19.7 RESISTIVITY
19.7.1 Resistivity
The electrical conduktivity $\sigma$ ia a measure of a material's ability to conduct electric current through the material. A material possessing a very large value of $\sigma$ can conduct electricity very well. This being the case, then the reciprocal of $\sigma$ must be a measure of a material's ability to oppose the flow of electric current throught the material. The reciprocal of $\sigma$ is known as the resistivity, $\rho$, of the material, that is
$\boxed{\rho=\frac{1}{\sigma}}$
19.7.2 Relationship between ρ and R
A p.d. V is applied across length L of an ohmic conductor of uniform cross-sectional area A and electrical conductivity σ. Current I flows in the conductor. For an ohmic conductor, we have the relationship
$V=\left(\frac{L}{\sigma A}\right)I$
But,
$V = IR$
Hence,
$R = \frac{L}{\sigma A}$
or,
$\boxed{R = \frac{\rho L}{A}}$
2. Definition of ρ
Let L = 1 m, A = 1 m². Then R = ρ.
The resistivity of a material is numerically equal to the resistance of a material of length 1 m and uniform cross-sectional area 1 m².
3. Unit of ρ
Unit of ρ is $\Omega \ m^2/m ⇒ \Omega m$
19.7.3 Effect of Temperature on Resistance
1. Metallic conductor
$R=\frac{L}{\sigma A}$
If L and A are constant, we will have
$R \propto \frac{1}{\sigma}$
The electrical conductivity σ is given by
$\sigma=\frac{ne^2\tau}{2m}$
For a metallic conductor,
(a) n does not change much when temperature increases. Hence, we may assume n to be a constant.
(b) As temperature increases, the frequency of collision of conduction electrons with atoms also increases, causing the mean free time τ to decrease. A decrease in τ will cause σ to decrease. Hence if the temperature of an ohmic conductor increases, its resistance will increase.
The graph in Figure 19.9 shows how the resistance of a platinum wire of length 2.0 m and diameter 0.1 mm varies with temperature over a temperature range of 100°C. The increase in resistance is linear.
2. Semiconductor
(a) For an intrinsic semiconductor, n increases rapidly when temperature increases.
(b) A large increase in the value of n makes n more predominant than τ, whose decrease in value is comparatively much smaller. The overall result is that σ effectively increases with temperature. This means that the resistance of a semiconductor decreases as temperature increases, as shown in Figure 19.10.
Worked Examples on Resistivity, Resistance, and Ohm’s Law with Step-by-Step Solutions
Example 19.9
If a current of 1.5 A flows in a constantan wire of length 1.0 m and cross-sectional area 0.50 mm², determine the p.d. across the two ends of the wire. (Resistivity of constantan = 4.9 × 10⁻⁷ Ωm)
Answer
The resistance of a wire is given by the formula:
$R = \frac{\rho L}{ A}$
Substitute the values:
$R = \frac{(4.9 \times 10^{-7})(1.0)}{0.50 \times 10^{-6}} \cong 0.98 \ \Omega$
Resistance depends on the resistivity of the material (), the length of the wire (longer wire → higher resistance), and the cross-sectional area (larger area → lower resistance).
Step 2: Apply Ohm’s Law to find potential difference
Ohm’s Law states:
$V = IR$
Substitute the values:
$R = (1.5)(0.98) = 1.47 \ V$
The potential difference is the voltage required to push the given current (I) through the wire with resistance (R).
Example 19.10
A p.d. of 10 V is applied across the two ends of a wire of length 50 cm and diameter 3.0 mm. If the resistivity of the material of the wire is 3.0 × 10⁻⁵ Ω m, determine the current which flows in the wire.The cross-sectional area of a wire is:
$A=\frac{1}{2}\pi d^2$
Step 1: Convert diameter to radius
The cross-sectional area of a wire is:
The resistance R of a uniform wire is given by:
$R = \frac{\rho L}{A}$
$R = \frac{(3.0 \times 10^{-5}) (0.5)}{7.07 \times 10^{-6}} \cong 2.12 \Omega$
Resistance depends on resistivity (material property), length (longer → more resistance), and cross-sectional area (larger → less resistance).
Step 4: Use Ohm’s Law to find current
Ohm’s Law states:
$I=\frac{V}{R}=\frac{10}{2.12}\cong 4.72 \ A$
Current is the flow of charge through the wire, and it is determined by the potential difference divided by the resistance.
Example 19.11
A 10.0-m length of wire consists of 5.0 m of copper
followed by 5.0 m of aluminum, both of diameter 1.4 mm.
A voltage difference of 95 mV is placed across the composite
wire.
(a) What is the total resistance (sum) of the two wires?
(b) What is the current through the wire?
(c) What are the
voltages across the aluminum part and across the copper part?
