In the case already considered, where the conductor is in the form of an elongated solid of revolution, let us measure $x$ along the axis, and let the radius of the section at any point be $b.$ Let one set of impermeable surfaces be the planes through the axis for each of which $\phi$ is constant, and let the other set be surfaces of revolution for which

$y^{2}=\psi b^{2},$

(9)

where $\psi$ is a numerical quantity between 0 and 1.

Let us consider a portion of one of the tubes bounded by the surfaces $\phi$ and $\phi +d\phi ,\psi$ and $\psi +d\psi ,x$ and $x+dx.$

The section of the tube taken perpendicular to the axis is

$y\,dy\,d\phi ={\frac {1}{2}}b^{2}\,d\psi \,d\phi .$

(10)

If $\theta$ be the angle which the tube makes with the axis

$\tan \theta =\psi ^{\frac {1}{2}}{\frac {db}{dx}}.$

(11)

The true length of the element of the tube is $dx\sec \theta ,$ and its true section is

${\frac {1}{2}}b^{2}\,d\psi \,d\phi \cos \theta ,$

so that its resistance is

$2\rho {\frac {dx}{b^{2}\,d\psi \,d\phi }}\sec ^{2}\theta =2\rho {\frac {dx}{b^{2}\,d\psi \,d\phi }}\left(1+\psi \left.{\frac {\overline {db}}{d}}\right|^{2}\right).$

(12)

Let

$A=\int \rho {\frac {dx}{b^{2}}},\quad \quad \mathrm {and} \quad \quad B=\int \rho {\frac {dx}{b^{2}}}\left.{\frac {\overline {db}}{dx}}\right|^{2},$

(13)