Chapter 4: Q59P (page 845)

CALC A material of resistivity $ρ$ is formed into a solid, truncatedcone of height $h$ and radii $r_{1}$ and $r_{2}$ at either end (Fig. P25.59). (a) Calculatethe resistance of the cone between the two flat end faces. (Hint: Imagineslicing the cone into very many thin disks, and calculate the resistance of one such disk.) (b) Show that your result agrees with Eq. (25.10) when . $r_{1} = r_{2}$ .

Short Answer

Expert verified

The resistance of the cone between two flat end faces is $R = \frac{ρ h}{π r_{1} r_{2}}$ .
Yes, the agrees with equation $R = \frac{ρ L}{A}$ when $r_{1} = r_{2}$ .

Step by step solution

Define the ohm’s law, resistance ( R ) and power ( P ) .

According to Ohm’s law the current flowing through conductor is directly proportional to the voltage across the two points.

$V = I R$

Where, $I$ is current in ampere $(A)$ , $R$ is resistance in ohms $(Ω)$ and $V$ is the potential difference volt $(V)$ .

The ratio of V to I for a particular conductor is called its resistance R :

$R = \frac{V}{I}$ or $\frac{ρ L}{A}$

Where, $ρ$ is resistivity $Ω \cdot m$ , $L$ is length in m and $A$ is area in $m^{2}$ .

Determine the resistance.

Consider for dasher disk $L = d y$

And $A = π r^{2}$ $A = π r^{2}$

Therefore, $d R = \frac{ρ d y}{π r^{2}}$

From the figure, $r = r_{2} + x$

The two right-angled triangle are similar. So,

$\frac{h}{y} = \frac{r_{1} - r_{2}}{x}$ ........( 1 )

Solve for x

role="math" localid="1664178294308" $x = \frac{y (r_{1} - r_{2})}{h} r = r_{2} + \frac{y (r_{1} - r_{2})}{h} (from (1))$

Consider role="math" localid="1664178332890" $c = \frac{(r_{1} - r_{2})}{h}$ as the term $\frac{(r_{1} - r_{2})}{h}$ is constant.

Therefore, $r = r_{2} + y c$ .............(2)

Also, role="math" localid="1664178459170" $d R = \frac{ρ d y}{π {[r_{2} + y c]}^{2}}$

Integrate both side of equation $d R = \frac{ρ d y}{π {[r_{2} + y c]}^{2}}$

$\int_{0}^{R} d R = \int_{0}^{h} \frac{ρ d y}{π {[r_{2} + y c]}^{2}} R = \frac{ρ}{π} \int_{0}^{h} \frac{d y}{{[r_{2} + y c]}^{2}} R = \frac{ρ}{π} {[\frac{- 1}{c [r_{2} + y c]}]}_{0}^{h} R = \frac{ρ}{c π} [\frac{- 1}{[r_{2} + h c]} + \frac{1}{r^{2}}]$

From equation $(2), h c = r_{1} - r_{2}$ and c is $\frac{r_{1} - r_{2}}{h}$

$R = \frac{ρ}{(\frac{r_{1} - r_{2}}{h}) π} [\frac{1}{r_{2} + r_{1} - r_{2}} + \frac{1}{r_{2}}] = - \frac{ρ}{(\frac{r_{1} - r_{2}}{h}) π} [\frac{1}{r_{1}} + \frac{1}{r_{2}}] = \frac{ρ}{π} \times \frac{h}{r_{1} - r_{2}} \times [\frac{r_{1} - r_{2}}{r_{1} r_{2}}] = \frac{ρ h}{π r_{1} r_{2}}$

Hence, the resistance of the cone between two flat end faces is $R = \frac{ρ h}{π r_{1} r_{2}}$ .

Find whether resistance agrees with .

If in equation of resistance $R = \frac{ρ h}{π r_{1} r_{2}}$ the radius $r_{1}$ and $r_{2}$ equal to $r$ , then the resistance of disk is $R = \frac{ρ h}{π r^{2}}$ where $h$ is the height of disk and $π r^{2}$ is the cross-sectional area of the disk.