Drift Velocity And Resistance

Introduction to the concept of drift velocity and resistance
Definition of drift velocity as the average velocity of charge carriers in a conductor
Explanation of resistance as the hindrance faced by charge carriers in their motion
Overview of how the drift velocity and resistance are related

Drift Velocity

Definition: the average velocity of charge carriers in a conductor under the influence of an electric field
Symbol: $ v_d $
Unit: m/s
Equation: $ v_d = \frac {I} {nAe} $ , where $ I $ is current, $ n $ is charge carrier density, $ A $ is the cross-sectional area, and $ e $ is the charge of an electron

Resistance

Definition: the opposition offered by a substance or a circuit to the flow of electric current
Symbol: $ R $
Unit: Ohm ( $ \Omega $ )
Equation: $ R = \frac {V} {I} $ , where $ V $ is the potential difference across a conductor and $ I $ is the current passing through it

Ohm’s Law

Ohm’s Law relates the current flowing through a conductor to the potential difference across it, given that the temperature and other physical conditions remain constant.
Equation: $ V = IR $ , where $ V $ is potential difference, $ I $ is current, and $ R $ is resistance
Ohm’s Law is applicable to a wide range of conducting materials under normal conditions

Relationship Between Drift Velocity and Resistance

The drift velocity of the charge carriers in a conductor is directly proportional to the applied electric field.
The higher the resistance of a conductor, the lower the drift velocity of charge carriers.
This relationship is explained by the equation: $ R = \rho \frac {L} {A} $ , where $ \rho $ is the resistivity of the material, $ L $ is the length of the conductor, and $ A $ is the cross-sectional area

Factors Affecting Resistance

Length ( $ L $ ) of the conductor: Longer the conductor, higher the resistance
Cross-sectional area ( $ A $ ) of the conductor: Smaller the cross-sectional area, higher the resistance
Resistivity ( $ \rho $ ) of the material: Higher the resistivity, higher the resistance
Temperature ( $ T $ ) of the conductor: Higher the temperature, higher the resistance (in most conductive materials)

Example 1

A copper wire of length 2 m and cross-sectional area 1.5 mm² is connected to a voltage source that produces a potential difference of 12 V across the wire. Calculate the resistance of the wire. Given:

Length ( $ L $ ) = 2 m
Cross-sectional area ( $ A $ ) = 1.5 mm² = $ 1.5 \times 10^{-6} $ m²
Potential difference ( $ V $ ) = 12 V Using the formula $ R = \rho \frac {L} {A} $ , we can calculate $ R $ : $ R = \rho \frac {L} {A} = \rho \frac {2} {1.5 \times 10^{-6}} $ Note: Copper has a resistivity of $ \rho = 1.7 \times 10^{-8} \Omega \cdot \text{m} $ Plugging in the values, we find: $ R \approx 22.67 , \Omega $ (approximately)

Example 2

A wire made of an unknown material has a length of 3 m, a cross-sectional area of 2 mm², and a resistance of 2.5 Ohms. Calculate the resistivity of the material. Given:

Length ( $ L $ ) = 3 m
Cross-sectional area ( $ A $ ) = 2 mm² = $ 2 \times 10^{-6} $ m²
Resistance ( $ R $ ) = 2.5 $ \Omega $ Using the formula $ R = \rho \frac {L} {A} $ , we can rearrange it to solve for resistivity $ \rho $ : $ \rho = R \frac {A} {L} $ Plugging in the values, we find: $ \rho = 2.5 \times \frac {2 \times 10^{-6}} {3} $ So, $ \rho \approx 1.67 \times 10^{-6} , \Omega \cdot \text{m} $ (approximately)

Conductivity

Conductivity ( $ \sigma $ ) is the reciprocal of resistivity ( $ \sigma = \frac {1} {\rho} $ )
It is a measure of how easily a material allows the flow of electric current
Symbol: $ \sigma $
Unit: Siemens per meter (S/m)

Relationship Between Resistance and Conductivity

The resistance ( $ R $ ) of a conductor is inversely proportional to its conductivity ( $ \sigma $ ).
The relationship can be given as: $ R = \frac {\rho} {A} $ , where $ \rho $ is resistivity and $ A $ is cross-sectional area.
Materials with high conductivity tend to have low resistance, and vice versa.

Drift Velocity and Resistance - Problem on Resistance

Example:
- A wire with a resistance of 4 Ohms is connected to a voltage source that produces a potential difference of 12 V across the wire. Determine the current flowing through the wire.
Given:
- Resistance ( $ R $ ) = 4 Ohms
- Potential difference ( $ V $ ) = 12 V
Using Ohm’s Law ( $ V = IR $ ), we can rearrange the equation to solve for current ( $ I $ ):
- $ I = \frac{V}{R} = \frac{12}{4} = 3 $ Amps
Therefore, the current flowing through the wire is 3 Amps.

