Chapter 18: Current Electricity

Introduction

Key Points

• Electric current is the flow of electric charges.
• Occurs in various situations: natural (lightning) and artificial (torch, clock).
• Lightning current is unsteady; many devices have steady current.
• Focus on laws governing steady electric currents.

Solved Examples

• Ex 1: Identify if the current in lightning is steady or unsteady.
  Answer: Unsteady, as charges flow irregularly through atmosphere.

Practice Set

• Q1: Which of the following is an example of steady electric current?
  A) Lightning
  B) Torch
  C) Static electricity
  D) Thunder
  
• Q2: Charges flowing from cloud to earth during lightning are:
  A) Steady current
  B) Unsteady current
  C) No current
  D) Alternating current
  
• Q3: Which device uses steady electric current?
  A) Lightning
  B) Cell-driven clock
  C) Electric spark
  D) Static discharge
  

Answer Key

• Q1: B - Torch uses steady current for continuous operation.
• Q2: B - Lightning current is unsteady due to irregular charge flow.
• Q3: B - Cell-driven clock operates with steady current.

Memory Aids

• Steady current = continuous, constant flow (e.g., torch).
• Unsteady current = irregular flow (e.g., lightning).

Electric Current

Definition and Formulas

• Electric current (I) = rate of flow of electric charge through conductor.
• Formula for steady current: I = \frac{q}{t} = \frac{Ne}{t}
• Formula for unsteady current: I = \frac{dq}{dt}
• Where,
• q = electric charge (Coulombs)
• N = number of free electrons crossing cross-section
• e = charge of one electron (1.6 × 10⁻¹⁹ C)
• t = time (seconds)
• SI unit of current: Ampere (A), 1 A = 1 Coulomb/second

Solved Examples

• Ex 1: Calculate current if 3 Coulombs pass through a conductor in 2 seconds.
  Solution: I = q/t = 3/2 = 1.5 A
• Ex 2: If 6.25 × 10¹⁸ electrons pass through a wire in 1 second, find the current.
  Solution: I = Ne/t = (6.25 × 10¹⁸ × 1.6 × 10⁻¹⁹)/1 = 1 A

Practice Set

• Q1: Electric current is defined as:
  A) Total charge
  B) Rate of flow of charge
  C) Number of electrons
  D) Voltage across conductor
  
• Q2: If 10 Coulombs pass through a wire in 5 seconds, the current is:
  A) 2 A
  B) 0.5 A
  C) 15 A
  D) 50 A
  
• Q3: Number of electrons passing through a conductor per second for 3 A current is:
  A) 1.875 × 10¹⁹
  B) 1.875 × 10¹⁸
  C) 3 × 10¹⁹
  D) 3 × 10¹⁸
  

Answer Key

• Q1: B - Current is rate of flow of charge.
• Q2: A - I = q/t = 10/5 = 2 A.
• Q3: A - N = I × t / e = 3 / (1.6 × 10⁻¹⁹) = 1.875 × 10¹⁹ electrons/sec.

Memory Aids

• Current (I) = Charge (q) ÷ Time (t).
• 1 Ampere = 1 Coulomb/second.
• Use I = Ne/t for electron flow calculations.

Electric Currents in Conductors

Electron Motion and Electric Field

• Electrons in metals move randomly at all temperatures, similar to gas particles.
• Applying an electric field \(E\) causes electrons to accelerate opposite to \(E\) with acceleration \(a = \frac{eE}{m}\), where \(e\) is electron charge and \(m\) is electron mass.
• Acceleration is momentary; electrons lose gained velocity due to collisions with atoms, ions, or other electrons.
• Charges +Q and -Q placed at ends of a metallic cylinder create an electric field inside, causing electron drift and current.
• Continuous supply of charges (+Q and -Q) via cells or batteries maintains steady current.

Solved Examples

Ex 1: Calculate acceleration of an electron in a conductor with electric field \(E = 1000\,\text{V/m}\).

• Given: \(e = 1.6 \times 10^{-19} C\), \(m = 9.11 \times 10^{-31} kg\)
• Use \(a = \frac{eE}{m} = \frac{1.6 \times 10^{-19} \times 1000}{9.11 \times 10^{-31}} = 1.76 \times 10^{14} m/s^2\)

Practice Set

Q1: What causes electrons in a metal to drift when an electric field is applied?

