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Resistance

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.

Concept of Resistance

  • 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.

Illustration Questions

Which one of the following wires will have the highest resistance? Wire 1 : Length = L m, Area = Am2, Resistivity =  \(\rho\;\Omega\) m Wire 2 : Length = 2 L m, Area = Am2, Resistivity =  \(\rho\;\Omega\) m Wire 3 : Length = L/2 m, Area = 2Am2, Resistivity = 2 \(\rho\;\Omega\) m Wire 4 : Length = L/4 m, Area = Am2, Resistivity =  2 \(\rho\;\Omega\) m

A Wire 1

B Wire 2

C Wire 3

D Wire 4

×

For resistance of Wire 1,

\(R_1=\dfrac {\rho_1\;L_1}{A_1}=\dfrac {\rho\;L}{A}\;\Omega\)

For resistance of Wire 2,

\(R_2=\dfrac {\rho_2\;L_2}{A_2}=\dfrac {\rho\;×2L}{A}\)

\(=\dfrac {2\;\rho\;L}{A}=2R_1\;\Omega\)

For resistance of Wire 3,

\(R_3=\dfrac {\rho_3\;L_3}{A_3}=\dfrac {2\rho\;×(L/2)}{2A}\)

\(=\dfrac {1}{2}\dfrac {\rho\;L}{A}=\dfrac {R_1}{2}\;\Omega\)

For resistance of Wire 4

\(R_4=\dfrac {\rho_4\;L_4}{A_4}=\dfrac {2\rho\;×(L/4)}{A}\)

\(=\dfrac {1}{2}\dfrac {\rho\;L}{A}=\dfrac {R_1}{2}\;\Omega\)

Hence, resistance of wire 2 is the highest.

Which one of the following wires will have the highest resistance? Wire 1 : Length = L m, Area = Am2, Resistivity =  \(\rho\;\Omega\) m Wire 2 : Length = 2 L m, Area = Am2, Resistivity =  \(\rho\;\Omega\) m Wire 3 : Length = L/2 m, Area = 2Am2, Resistivity = 2 \(\rho\;\Omega\) m Wire 4 : Length = L/4 m, Area = Am2, Resistivity =  2 \(\rho\;\Omega\) m

A

Wire 1

.

B

Wire 2

C

Wire 3

D

Wire 4

Option B is Correct

Resistance of Hollow Cylinder

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

Illustration Questions

Consider a hollow cylinder of inner radius \(r_1=a\) and outer radius \(r_2=b\) respectively. The length of the cylinder is \(L\)= \(\ell\). Determine the resistance of the hollow cylinder, if the resistivity is \(\rho\).

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

×

The area of hollow cylinder = Area of shaded region

\(A=\pi\;r_2^2-\pi\;r_1^2\)

\(A=\pi(r_2^2-r_1^2)\)

image

Given, r1 = a , r2 = b

\(A=\pi(b^2-a^2)\)

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

image

Consider a hollow cylinder of inner radius \(r_1=a\) and outer radius \(r_2=b\) respectively. The length of the cylinder is \(L\)= \(\ell\). Determine the resistance of the hollow cylinder, if the resistivity is \(\rho\).

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

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

Option A is Correct

Resistance of Cuboid

  • 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 = \(b\;\ell \)

 

Illustration Questions

A cuboid of length \(\ell =\)3 m, height h = 2m and breadth b = 5m is placed in x-y-z plane in such a way that current I is passing through it as shown in figure. If the resistance in the direction is R = \(4\;\Omega\), then calculate resistivity.

A 20 \(\Omega\;m\)

B 15 \(\Omega\;m\)

C 30 \(\Omega\;m\)

D 40 \(\Omega\;m\)

×

The current is passing through cuboid as shown in figure.

Length  \(\ell\,'=2\;m\)

Area = 5 × 3 = 15 m2

image

Resistance, 

\(R=\dfrac {\rho\;\ell'}{A}\)

\(4=\dfrac {\rho\;×2}{15}\)

\(6\,Q=2\;\rho\)

\(\rho=30\;\Omega\;m\)

A cuboid of length \(\ell =\)3 m, height h = 2m and breadth b = 5m is placed in x-y-z plane in such a way that current I is passing through it as shown in figure. If the resistance in the direction is R = \(4\;\Omega\), then calculate resistivity.

image
A

20 \(\Omega\;m\)

.

B

15 \(\Omega\;m\)

C

30 \(\Omega\;m\)

D

40 \(\Omega\;m\)

Option C is Correct

Resistance of a Rod whose Length Varies Linearly as a Function

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

Illustration Questions

Calculate the value of resistance of a rod of length \(\ell =\)5m and cross sectional area A = 3m2 whose resistivity is varied linearly along its length as \(\rho (x)=2+5x\)  where \(x\) is in meter.

A 30 \(\Omega\)

B 20 \(\Omega\)

C 21 \(\Omega\)

D 24.16 \(\Omega\)

×

Given : \(\rho_{(x=0)}=\) \(2\)  \(\rho_{(x=5)}=2+5×5\)

\(\rho_{avg}=\dfrac {[\rho_{(x=0)}+\rho_{(x=5)} ]}{2}\)

\(\rho_{avg}=\dfrac {2+27}{2}=\dfrac {29}{2}\)

\(\rho_{avg}=14.5\,\Omega\,m\)

image

Resistance of rod

\(R=\dfrac {\rho_{avg}\,L}{A}=\dfrac {14.5×5}{3}\\=\dfrac {72.5}{3}=24.16\,\Omega\)

image

Calculate the value of resistance of a rod of length \(\ell =\)5m and cross sectional area A = 3m2 whose resistivity is varied linearly along its length as \(\rho (x)=2+5x\)  where \(x\) is in meter.

