• Consider a conductor of length \(\ell\) and uniform cross sectional area A.

• Resistance of conductor is given as \(R=\dfrac {\rho L}{A}\)

• Resistance depends on length of conductor \((\ell)\) and area of conductor (A).
• Resistivity \((\rho)\) is a constant of proportionality and depends only on material.

• Consider a cuboid of length \(\ell\), height \(h\) and breadth \(b\) as shown in figure, of resistivity \(\rho\).

Case 1 

• Current is passing through face (1) (shaded one) to face (4).

For face (1) or (4)

Length = \(\ell\) , Area = bh

Case 2 

• Current is passing through face (2) and face (5)

For Face (2) or face (5)

Length = b, Area = \(\ell h\)

Case 3 

• Current is passing through face (3) to (6)

For Face (3) or face (6)

Length = h, Area = \(b\;\ell \)

Case 1 : Hollow Cylinder

• Consider a hollow cylinder of length L and inner and outer radius be 'a' and 'b' respectively, as shown in figure.

The resistivity of material is \(\rho\).

• To calculate the resistance of a hollow cylinder, select a small element made up of thin cylinder of inner radius 'r' and outer radius 'r+dr'. The length will be 'dr' .
• Summing up all three small elements made up of thin cylinder give the entire cylinder.
• The resistance of the whole cylinder will be sum of resistance due to all these small elements made of thin cylinder.

Differential Element

Case 2 : Hollow Sphere

• Consider a hollow sphere of inner radius 'a' and outer radius 'b' for which direction of current is radially outward.

• To calculate the resistance of a hollow sphere, choose a differential element of thin hollow sphere of inner radius 'x' and outer radius 'x+dx'. When current is radially outward, the length of differential element is 'dx'.
• Summing of all these elements made of thin sphere give the entire sphere.
• Resistance of whole sphere is the sum of resistance due to all these elements made of thin sphere.

Differential Element

Case 3 : Truncated Cone

• Consider a cone of radius 'a' and 'b' for the right and left end and height 'h'.

• To calculate the resistance of a truncated cone, choose a differential element of cone made of thin disk of radius 'r' at a distance 'x' from the left end.
• Summing up all these small elements of cone made of thin disk give the entire cone.
• Resistance of whole cone is the sum of resistance due to all these small elements of thin disk.

Differential Element

• Consider a hollow cylinder of length L and inner and outer radii be a and b respectively, as shown in figure.
• The material has resistivity \(\rho\).

• Potential difference is applied between the inner and outer surface such that current flows radially outward.
• To calculate resistance, consider a differential element made up of thin hollow cylinder of inner radius 'r' and outer radius 'r+dr'.

• Resistance of the differential element (dR) is given as 

\(dR=\dfrac {\rho\,dr}{2\pi\,rL}\)

Integrating both sides,

or, \(\int\limits_0^RdR=\displaystyle\int\limits_a^b\dfrac {\rho\,dr}{2\pi\, r\,L}\)

or, \(R=\dfrac {\rho}{2\pi\, L}\displaystyle\int\limits_a^b\dfrac {1}{r}\,dr\)

\(R=\dfrac {\rho}{2\pi\, L}\ell n \left (\dfrac {b}{a} \right)\)

Consider a hollow cylinder of inner radius \(r_1\) and outer radius \(r_2\) of length \(L\) as shown in figure.

The area of hollow cylinder = Area of shaded region

\(A=\pi\,\left(r_2^2-r_1^2\right)\)

Resistance, \(R=\dfrac{\rho L}{A}=\dfrac{\rho L}{\pi\,\left(r_2^2-r_1^2\right)}\,\)

• Consider a rod of length \(\ell\), cross-sectional area \(A\), its resistivity is varied linearly as a function.

\(\rho=\rho_0+ax\)

where,

\(\rho_0\) and \(a\) are constant

\(\rho_{(x=0)}=\rho_0\)

\(\rho_{(x=\ell)}=\rho_0+a\ell\)

\(\rho_{avg}=\dfrac {\rho_{(x=0)}+\rho_{(x=\ell)}}{2}\)

\(\rho_{avg}=\dfrac {\rho_0+\rho_0+a\ell}{2}\)

\(\rho_{avg}=\rho_0+\dfrac {a\ell}{2}\)

• Average method is applied only when resistivity is varied linearly.