Effect of Temperature on Resistance

Temperature greatly affects the resistance of a conductor.
Most conductive materials show an increase in resistance with an increase in temperature, known as positive temperature coefficient (PTC).
The relationship between resistance ( $ R $ ) and temperature ( $ T $ ) can be approximated by the equation: $ R = R_0 \left(1 + \alpha(T - T_0)\right) $
- $ R_0 $ is the resistance at a reference temperature ( $ T_0 $ )
- $ \alpha $ is the temperature coefficient of resistance (given in $ ^{\circ}\text{C}^{-1} $ )

Example 3

A wire has a resistance of 100 Ohms at 20 $ ^\circ $ C and 110 Ohms at 50 $ ^\circ $ C. Calculate the temperature coefficient of resistance ( $ \alpha $ ) for the wire.
Given:
- Resistance at 20 $ ^\circ $ C ( $ R_0 $ ) = 100 Ohms
- Resistance at 50 $ ^\circ $ C = 110 Ohms
Using the equation $ R = R_0 \left(1 + \alpha(T - T_0)\right) $ , we can solve for $ \alpha $ :
- $ 110 = 100 \left(1 + \alpha(50 - 20)\right) $
- $ \alpha = \frac{110 - 100}{100 \times (50 - 20)} $
Therefore, the temperature coefficient of resistance is approximately 0.0333 $ ^{\circ}\text{C}^{-1} $ .

Series and Parallel Combination of Resistors

Resistors can be connected in series or parallel to create different resistance values.
Series Combination:
- When resistors are connected in series, the total resistance ( $ R_{\text{total}} $ ) is the sum of individual resistances ( $ R_1, R_2, R_3, \ldots $ ):
  - $ R_{\text{total}} = R_1 + R_2 + R_3 + \ldots $
Parallel Combination:
- When resistors are connected in parallel, the reciprocal of the total resistance ( $ \frac{1}{R_{\text{total}}} $ ) is the sum of the reciprocals of individual resistances:
  - $ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \ldots $
These combinations are commonly used in electrical circuits.

Example 4

Three resistors with values 6 Ohms, 4 Ohms, and 8 Ohms are connected in series. Calculate the total resistance of the combination.
Given:
- Resistor 1 ( $ R_1 $ ) = 6 Ohms
- Resistor 2 ( $ R_2 $ ) = 4 Ohms
- Resistor 3 ( $ R_3 $ ) = 8 Ohms
Using the series combination equation $ R_{\text{total}} = R_1 + R_2 + R_3 $ , we can find:
- $ R_{\text{total}} = 6 + 4 + 8 = 18 $ Ohms

Example 5

Two resistors, 12 Ohms and 6 Ohms, are connected in parallel. Calculate the total resistance of the combination.
Given:
- Resistor 1 ( $ R_1 $ ) = 12 Ohms
- Resistor 2 ( $ R_2 $ ) = 6 Ohms
Using the parallel combination equation $ \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} $ , we can find:
- $ \frac{1}{R_{\text{total}}} = \frac{1}{12} + \frac{1}{6} = \frac{1}{12} + \frac{2}{12} = \frac{3}{12} = \frac{1}{4} $
- Taking the reciprocal of both sides, we get $ R_{\text{total}} = 4 $ Ohms

Kirchhoff’s Junction Rule

Also known as Kirchhoff’s First Law or the Conservation of Charge.
At a junction point in an electrical circuit, the sum of currents entering the junction is equal to the sum of currents leaving the junction.
Mathematical expression: $ \sum I_{\text{in}} = \sum I_{\text{out}} $
This principle follows the law of conservation of charge.

Kirchhoff’s Loop Rule

Also known as Kirchhoff’s Second Law or the Conservation of Energy.
Around any closed loop in an electrical circuit, the sum of potential differences (voltages) across all elements is equal to the sum of electromotive forces (emfs).
Mathematical expression: $ \sum \Delta V_{\text{elements}} = \sum \varepsilon $
This principle follows the law of conservation of energy.

Example 6

Consider the circuit shown, where three resistors ( $ R_1 $ , $ R_2 $ , and $ R_3 $ ) and two sources of emf ( $ \varepsilon_1 $ and $ \varepsilon_2 $ ) are connected in series.
Apply Kirchhoff’s Loop Rule to determine the potential difference across each element:
- $ \Delta V_1 + \Delta V_2 + \Delta V_3 = \varepsilon_1 + \varepsilon_2 $
- $ I \cdot R_1 + I \cdot R_2 + I \cdot R_3 = \varepsilon_1 + \varepsilon_2 $ (using Ohm’s Law $ V = IR $ )
- $ I \cdot (R_1 + R_2 + R_3) = \varepsilon_1 + \varepsilon_2 $
Therefore, the potential difference across each element depends on the emfs and the resistances.