A

30 \(\Omega\)

.

B

20 \(\Omega\)

C

21 \(\Omega\)

D

24.16 \(\Omega\)

Option D is Correct

Selection of Differential Element to Calculate Resistance

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

Illustration Questions

To calculate the resistance of a hollow sphere of some inner and outer radius, the correct choice of differential element will be (when current is radially outward).

A

B

C

D

×

The correct choice of differential element of a sphere will be a thin hollow sphere.

So, option (C) is correct.

To calculate the resistance of a hollow sphere of some inner and outer radius, the correct choice of differential element will be (when current is radially outward).

A image
B image
C image
D image

Option C is Correct

Resistance of a Hollow Sphere Along Radial Direction

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

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

Illustration Questions

Determine the value of resistance of a hollow sphere of resistivity \(\rho=8\pi\;\Omega \,m\), whose inner radius a = 4m and outer radius b = 8 m, if the current is flowing radially outward.

A 0.25 \(\Omega\)

B 5 \(\Omega\)

C 1 \(\Omega\)

D 2 \(\Omega\)

×

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

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

image

Resistance of a hollow sphere is given by

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

where,

a and b are inner and outer radius and  \(\rho=\) resistivity

image

Given : \(\rho=8\pi\;\Omega\, m\)     \(a = 4 \,m\,\,\,\,\,\,\, b = 8\, m\)

\(R=\dfrac {8\pi}{4\pi} \left [ \dfrac {8-4}{8×4} \right]=\dfrac {1}{4}\\=0.25\,\Omega\)

image

Determine the value of resistance of a hollow sphere of resistivity \(\rho=8\pi\;\Omega \,m\), whose inner radius a = 4m and outer radius b = 8 m, if the current is flowing radially outward.

A

0.25 \(\Omega\)

.

B

\(\Omega\)

C

\(\Omega\)

D

\(\Omega\)

Option A is Correct

Resistance of Hollow Cylinder along Radial direction 

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

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

Illustration Questions

Consider a hollow cylinder of length \(\ell=4\,m\) and inner and outer radii \(a = 2\,m\) and \(b = 4\,m\) respectively. Find the value of resistance given, the material has resistivity \(\rho=4\,\pi\,\Omega\, m\). (Assume flow of current is radially outward)

A \(0.5\) \(\ell n\,(2)\,\Omega\)

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

C \(3\) \(\ell n\,(2)\,\Omega\)

D \(8\) \(\ell n\,(2)\,\Omega\)

×

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

image

Resistance of hollow cylinder of length \(\ell\) and inner and outer radii a and b respectively. is given as 

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

Given: \(L = 4\,m\,\,\,\,\,\,\, a = 2\,m\,\,\,\,\,\,\,b = 4\,m\)   \(\rho=4\pi\,\Omega m\)

\(R=\dfrac {4\pi}{2\pi\, ×4}\;\ell n \left (\dfrac {4}{2} \right)\)

= \(0.5\) \(\ell n\,(2)\,\Omega\)

Consider a hollow cylinder of length \(\ell=4\,m\) and inner and outer radii \(a = 2\,m\) and \(b = 4\,m\) respectively. Find the value of resistance given, the material has resistivity \(\rho=4\,\pi\,\Omega\, m\). (Assume flow of current is radially outward)

A

\(0.5\) \(\ell n\,(2)\,\Omega\)

.

B

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

C

\(3\) \(\ell n\,(2)\,\Omega\)

D

\(8\) \(\ell n\,(2)\,\Omega\)

Option A is Correct

Resistance of Truncated Cone

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

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

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

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

Illustration Questions

A truncated cone has an altitude \(h = 8 \,m\) and radii \(a = 2\,m\) and \(b = 4\,m\). Calculate the value of resistance, if the value of resistivity is \(\rho=\pi\;\Omega m\).

A 1 \(\Omega\)

B 3 \(\Omega\)

C 2 \(\Omega\)

D 4 \(\Omega\)

×

Consider a truncated cone of height 'h' and radii 'a' and 'b' for the right and left end respectively, as shown in figure.

image

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

image image

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, \(r=(a-b)\dfrac {x}{h}+b\)

image image

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)

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

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

image

Given: \(h = 8 \,m\,\,\,\,\,\,\,\,a = 2 \,m\,\,\,\,\,\,\,b = 4 \,m\)    \(\rho=\,\pi\,\Omega\,m\)

\(R\) = \(\dfrac{\pi×8}{\pi×4×2}\)

= 1 \(\Omega\)

image

A truncated cone has an altitude \(h = 8 \,m\) and radii \(a = 2\,m\) and \(b = 4\,m\). Calculate the value of resistance, if the value of resistivity is \(\rho=\pi\;\Omega m\).

A

\(\Omega\)

.

B

\(\Omega\)

C

\(\Omega\)

D

\(\Omega\)

Option A is Correct

Practice Now