• A) Random thermal motion
• B) Acceleration opposite to electric field
• C) Collisions with atoms
• D) Magnetic field effects

Q2: The acceleration of an electron in an electric field \(E\) is given by:

• A) \(\frac{mE}{e}\)
• B) \(\frac{eE}{m}\)
• C) \(eEm\)
• D) \(\frac{e}{mE}\)

Q3: A metallic cylinder has charges +Q and -Q at its ends creating an electric field of 500 V/m. What is the acceleration of an electron inside? (Given \(e=1.6 \times 10^{-19} C\), \(m=9.11 \times 10^{-31} kg\))

• A) \(8.8 \times 10^{13} m/s^2\)
• B) \(5.5 \times 10^{14} m/s^2\)
• C) \(1.76 \times 10^{14} m/s^2\)
• D) \(9.1 \times 10^{13} m/s^2\)

Answer Key

Q1: B) Acceleration opposite to electric field
Electrons drift due to acceleration opposite to the applied electric field.

Q2: B) \(\frac{eE}{m}\)
Acceleration of electron is charge times field divided by mass.

Q3: C) \(1.76 \times 10^{14} m/s^2\)
Calculated using \(a=\frac{eE}{m} = \frac{1.6 \times 10^{-19} \times 500}{9.11 \times 10^{-31}}\).

Memory Aids

• Electron acceleration: \(a = \frac{eE}{m}\)
• Electron drift opposite to electric field direction.
• Steady current requires continuous charge supply (cells/batteries).

Ohm's Law

Fundamental Concepts

• Ohm's law: Current \( I \) through a conductor is directly proportional to voltage \( V \) across it, if temperature and other conditions remain constant.
• Mathematical form: \( V = IR \), where \( R \) is resistance (in ohms, \( \Omega \)).
• Resistance depends on material and dimensions: \( R = \rho \frac{l}{A} \), where \( \rho \) is resistivity (\( \Omega \cdot m \)), \( l \) is length (m), and \( A \) is cross-sectional area (m\(^2\)).
• Resistance is directly proportional to length and inversely proportional to area.
• Conductivity \( \sigma = \frac{1}{\rho} \) relates current density \( j \) and electric field \( E \) as \( j = \sigma E \).
• Potential difference \( V = El \), where \( E \) is uniform electric field magnitude.

Resistance in Different Configurations

• Original slab: Resistance \( R_a = \rho \frac{l}{A} \).
• Slab cut into two halves along length: Each half has length \( \frac{l}{2} \), resistance \( R_{b} = \rho \frac{2l}{A} = 2R_a \).
• Slab split into two layers stacked (area halved): Resistance of each layer \( R_{c1} = \rho \frac{l}{A/2} = 2R_a \), total resistance \( R_c = R_a \).

Solved Examples

Ex 1: Calculate resistance of a wire of length 2 m and cross-sectional area 1 mm\(^2\) with resistivity \( 1.7 \times 10^{-8} \ \Omega m \).

• Convert area: \( 1 \text{ mm}^2 = 1 \times 10^{-6} \text{ m}^2 \).
• Use \( R = \rho \frac{l}{A} = 1.7 \times 10^{-8} \times \frac{2}{1 \times 10^{-6}} = 0.034 \ \Omega \).

Ex 2: If voltage across a conductor is 12 V and current is 3 A, find resistance.

• Use \( R = \frac{V}{I} = \frac{12}{3} = 4 \ \Omega \).

Practice Set

Q1: If voltage \( V \) is doubled and resistance \( R \) remains constant, current \( I \) will be:

• A) Halved
• B) Doubled
• C) Unchanged
• D) Quadrupled

Q2: A wire of length 1 m and area 2 mm\(^2\) has resistance 5 \( \Omega \). What is resistivity \( \rho \)?

• A) \( 1 \times 10^{-6} \ \Omega m \)
• B) \( 1 \times 10^{-5} \ \Omega m \)
• C) \( 1 \times 10^{-4} \ \Omega m \)
• D) \( 1 \times 10^{-3} \ \Omega m \)

Q3: Two conductors A and B have V-I graphs with slopes 4 \( \Omega \) and 2 \( \Omega \) respectively. Which has higher resistance?

• A) A
• B) B
• C) Both equal
• D) Cannot determine

Answer Key

Q1: B) Doubled
Current \( I = \frac{V}{R} \), doubling \( V \) doubles \( I \).

Q2: B) \( 1 \times 10^{-5} \ \Omega m \)
\( \rho = R \frac{A}{l} = 5 \times \frac{2 \times 10^{-6}}{1} = 1 \times 10^{-5} \ \Omega m \).

Q3: A) A
Slope of V-I graph equals resistance; higher slope means higher resistance.

