Current and Resistance: HRK Physics Complete Guide to Electric Current, Ohm's Law & Resistivity

House of Physics September 30, 2025
Current and Resistance: Complete Physics Guide to Electric Current, Ohm's Law & Resistivity

Chapter 32: Current and Resistance

Complete Physics Notes Based on Halliday, Resnick and Krane - B.Sc. Physics Edition 2015-16
Electric Current Resistance Ohm's Law Current Density Reading Time: 25 min

📋 Table of Contents

Introduction to Electric Current

⚡ What is Electric Current?

The time rate of flow of charge through a conductor is called current. If a charge 'dq' flows through any cross-section of a conductor in time 'dt', then the current 'I' is given by:

\[ I = \frac{dq}{dt} \]

The SI unit of current is Ampere, which can be defined as: "when one coulomb of charge flows through a cross-section in one second, then the current flowing is one ampere".

🔬 Examples of Electric Currents

  • Large currents: Lightning strokes
  • Tiny currents: Nerve currents that regulate muscular activity
  • Solid conductors: Household wiring, light bulbs
  • Semiconductors: Integrated circuits
  • Gases: Fluorescent lamps
  • Liquids: Automobile batteries
  • Evacuated spaces: TV picture tubes
Problem 1: Charge and Electron Calculation

Question: A current of 4.82 A exists in a 12.42 Ω resistor for 4.6 minutes. (a) Find out charge, (b) How many electrons pass through resistor in this time?

Solution:

Given: \( I = 4.82 \, A \), \( R = 12.42 \, \Omega \), \( t = 4.6 \, min = 4.6 \times 60 \, s \)
(a) Charge calculation:
\[ I = \frac{q}{t} \]
\[ \Rightarrow q = It \]
\[ = 4.82 \times 4.6 \times 60 \]
\[ = 1.33 \times 10^3 \, C \]
(b) Number of electrons:
\[ q = ne \Rightarrow n = \frac{q}{e} \]
\[ = \frac{1.33 \times 10^3}{1.6 \times 10^{-19}} \]
\[ = 8.3 \times 10^{21} \, \text{electrons} \]
Problem 2: Electron Beam Current

Question: The current in the electron beam of a typical video display terminal is 200 μA. How many electrons strike the screen each minute?

Solution:

Given: \( I = 200 \, \mu A = 200 \times 10^{-6} \, A \), \( t = 1 \, min = 60 \, s \)
\[ I = \frac{q}{t} \Rightarrow q = It \]
\[ = 200 \times 10^{-6} \times 60 \]
\[ = 12000 \times 10^{-6} \, C \]
\[ q = ne \Rightarrow n = \frac{q}{e} \]
\[ = \frac{12000 \times 10^{-6}}{1.6 \times 10^{-19}} \]
\[ = 7.5 \times 10^{16} \, \text{electrons} \]

Current Flow and Fluid Flow Comparison

💧 Analogy Between Current and Fluid Flow

When a steady current is flowing through an idealized conducting wire, the electric current remains the same for all cross-sections, even though the cross-sectional area may be different at different points.

🔁 Comparison with Incompressible Fluid

The condition of steady current flow is similar to the motion of incompressible fluid:

  • The fluid that flows through any cross-section of the pipe is the same even if the cross-section varies
  • The fluid flows faster where the cross-section of the pipe is smaller and slower where it is larger
  • The volume rate of flow remains constant

Direction of Current

🧭 Conventional Current Direction

In metals, the charge carriers are electrons. But in electrolytes, the current flows due to motion of both negative and positive ions. A positive charge moving in one direction is equivalent in all external effects to a negative charge moving in the opposite direction.

📏 Current Direction Convention

For simplicity and algebraic consistency, we adopt the following convention:

The direction of current is the direction that positive charges would move, even if the actual charge carriers are negative. Thus, the direction of current is taken from the point of higher potential to the point of lower potential.

⚠️ Important Note: Current is a Scalar

Even though we assign a direction, current is a scalar quantity not a vector. The arrow that we draw to indicate the direction of current merely shows the sense of charge flow through the wire and is not to be taken as a vector. Current does not obey the law of vector addition. Changing the direction of wires does not change the way the currents are added.

Current Density

📊 Definition of Current Density

The current flowing per unit area is called the current density. It is a vector quantity and the SI unit of this quantity is Ampere per square meter \( \left( \frac{A}{m^2} \right) \).