Answer
Given
Length of copper wire, $L_{Cu}=5.0 \ m$
Length of aluminum wire, $L_{Al}=5.0 \ m$
Diameter of wire, $d = 1.4 \ mm = 1.4 \times 10^{-3} \ m $
Voltage across composite wire, $V_{total} = 95 \ mV = 95 \times 10^{-3} \ V$
Resistivity of copper, $\rho_{Cu}=1.68 \times 10^{-8} \Omega m$
Resistivity of aluminium, $\rho_{Al}=2.82 \times 10^{-8} \Omega m$
Step 1: Compute the cross-sectional area
$A=\frac{1}{4}\pi d^2 = \frac{1}{4}\pi (1.4 \times 10^{-3})^2=1.54 \times 10^{-6} \ m^2$
The area is the same for both wires, which affects their individual resistances
(a) Step 2: Compute the resistance of each wire
Copper:
$R_{Cu} = \frac{\rho_{Cu} L_{Cu}}{A}$
$R_{Cu} = \frac{(1.68 \times 10^{-8}) (5.0)}{1.54 \times 10^{-6}} \cong 0.0545 \Omega$
Aluminum:
$R_{Al} = \frac{\rho_{Al} L_{Al}}{A}$
$R_{Al} = \frac{(2.82 \times 10^{-8}) (5.0)}{1.54 \times 10^{-6}} \cong 0.0916 \Omega$
Resistance is proportional to resistivity and length, and inversely proportional to cross-sectional area.
Step 3: Compute total resistance
The wires are in series, so resistances add up:
$R_{total}=R_{Cu}+R_{Al}=0.0545 +0.0916=0.146 \Omega$
(b) Step 4: Compute the current through the wire
Using Ohm’s Law:
$I=\frac{V_{total}}{R_{total}}=\frac{0.095}{0.146}\cong 0.651 \ A$
Explanation: In series, the same current flows through both wires.
(c) Step 5: Compute voltage across each wire
Voltage across a wire in series:
$V=IR$
Copper:
$V_{Cu}=IR_{Cu}=(0.651)(0.0545) \cong 0.0355 \ V = 35.5 \ mV.$
Aluminium:
$V_{Al}=IR_{Al}=(0.651)(0.0916) \cong 0.0596 \ V = 59.6 \ mV.$
Check: $V_{Cu}+V_{Al}=0.0355+0.0596 \cong 0.095 \ V$
Example 19.12
A solid cube of silver (density $\rho =10.5 \ g/cm^3$) has a mass of 90.0 g. (a) What is the resistance between opposite faces of the cube? (b) Assume each silver atom contributes one conduction electron. Find the average drift speed of electrons when a potential difference of $1.00 \times 10^5 \ V$ is applied to opposite faces. The atomic number of silver is 47, and its molar mass is 107.87 g/mol.
Answer
Given
Mass of silver cube, $m = 90 \ g = 90 \times 10^{-3} \ kg$
Molar mass of silver, $\rho_m=10.5 \ g/cm^3=10.500 \ kg/m^3$
Applied potential difference, $V=1.00 \times 10^5 \ V$
Molar mass of silver, $M = 107.87 \ g/mol$
Atomic number, $Z=47$
Resistivity of silver, $\rho=1.59 \times 10^{-8} \ \Omega m$
Electron charge, $e = 1.6 \times 10^{-19} \ C$
Avogadro’s number, $N_A=6.022 \times 10^{23}/mol$
Step (a) – Resistance between opposite faces
Step 1: Find the volume of the cube
$V_{cube}=\frac{m}{\rho_m}$
$V_{cube}==\frac{0.090}{10500}=8.57 \times 10^{-6} \ m^3$
Step 2: Find the cube side length
$L=V^{1/3}=(8.57 \times 10^{-6})^{1/3}=0.0205 \ m = 2.05 \ cm$
Step 3: Compute resistance
For a cube, resistance between opposite faces:
$R=\rho \frac{L}{A}$
Cross-sectional area of one face:
$A=L^2 = (0.0205)^2=4.20 \times 10^{-4} \ m^2$
$R=(1.59 \times 10^{-8}) \frac{0.0205}{4.20 \times 10^{-4}}\cong 7.76 \times 10^{-7} \ \Omega$
Silver is highly conductive, so the resistance is extremely small.
