Learn definition and concept of resistance, practice current and resistance equation, and calculate resistance of truncated cone, hollow sphere & cylinder and resistance of a rod whose length varies linearly as a function.

- 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 \(\rho\).

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

For face (1) or (4)

Length = \(\ell\) , Area = bh

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

For Face (2) or face (5)

Length = b, Area = \(\ell h\)

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

For Face (3) or face (6)

Length = h, Area = \(b\;\ell \)

A 20 \(\Omega\;m\)

B 15 \(\Omega\;m\)

C 30 \(\Omega\;m\)

D 40 \(\Omega\;m\)

- 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.

- 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.

- 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.

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

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

A \(0.5\) \(\ell n\,(2)\,\Omega\)

B \(\ell n\,(2)\,\Omega\)

C \(3\) \(\ell n\,(2)\,\Omega\)

D \(8\) \(\ell n\,(2)\,\Omega\)

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

A \(R=\dfrac {\rho\;L}{\pi(b^2-a^2)}\)

B \(R=\dfrac {\rho\;L}{\pi\; b^2}\)

C \(R=\dfrac {\rho\;L}{\pi\; a^2}\)

D \(R=\dfrac {\rho(b^2-a^2)}{L}\)

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

A 30 \(\Omega\)

B 20 \(\Omega\)

C 21 \(\Omega\)

D 24.16 \(\Omega\)

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

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

A 0.25 \(\Omega\)

B 5 \(\Omega\)

C 1 \(\Omega\)

D 2 \(\Omega\)

**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'.

- \(For\) \(\Delta PQC\)

\(tan\,\theta=\dfrac {b-r}{x}\) ...(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, \(R=\dfrac {\rho}{\pi} \left [ \dfrac {h}{(\alpha h + \beta)\beta} \right ]\)

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

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

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

A 1 \(\Omega\)

B 3 \(\Omega\)

C 2 \(\Omega\)

D 4 \(\Omega\)