Mathematical Definition

"The scalar product of current density 'J' and vector area 'A' is called the electric current".

\[ I = \int \vec{J} \cdot d\vec{A} \]

Macroscopic vs Microscopic

The electric current is a macroscopic quantity, while the current density is its corresponding microscopic quantity.

Sample Problem 1: Current Density in Composite Wire

Question: One end of an aluminum wire whose diameter is 2.5 mm is welded to one end of copper wire whose diameter is 1.8 mm. The composite wire carries a steady current of 1.3 A. What is current density in each wire?

Solution:

Current flowing through composite wire \( I = 1.3 \, A \)
For Aluminum:
Diameter of wire \( d = 2.5 \, mm \)
Radius \( r = 1.25 \, mm = 1.25 \times 10^{-3} \, m \)
Area \( A = \pi r^2 = 3.14 \times (1.25 \times 10^{-3})^2 = 4.91 \times 10^{-6} \, m^2 \)
Current Density \( J = \frac{I}{A} = \frac{1.3}{4.91 \times 10^{-6}} = 2.65 \times 10^5 \, A/m^2 \)
For Copper:
Diameter of wire \( d = 1.8 \, mm \)
Radius \( r = 0.9 \, mm = 0.9 \times 10^{-3} \, m \)
Area \( A = \pi r^2 = 3.14 \times (0.9 \times 10^{-3})^2 = 2.54 \times 10^{-6} \, m^2 \)
Current Density \( J = \frac{I}{A} = \frac{1.3}{2.54 \times 10^{-6}} = 5.12 \times 10^5 \, A/m^2 \)

Determination of Current Density

🔍 Microscopic View of Current Flow

The flow of current through a conductor is due to motion of electrons in the direction opposite to electric field E. The force on one electron due to electric field is eE. But this force does not produce any acceleration in the motion of electrons, because the conduction electrons keep on colliding with the lattice ions of conductor. Instead, the electrons acquire a constant drift speed v_d in the direction of -E.

Define Variables

Let:

\( n \) = Number of free electrons per unit volume of the conductor
\( N \) = Number of free electrons in the conductor = \( nAL \)
\( e \) = Charge on one electron
\( Q \) = Total charge flowing in conductor = \( Ne = nALe \)

Time for Charge to Pass

If the charge Q passes through conductor in time t, then:

\[ t = \frac{L}{v_d} \]

Current Calculation

\[ I = \frac{Q}{t} = \frac{nALe}{L/v_d} \]
\[ = nAev_d \]

Current Density

\[ J = \frac{I}{A} = \frac{nAev_d}{A} \]
\[ = nev_d \]

Vector Form

\[ \vec{J} = -ne\vec{v_d} \]

The direction of \( \vec{J} \) is opposite to the direction of flow of electrons.

📏 Drift Velocity Magnitude

The mean drift velocity of electrons is very small i.e., of the order of \( \frac{cm}{s} \). While in random motion, the speed of electrons has a typical value of \( 10^6 \, m/s \) in metals.

Sample Problem 2: Drift Speed in Copper Wire

Question: What is drift speed of the conduction electron in copper wire of sample problem 1? The electron density in copper is \( 8.49 \times 10^{28} \, \frac{electrons}{m^3} \)

Solution:

Given: \( n = 8.49 \times 10^{28} \, \frac{electrons}{m^3} \), \( J_{cu} = 5.1 \times 10^5 \frac{A}{m^2} \)
\[ v_d = \frac{J}{ne} \]
\[ = \frac{5.1 \times 10^5}{8.49 \times 10^{28} \times 1.6 \times 10^{-19}} \]
\[ = 0.37 \times 10^{-4} \frac{m}{s} \]
Sample Problem 3: Silicon Strip Current

Question: A strip of Si of width 3.2 cm and thickness \( d = 250 \, \mu m \) carries a current of 190 mA and \( n = 8 \times 10^{21} \, m^{-3} \), (a) Find current density. (b) Find drift speed

Solution:

(a) Current Density:
Current flowing through strip \( I = 190 \, mA = 190 \times 10^{-3} A \)
Width \( w = 3.2 \, cm = 3.2 \times 10^{-2} m \)
Thickness \( d = 250 \, \mu m = 250 \times 10^{-6} m \)
Area \( A = w \times d = 3.2 \times 10^{-2} \times 250 \times 10^{-6} = 8 \times 10^{-6} m^2 \)
Current Density \( J_{Si} = \frac{I}{A} = \frac{190 \times 10^{-3}}{8 \times 10^{-6}} = 2.375 \times 10^4 \frac{A}{m^2} \)
(b) Drift Speed:
Given \( n = 8 \times 10^{21} \, m^{-3} \)
\[ v_d = \frac{J}{ne} = \frac{2.375 \times 10^4}{8 \times 10^{21} \times 1.6 \times 10^{-19}} \]
\[ = \frac{2.375 \times 10^4}{1.28 \times 10^3} \]
\[ = 18.55 \, m/s \]
Problem 3: Current Density of Ions

Question: Suppose that we have \( 2.1 \times 10^8 \, \text{doubly charge positive ions per cubic centimeter}, \) all moving north with a speed of \( 1 \times 10^5 \, m/s \). What is current density?

Solution:

Given: \( n = 2.1 \times 10^8 \, \text{ions/cm}^3 = 2.1 \times 10^{14} \, \text{ions/m}^3 \)
Charge on each ion \( q = 2e = 3.2 \times 10^{-19} C \)
Drift velocity \( v_d = 1 \times 10^5 \, m/s \)
\[ J = nqv_d \]
\[ = 2.1 \times 10^{14} \times 3.2 \times 10^{-19} \times 1 \times 10^5 \]
\[ = 6.72 \, A/m^2 \]

Ohm's Law

📏 Statement of Ohm's Law

"The current flowing through a conductor is directly proportional to the potential difference applied across its ends, provided the physical conditions such as temperature, pressure, etc., remain unchanged."

\[ I \propto V \]
↓
\[ V = IR \]

where R is the constant of proportionality called resistance.

🔬 Experimental Verification

Ohm's law is not a fundamental law of nature but an empirical relationship valid only for certain materials. Materials that obey Ohm's law are called ohmic materials, while those that don't are called non-ohmic materials.

📊 Definition of Resistance

The resistance R of a conductor is defined as the ratio of potential difference V across it to the current I flowing through it:

\[ R = \frac{V}{I} \]

The SI unit of resistance is Ohm (Ω).

Problem 4: Resistance Calculation

Question: A potential difference of 1.2 V is applied across a conductor of length 2.4 m. The resulting current through conductor is 5.6 A. What is resistance of conductor?

Solution:

Given: \( V = 1.2 \, V \), \( I = 5.6 \, A \)
\[ R = \frac{V}{I} = \frac{1.2}{5.6} = 0.214 \, \Omega \]

Resistivity

📏 Definition of Resistivity

The resistance of a conductor depends on:

  • Length of conductor (L)
  • Cross-sectional area (A)
  • Nature of material

Experimentally: \( R \propto L \) and \( R \propto \frac{1}{A} \)

\[ R = \rho \frac{L}{A} \]

where \( \rho \) is the resistivity of the material.

🔍 Resistivity as a Material Property

Resistivity is an intrinsic property of a material that depends only on the nature of the material and its temperature, not on the dimensions of the conductor.

💡 Practical Insight

The reciprocal of resistivity is called conductivity (σ):

\[ \sigma = \frac{1}{\rho} \]
Problem 5: Resistivity of Copper

Question: A copper wire has a resistance of 10 Ω. It is stretched to double its length. If volume and resistivity remain constant, find new resistance.

Solution:

Given: \( R_1 = 10 \, \Omega \), \( L_2 = 2L_1 \)
Volume remains constant: \( A_1L_1 = A_2L_2 \)
\[ A_1L_1 = A_2(2L_1) \Rightarrow A_2 = \frac{A_1}{2} \]
\[ R = \rho \frac{L}{A} \]
\[ R_2 = \rho \frac{L_2}{A_2} = \rho \frac{2L_1}{A_1/2} = 4\rho \frac{L_1}{A_1} \]
\[ R_2 = 4R_1 = 4 \times 10 = 40 \, \Omega \]

Microscopic Form of Ohm's Law

🔬 Microscopic View of Resistance

The resistance of a conductor arises due to collisions between conduction electrons and the lattice ions of the conductor. The mean free path between collisions is typically of the order of \( 10^{-8} \, m \) in metals at room temperature.

