Electrical Resistivity (ρ) is the fundamental property of a material that quantifies how strongly it opposes the flow of electric current. It is defined as the resistance of a conductor having unit length and unit cross-sectional area.
\rho = \frac{RA}{L} where,
- ρ = resistivity of the material =
- R = resistance
- l = length
- A = cross-sectional area
It depends on factors like temperature, the nature of the material, and impurities, but not on the size or shape of the conductor. Materials with low resistivity (like metals) are excellent conductors, while those with high resistivity (like insulators) resist the flow of current.
We can also determine the resistivity of a material using the electric field and current density as follows:
\rho = \frac{E}{J} where,
- ρ = resistivity of the material (Ω·m)
- E = electric field (V/m)
- J = current density (A/m²)
Learn more about, Resistance Formula
Unit of Resistivity
- The SI unit of resistivity is ohm-meter (Ω·m).
- In the CGS system, the unit is ohm-centimeter (Ω·cm).
- The dimensional formula of resistivity is [M1L3T-3A-2 ]
Resistivity of Different Materials
The table for resistivity and conductivity for various materials is given as follows:
Material | Resistivity, ρ (Ω⋅m) | Conductivity, σ (Ω⋅m−1) |
|---|---|---|
Conductors | ||
| Silver | 1.59×10−8 | 6.29×107 |
| Copper | 1.68×10−8 | 5.95×107 |
| Aluminum | 2.65×10−8 | 3.77×107 |
| Iron | 9.71×10−8 | 1.03×107 |
| Nichrome (Ni, Fe, Cr alloy) | 100.00×10−8 | 0.10×107 |
| Platinum | 10.60×10−8 | 0.94×107 |
| Tungsten | 5.60×10−8 | 1.79×107 |
Semiconductors | ||
| Carbon | (3.5−60)×10−5 | (2.86−1.67)×10⁻⁶ |
| Silicon | 0.1 - 2300 | - |
Insulators | ||
| Glass | 109−1014 | 10−9−19−14 |
| Mica | 1011−1015 | 10−11−10−15 |
| Wood | 108−1011 | 10−8−10−11 |
Relation of Resistance and Resistivity
Resistance in a conductor can be understood using the analogy of water flow in a pipe. Just as it is harder for water to flow through a longer pipe, the resistance increases with the length of the conductor. Similarly, a wider pipe allows more water to flow easily, so resistance decreases with an increase in the cross-sectional area.
In the same way, the resistance offered by a conductor is directly proportional to its length and inversely proportional to its cross-sectional area.
Thus, the resistance of the conductor is given by,
R \propto \frac{l}{A}
R = \rho \frac{l}{A} where:
- R = resistance of the conductor (Ω)
- l = length of the wire (m)
- A = cross-sectional area of the wire (m²)
- ρ = resistivity of the material (Ω·m)
Specific Resistivity
Specific resistivity is the resistivity of a material when its length and cross-sectional area are both unity. It represents how strongly a material opposes the flow of electric current independent of its dimensions.
In other words, it is the resistance of a conductor of unit length and unit cross-sectional area.
From the relation:
R = \frac{\rho l}{A} If l = 1 and A = 1 then:
R = \rho
Resistor Colour Coding
Resistors are used in electrical circuits to control and limit the flow of current. The value of resistance in a resistor is indicated by a series of colored bands printed on its surface. Each color corresponds to a specific numerical value, and together these bands determine the resistance and its accuracy.
In a typical four-band resistor, the first two bands represent the significant digits, the third band indicates the multiplier, and the fourth band shows the tolerance. For example, black represents 0, brown represents 1 (or acts as a multiplier of 10), and a brown band also indicates a tolerance of ±1%.
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Sample Problems
Question 1: A battery of 5 Volts connected to a conductor induces a current of 10 mA in the conductor. Find the resistance of the conductor.
Solution: Resistance of conductor is given by the relation,
R = V/I
Given:
V = 5 V
I = 10 mA = 0.01 A
Plugging in the values inside the relation,
R = V/I
⇒ R = (5)/(0.01)
⇒ R= 500 Ω
Question 2: A battery of 20 Volts connected to a conductor induces a current of 50 mA in the conductor. Find the resistance of the conductor.
Solution: Resistance of conductor is given by the relation,
R = V/I
Given:
V = 20 V
I = 50 mA = 0.05 A
Plugging in the values inside the relation,
R = V/I
⇒ R = (20)/(0.05)
⇒ R= 400 Ω
Question 3: A battery of 100 Volts connected to a material induces a current of 0 mA in the conductor. Find the resistance of the material.
Solution: Resistance of conductor is given by the relation,
R = V/I
Given:
V = 100 V
I = 0 mA = 0 A
Plugging in the values inside the relation,
⇒ R = (100)/(0) = Not defined
Resistance approaches to infinity, which means the material is an insulator.
Question 4: A cylindrical conductor of length 0.5m and cross-sectional area 0.01 m2 and resistivity = 2 x 10 -8 ohm-m. Find the resistance of the material.
Solution: Resistance of a conductor is given by,
ρ = (R×A)/l
Given:
l = 0.5m
A = 0.01 m2
ρ = 2 x 10-8
Plugging the values in the relation given above,
⇒ R =
\frac{2 \times 10^{-8} 0.5}{0.01}
R = \frac{1 \times 10^{-8}}{10^{-2}} ⇒
R = 1 \times 10^{-6}\,\Omega
Question 5: A cylindrical conductor of length 2 m and cross-sectional area 0.05 m2 and resistivity = 4 x 10 -8 ohm-m. Find the resistance of the material.
Solution: Resistance of a conductor is given by,
ρ = (R×A)/l
Given:
l = 2m
A = 0.05 m2
ρ = 4 × 10-8
Plugging the values in the relation given above,
R = \frac{4 \times 10^{-8} 2}{0.05}
R = \frac{8 \times 10^{-8}}{5 \times 10^{-2}}
R = 1.6 \times 10^{-6}\,\Omega
Unsolved Problems
Question 1: A wire carries a current of 2 A when a potential difference of 10 V is applied across it. Find the resistance of the wire.
Question 2: A conductor has a resistance of 50 Ω and a cross-sectional area of 0.02 m². If its length is 1 m, find the resistivity of the material.
Question 3: A wire of length 3 m and cross-sectional area 0.01 m² has a resistivity of 1.5 × 10⁻⁸Ω·m.-8Ω·m. Find its resistance.
Question 4: If the electric field in a material is 5 V/m and the current density is 2 A/m², find the resistivity of the material.
Question 5: A conductor of resistance 20 Ω is stretched to double its original length. What will be the new resistance (assuming volume remains constant)?