Step (b) – Average drift speed of electrons
Step 1: Find number of conduction electrons per unit volume
$n=\frac{number \ of \ atoms}{volume} \times electrons \ per \ atom$
Number of moles:
$n_{mol}=\frac{m}{M}=\frac{90.0}{107.87}=0.834 \ mol$
Number of atoms:
$N_{atoms}=n_{mol}.N_A=(0.834)(6.022 \times 10^{23}) \cong 5.02 \times 10^{23} \ atoms$
Number density:
$n=\frac{5.02 \times 10^{23}}{8.57 \times 10^{-6}}\cong 5.86 \times 10^{28} \ electrons/m^3$
Step 2: Compute drift velocity
$v_d=\frac{I}{neA}$
First, compute current using Ohm’s law:
$I=\frac{V}{R}=\frac{1.00 \times 10^5}{7.76 \times 10^{-7}}=1.29 \times 10^{11} \ A$
This is extremely high because the resistance is tiny – idealized calculation.
$v_d=\frac{1.29 \times 10^{11}}{(5.86 \times 10^{28})(1.6 \times 10^{-19})(4.20 \times 10^{-4})}$
$v_d \cong 0.336 \ m/s$
The drift velocity is the average speed of electrons in the direction of the electric field, which is much slower than the actual electron motion.
Applications of Resistivity in Daily Life and Technology
Resistivity is an important physical property that determines how strongly a material opposes the flow of electric current. It plays a crucial role in electrical engineering, electronics, and modern technology. Understanding resistivity helps engineers design efficient systems and choose suitable materials for specific applications.
What is Resistivity?
Resistivity is a measure of how much a material resists the flow of electric current. It is represented by the symbol ρ (rho) and is measured in ohm-meter (Ω·m). Materials with low resistivity, such as copper and aluminum, are good conductors, while materials with high resistivity, such as rubber and glass, are insulators.
Key Applications of Resistivity
1. Electrical Wiring and Power Transmission
Resistivity is used to select materials for electrical wiring. Low-resistivity materials like copper and aluminum are used in power transmission lines to minimize energy loss and improve efficiency over long distances.
2. Heating Devices
High-resistivity materials are used in heating appliances such as electric heaters, irons, and toasters. When electric current flows through these materials, electrical energy is converted into heat energy due to resistance.
3. Electronic Components
Resistivity is essential in designing resistors and other electronic components. Engineers use materials with specific resistivity values to control current flow in circuits, ensuring proper functioning of devices like smartphones, computers, and televisions.
4. Temperature Sensors (Thermistors)
Certain materials change their resistivity with temperature. This property is used in thermistors to measure and control temperature in devices such as air conditioners, refrigerators, and industrial systems.
5. Superconductors
At very low temperatures, some materials exhibit zero resistivity and become superconductors. These materials are used in advanced technologies such as MRI machines, maglev trains, and particle accelerators.
6. Material Identification and Quality Control
Resistivity is used to identify materials and check their purity. Different materials have unique resistivity values, making it useful in laboratory testing and industrial quality control.
Importance of Resistivity in Engineering
Understanding resistivity allows engineers to design safe and efficient electrical systems. It helps in minimizing energy loss, preventing overheating, and improving the performance of electrical and electronic devices.
Conclusion
Resistivity is a fundamental property of materials that determines how strongly they oppose the flow of electric current. It provides a deeper understanding of how electrical conduction occurs at the microscopic level and explains why different materials behave differently in electrical circuits.
Materials with low resistivity, such as metals, are excellent conductors and are widely used in electrical wiring and power transmission. In contrast, materials with high resistivity act as insulators and are essential for ensuring safety in electrical systems. The ability to control and manipulate resistivity is crucial in designing efficient and reliable devices.
Overall, resistivity plays a vital role in physics, engineering, and modern technology. From everyday electrical appliances to advanced systems like semiconductors and superconductors, understanding resistivity is key to optimizing performance and energy efficiency in electrical applications.
Frequently Asked Questions (FAQ) about Resistivity
What is resistivity in physics?
Resistivity is a physical property of a material that measures how strongly it opposes the flow of electric current. It is represented by the symbol ρ (rho) and is measured in ohm-meter (Ω·m).
What is the formula for resistivity?
The formula for resistivity is ρ = RA / L, where R is resistance, A is the cross-sectional area, and L is the length of the conductor.
What is the unit of resistivity?
The SI unit of resistivity is ohm-meter (Ω·m).
Why do different materials have different resistivity?
Different materials have different resistivity because of variations in their atomic structure, number of free electrons, and how easily electrons can move through the material.
What are examples of low and high resistivity materials?
Metals such as copper and aluminum have low resistivity and are good conductors, while materials like rubber, glass, and plastic have high resistivity and act as insulators.
How does temperature affect resistivity?
For most conductors, resistivity increases with temperature because increased atomic vibrations make it harder for electrons to move. For semiconductors, resistivity usually decreases as temperature increases.
What are the applications of resistivity?
Resistivity is used in electrical wiring, heating devices, electronic components, temperature sensors, and advanced technologies such as superconductors.
Keywords: resistivity FAQ, what is resistivity, resistivity formula, electrical resistivity explanation, resistivity applications
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