Informative line

### 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.

#### 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 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 = $$b\;\ell$$

#### 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

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.

A

20 $$\Omega\;m$$

.

B

15 $$\Omega\;m$$

C

30 $$\Omega\;m$$

D

40 $$\Omega\;m$$

Option C 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.

## 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.

## 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

#### 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
B
C
D

Option C 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, $$R=\dfrac {\rho}{2\pi\, L}\displaystyle\int\limits_a^b\dfrac {1}{r}\,dr$$

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

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

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

#### 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

$$A=\pi\;r_2^2-\pi\;r_1^2$$

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

Given, r1 = a , r2 = b

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

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

### 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 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, $$R=\dfrac {2\rho_0\ell+a\ell^2} {2A}$$

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

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

$$\rho_{avg}=\dfrac {2+27}{2}=\dfrac {29}{2}$$

$$\rho_{avg}=14.5\,\Omega\,m$$

Resistance of rod

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

### 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

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

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

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

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

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

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

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

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

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

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

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

#### 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.

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'.

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)

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

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

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

$$R$$ = $$\dfrac{\pi×8}{\pi×4×2}$$

= 1 $$\Omega$$

### 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