\(R=\dfrac {\rho_{avg}\cdot \ell}{A}\)

or, \(R=\dfrac { \left ( \rho_0+\dfrac {a\ell}{2} \right)\ell} {A}\)

or, \(R=\dfrac {2\rho_0\ell+a\ell^2} {2A}\)

• Consider a hollow sphere of inner radius 'a' and outer radius 'b', as shown in figure.

• Potential difference is applied between the inner and outer surface such that the current flows radially outward. The material has resistivity \(\rho\).
• To calculate the resistance, consider a differential element made up of a thin hollow sphere of inner radius 'x' and outer radius 'x+dx'

• The resistance of differential element (dR) is given by

\(dR=\dfrac {\rho\;dx}{4\pi\,x^2}\)                \(\left [ \therefore R=\dfrac {\rho\,L}{A} \right ]\)

Integrating both sides,

\(\int dR=\displaystyle \int \dfrac {\rho\,dx}{4\pi\,x^2}\)

or,  \(\int\limits_0^R dR=\displaystyle\int\limits_a^b \dfrac {\rho\,dx}{4\pi\,x^2}\)

or, \(R=\dfrac {\rho}{4\pi}\displaystyle\int\limits_a^b \dfrac {1}{x^2}dx\)

or, \(R=\dfrac {\rho}{4\pi} \left [ -\dfrac {1}{x} \right]_a^b \)

or, \(R=\dfrac {\rho}{4\pi} \left [ \dfrac {1}{a} -\dfrac {1}{b} \right]\)

\(R=\dfrac {\rho}{4\pi} \left [ \dfrac {b-a}{ab} \right]\)

Truncated Cone: The truncated cone is a solid, similar to cylinder except that the circular and planes are not of same size.

• Center points are aligned along a straight line.
• Consider a truncated cone of height 'h' and radii 'a' and 'b' for the right and left end respectively, as shown in figure

• Consider a differential element of a truncated cone in a form of disk of radius 'r' at a distance 'x' from the left end and of thickness 'dx'.

Calculation of Resistance

• \(For\) \(\Delta PQC\)

\(tan\,\theta=\dfrac {b-r}{x}\) ...(i)

• \(For\) \(\Delta RSC\)

\(tan\,\theta=\dfrac {b-a}{h}\) ...(ii)

\(From\, (i) \,and \,(ii)\)

\(\dfrac {b-r}{x}=\dfrac {b-a}{h}\)

or, \(r=(a-b)\dfrac {x}{h}+b\)

• The resistance of small element is given by, 

               \(dR=\dfrac {\rho\, dx}{\pi\;r^2}\)

Integrating both sides,

\(\int\limits_0^R{dR}=\displaystyle\int\limits_0^h \dfrac {\rho\;dx} {\pi \left [ (a-b)\dfrac {x}{h}+b \right]^2} \)

Let \(\dfrac {a-b}{h}=\alpha,\; b=\beta\)   ...(iii)

\(R=\dfrac {\rho}{\pi} \left [ \;\displaystyle\int\limits_0^h\dfrac {dx}{(\alpha x+\beta)^2} \;\right]\)

\(\therefore\) As we know,  \(\displaystyle\int\dfrac {dx}{(\alpha x+\beta)^2} =\dfrac {-1}{\alpha(\alpha x+\beta)}\)

\(R=\dfrac {\rho}{\pi} \left [ \dfrac {-1}{\alpha(\alpha x + \beta)} \right]_0^h\)

or, \(R=\dfrac {\rho}{\pi} \left [ \dfrac {-1}{\alpha(\alpha h + \beta)} +\dfrac {1}{\alpha \beta}\right]\)

or, \(R=\dfrac {\rho}{\pi} \left [ \dfrac {-\beta+\alpha h+\beta}{\alpha(\alpha h + \beta)\beta} \right ]\)

or, \(R=\dfrac {\rho}{\pi} \left [ \dfrac {h}{(\alpha h + \beta)\beta} \right ]\)

\(From\, equation\, (iii)\)

\(R=\dfrac {\rho\,h}{\pi} \left [ \dfrac {1}{[a-b+b]b} \right ]\)

\(R=\dfrac {\rho\,h}{\pi\;ab}\)