Relate Current Density and Electric Field

From earlier derivation:

\[ J = nev_d \]

Relate Drift Velocity and Electric Field

\[ v_d = \frac{eE\tau}{m} \]

where τ is the average time between collisions.

Combine Equations

\[ J = ne \left( \frac{eE\tau}{m} \right) \]
\[ = \frac{ne^2\tau}{m} E \]

Define Conductivity

\[ J = \sigma E \]

where \( \sigma = \frac{ne^2\tau}{m} \) is the conductivity.

Define Resistivity

\[ \rho = \frac{1}{\sigma} = \frac{m}{ne^2\tau} \]

📊 Microscopic Ohm's Law

The microscopic form of Ohm's law states that the current density J is proportional to the electric field E:

\[ \vec{J} = \sigma \vec{E} \]

This is valid for ohmic materials where σ is constant.

Temperature Variation of Resistivity

🌡️ Temperature Dependence

The resistivity of most materials changes with temperature. For metals, resistivity increases with increasing temperature, while for semiconductors, resistivity decreases with increasing temperature.

Empirical Relationship

For many materials, the temperature dependence of resistivity can be approximated by:

\[ \rho(T) = \rho_0 [1 + \alpha (T - T_0)] \]

where:

\( \rho_0 \) = resistivity at reference temperature \( T_0 \)
\( \alpha \) = temperature coefficient of resistivity

Resistance Variation

Since \( R = \rho \frac{L}{A} \), and L and A also change with temperature:

\[ R(T) = R_0 [1 + \alpha (T - T_0)] \]

📊 Temperature Coefficients

  • Metals: α is positive (resistance increases with temperature)
  • Semiconductors: α is negative (resistance decreases with temperature)
  • Alloys: Small α values (used in precision resistors)
  • Superconductors: Zero resistance below critical temperature
Problem 6: Resistance Temperature Dependence

Question: A copper wire has a resistance of 10 Ω at 20°C. What is its resistance at 80°C? (α for copper = 3.9 × 10⁻³ /°C)

Solution:

Given: \( R_0 = 10 \, \Omega \), \( T_0 = 20°C \), \( T = 80°C \), \( \alpha = 3.9 \times 10^{-3} \, /°C \)
\[ R(T) = R_0 [1 + \alpha (T - T_0)] \]
\[ = 10 [1 + 3.9 \times 10^{-3} (80 - 20)] \]
\[ = 10 [1 + 3.9 \times 10^{-3} \times 60] \]
\[ = 10 [1 + 0.234] \]
\[ = 10 \times 1.234 = 12.34 \, \Omega \]

Energy Transfer and Power Dissipation

⚡ Power in Electric Circuits

When a current flows through a resistor, electrical energy is converted to thermal energy. The rate at which energy is delivered to a circuit element is called power.

Power Definition

The power P delivered to a circuit element is:

\[ P = \frac{dU}{dt} \]

where U is the energy.

Relate to Voltage and Current

Since voltage V = dU/dq (energy per unit charge):

\[ dU = V dq \]
\[ P = \frac{dU}{dt} = V \frac{dq}{dt} = VI \]

Using Ohm's Law

For resistors obeying Ohm's law (V = IR):

\[ P = VI = I^2R = \frac{V^2}{R} \]

📊 Power Formulas Summary

The power dissipated in a resistor can be calculated using any of these equivalent formulas:

\[ P = VI = I^2R = \frac{V^2}{R} \]

The SI unit of power is Watt (W).

Problem 7: Power Calculation

Question: A 100 W light bulb operates at 120 V. What is its resistance?

Solution:

Given: \( P = 100 \, W \), \( V = 120 \, V \)
\[ P = \frac{V^2}{R} \Rightarrow R = \frac{V^2}{P} \]
\[ = \frac{(120)^2}{100} = \frac{14400}{100} = 144 \, \Omega \]

Joule Heating

🔥 Joule's Law

When current flows through a resistor, the electrical energy is converted to heat energy. This phenomenon is called Joule heating or resistive heating.