Recap

Drift velocity: Average velocity of charge carriers in a conductor
Resistance: Hindrance offered by a substance or circuit to the flow of electric current
Ohm’s Law: Relationship between current, potential difference, and resistance
Factors affecting resistance: Length, cross-sectional area, resistivity, temperature
Effect of temperature on resistance: Positive temperature coefficient
Series and parallel combination of resistors
Kirchhoff’s Junction Rule: Sum of currents entering the junction equals sum of currents leaving
Kirchhoff’s Loop Rule: Sum of potential differences across elements equals sum of electromotive forces

Example 7

A circuit consists of two resistors, $ R_1 $ and $ R_2 $ , connected in parallel to a voltage source of 12 V. The resistance of $ R_1 $ is 2 Ohms and the resistance of $ R_2 $ is 4 Ohms. Calculate the current flowing through each resistor. Given:
Voltage ( $ V $ ) = 12 V
Resistance $ R_1 $ = 2 Ohms
Resistance $ R_2 $ = 4 Ohms Using Ohm’s Law, we can calculate the current ( $ I_1 $ and $ I_2 $ ) flowing through each resistor: For $ R_1 $ : $ I_1 = \frac {V} {R_1} = \frac {12} {2} = 6 $ Amps For $ R_2 $ : $ I_2 = \frac {V} {R_2} = \frac {12} {4} = 3 $ Amps Therefore, the current flowing through $ R_1 $ is 6 Amps, and the current flowing through $ R_2 $ is 3 Amps.

Factors Affecting Resistance (Recap)

Length ( $ L $ ) of the conductor: Longer the conductor, higher the resistance
Cross-sectional area ( $ A $ ) of the conductor: Smaller the cross-sectional area, higher the resistance
Resistivity ( $ \rho $ ) of the material: Higher the resistivity, higher the resistance
Temperature ( $ T $ ) of the conductor: Higher the temperature, higher the resistance in most conductive materials

Resistivity of Common Materials

Resistivity ( $ \rho $ ) is a material property that determines the resistance of a given volume of material.
Some examples of resistivity values at room temperature (approximately 20 $ ^\circ $ C):
- Copper: $ 1.7 \times 10^{-8} $ $ \Omega \cdot $ m
- Aluminum: $ 2.82 \times 10^{-8} $ $ \Omega \cdot $ m
- Carbon (graphite): $ 3.5 \times 10^{-5} $ $ \Omega \cdot $ m
- Nichrome: $ 1.10 \times 10^{-6} $ $ \Omega \cdot $ m

Example 8

A wire made of an unknown material has a resistivity of $ 2.5 \times 10^{-6} $ $ \Omega \cdot $ m. The wire has a length of 5 m and a cross-sectional area of $ 2 \times 10^{-6} $ m². Calculate the resistance of the wire. Given:
Resistivity ( $ \rho $ ) = $ 2.5 \times 10^{-6} $ $ \Omega \cdot $ m
Length ( $ L $ ) = 5 m
Cross-sectional area ( $ A $ ) = $ 2 \times 10^{-6} $ m² Using the formula $ R = \rho \frac {L} {A} $ , we can calculate the resistance ( $ R $ ): $ R = (2.5 \times 10^{-6}) \frac {5} {2 \times 10^{-6}} = 6.25 $ Ohms Therefore, the resistance of the wire is 6.25 Ohms.

Electric Power

Electric power ( $ P $ ) is the rate at which electric energy is transferred or consumed in a circuit.
Unit: Watt (W)
Equation: $ P = IV $ , where $ I $ is the current flowing through the circuit and $ V $ is the potential difference across the circuit.

Example 9

A circuit has a current of 2 Amps passing through it, and a potential difference of 10 Volts is applied across it. Calculate the electric power consumed by the circuit. Given:
Current ( $ I $ ) = 2 Amps
Potential difference ( $ V $ ) = 10 Volts Using the equation $ P = IV $ , we can calculate the power ( $ P $ ) consumed by the circuit: $ P = (2)(10) = 20 $ Watts Therefore, the electric power consumed by the circuit is 20 Watts.

Electric Energy

Electric energy ( $ E $ ) is the amount of work done or energy transferred by an electric circuit over a period of time.
Unit: Joule (J)
Equation: $ E = Pt $ , where $ P $ is the power and $ t $ is the time.

Example 10

A device with a power rating of 50 Watts is operated for 4 hours. Calculate the electric energy consumed by the device. Given:
Power ( $ P $ ) = 50 Watts
Time ( $ t $ ) = 4 hours Using the equation $ E = Pt $ , we can calculate the electric energy ( $ E $ ) consumed by the device: $ E = (50)(4) = 200 $ Joules Therefore, the electric energy consumed by the device is 200 Joules.

Electric Power and Energy (Recap)

Electric power ( $ P $ ) is the rate at which electric energy ( $ E $ )