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 ( $E$ )