Heat Produced

The heat energy H produced in time t is:

\[ H = Pt = I^2Rt \]

In Calories

Since 1 calorie = 4.184 Joules:

\[ H = \frac{I^2Rt}{4.184} \, \text{calories} \]

🔬 Applications of Joule Heating

  • Incandescent light bulbs
  • Electric heaters
  • Electric stoves and ovens
  • Fuses (overcurrent protection)
Problem 8: Joule Heating Calculation

Question: A current of 5 A flows through a 20 Ω resistor for 10 minutes. Calculate the heat produced in (a) joules, (b) calories.

Solution:

Given: \( I = 5 \, A \), \( R = 20 \, \Omega \), \( t = 10 \, min = 600 \, s \)
(a) Heat in joules:
\[ H = I^2Rt = (5)^2 \times 20 \times 600 \]
\[ = 25 \times 20 \times 600 = 300,000 \, J \]
(b) Heat in calories:
\[ H = \frac{300,000}{4.184} = 71,700 \, \text{calories} \]

Energy Band Theory

🔬 Classification of Materials

Based on their electrical properties, materials can be classified into three categories:

Conductors

Materials with high electrical conductivity. In conductors, the valence band and conduction band overlap, allowing electrons to move freely. Examples: metals like copper, aluminum, silver.

Insulators

Materials with very low electrical conductivity. In insulators, there is a large energy gap between the valence band and conduction band. Examples: glass, rubber, plastic.

Semiconductors

Materials with conductivity between conductors and insulators. In semiconductors, there is a small energy gap between valence and conduction bands. Examples: silicon, germanium.

Energy Band Gaps

The energy gap between valence and conduction bands determines the electrical properties:

  • Conductors: No gap or overlapping bands
  • Semiconductors: Small gap (0.1-2 eV)
  • Insulators: Large gap (>3 eV)

Superconductors

❄️ Superconductivity

Superconductivity is a phenomenon where certain materials exhibit zero electrical resistance when cooled below a critical temperature (T_c).

🔬 Key Properties of Superconductors

  • Zero resistance: Below T_c, resistance drops abruptly to zero
  • Meissner effect: Perfect diamagnetism - expulsion of magnetic field
  • Critical temperature: Temperature below which superconductivity occurs
  • Critical field: Magnetic field strength that destroys superconductivity

💡 Applications of Superconductors

  • MRI machines
  • Maglev trains
  • Particle accelerators
  • High-field magnets
  • Quantum computing

Frequently Asked Questions

❓ Why is current considered a scalar quantity even though it has direction?

Current is considered a scalar quantity because it doesn't obey the laws of vector addition. The direction associated with current is merely conventional and indicates the sense of charge flow, not a true vector direction. Currents add algebraically, not vectorially.

❓ What is the difference between resistance and resistivity?

Resistance (R) is a property of a specific object that depends on its material, length, and cross-sectional area. Resistivity (ρ) is an intrinsic property of the material itself, independent of the object's dimensions. Resistance can be calculated from resistivity using R = ρL/A.

❓ Why does the resistance of metals increase with temperature while that of semiconductors decreases?

In metals, increased temperature causes more lattice vibrations, which scatter electrons more effectively, increasing resistance. In semiconductors, increased temperature provides more energy for electrons to jump from the valence band to the conduction band, increasing the number of charge carriers and thus decreasing resistance.

❓ What is the physical significance of drift velocity?

Drift velocity represents the average velocity of charge carriers in the direction of the electric field. It's much smaller than the random thermal velocity of electrons but is responsible for the net transport of charge that constitutes electric current.

❓ How does Ohm's law relate to the microscopic properties of materials?

Ohm's law (V = IR) at the macroscopic level corresponds to J = σE at the microscopic level, where σ = ne²τ/m. The conductivity σ depends on the number density of charge carriers (n), their charge (e), mass (m), and the average time between collisions (τ).

📚 Continue Your Physics Journey

Mastering Current and Resistance is fundamental to understanding electromagnetism and circuit theory. These comprehensive notes based on Halliday, Resnick and Krane provide a solid foundation for further studies in physics including DC circuits, AC circuits, and electromagnetic theory.

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© 2025 Physics Education Initiative | Chapter 32: Current and Resistance

These comprehensive notes are designed to help B.Sc. Physics students understand fundamental concepts of Current and Resistance based on Halliday, Resnick and Krane

Author: Muhammad Ali Malik | Contact: +923016775811 | Email: aliphy2008@gmail.com

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