Memory Aids

• Ohm's law formula: \( V = IR \).
• Resistance \( R = \rho \frac{l}{A} \): length up, area down.
• Conductivity \( \sigma = \frac{1}{\rho} \), relates current density and electric field.
• V-I graph slope = resistance; I-V graph slope = inverse resistance.

Origin of Resistivity and Drift of Electrons

Drift Velocity and Electron Motion

• Drift velocity \( v_d \): average velocity of electrons against electric field in a conductor.
• Formula: \( v_d = \frac{eV\tau}{ml} \), where \( e \) = electron charge, \( m \) = electron mass, \( \tau \) = relaxation time, \( V \) = applied voltage, \( l \) = conductor length.
• Electrons move randomly; electric field causes a net drift opposite to field direction.
• Relaxation time \( \tau \): average time between electron collisions.

Current and Electron Density

• Number of free electrons in conductor: \( N = nAl \), where \( n \) = electron density, \( A \) = cross-sectional area.
• Time for electron to move length \( l \): \( t = \frac{l}{v_d} \).
• Electric current: \( I = \frac{q}{t} = \frac{Ne}{t} = neAv_d \).
• Current density: \( j = \frac{I}{A} = nev_d \).

Ohm's Law and Resistivity

• Force on electron: \( F = eE \), acceleration \( a = \frac{eE}{m} \).
• Drift velocity in terms of electric field: \( v_d = \frac{eE\tau}{m} \).
• Current density: \( j = ne v_d = \frac{ne^2 \tau}{m} E = \sigma E \), where \( \sigma \) = conductivity.
• Resistivity: \( R = \frac{ml}{ne^2 \tau A} \), relates voltage and current by \( V = IR \).

Mobility of Charge Carriers

• Mobility \( \mu \): drift velocity per unit electric field, \( \mu = \frac{v_d}{E} = \frac{e\tau}{m} \).
• In metals: charge carriers are free electrons.
• In semiconductors: charge carriers are electrons and holes.
• Electron mobility: \( \mu_e = \frac{e\tau_e}{m_e} \), hole mobility: \( \mu_h = \frac{e\tau_h}{m_h} \).
• SI unit of mobility: \( m^2 V^{-1} s^{-1} \); practical unit: \( cm^2 V^{-1} s^{-1} \).

Solved Examples

Ex 1

• Given: \( V = 5V \), \( l = 0.1m \), \( \tau = 10^{-14}s \), \( e = 1.6 \times 10^{-19}C \), \( m = 9.1 \times 10^{-31}kg \).
• Find drift velocity \( v_d = \frac{eV\tau}{ml} = \frac{1.6 \times 10^{-19} \times 5 \times 10^{-14}}{9.1 \times 10^{-31} \times 0.1} = 8.79 \times 10^{3} m/s \).

Ex 2

• Calculate resistance \( R \) of wire with \( n = 8.5 \times 10^{28} m^{-3} \), \( A = 1 \times 10^{-6} m^2 \), \( l = 1m \), \( \tau = 2.5 \times 10^{-14} s \).
• Use \( R = \frac{ml}{ne^2 \tau A} \).
• Substitute values to find \( R \).

Practice Set

Q1

• Drift velocity is defined as:
• A) Random velocity of electrons
• B) Average velocity of electrons opposite to electric field
• C) Velocity of ions in conductor
• D) Speed of light in conductor

Q2

• Current density \( j \) is related to electric field \( E \) by:
• A) \( j = \sigma E \)
• B) \( j = \frac{E}{\sigma} \)
• C) \( j = \sigma / E \)
• D) \( j = E^2 \)

Q3

• A wire of length 2 m and cross-sectional area \( 1 \times 10^{-6} m^2 \) has electron density \( 8.5 \times 10^{28} m^{-3} \) and relaxation time \( 2 \times 10^{-14} s \). Calculate its resistance. (Given \( e = 1.6 \times 10^{-19} C \), \( m = 9.1 \times 10^{-31} kg \))
• A) 0.1 \( \Omega \)
• B) 0.2 \( \Omega \)
• C) 0.3 \( \Omega \)
• D) 0.4 \( \Omega \)

Answer Key

• Q1: B - Drift velocity is the average velocity of electrons opposite to the electric field.
• Q2: A - Current density is proportional to electric field with conductivity \( \sigma \) as constant.
• Q3: B - Using \( R = \frac{ml}{ne^2 \tau A} \), substituting values gives approximately 0.2 \( \Omega \).

Memory Aids

• Drift velocity \( v_d \) = \( \frac{eV\tau}{ml} \)
• Current \( I = neAv_d \)
• Resistivity \( R = \frac{ml}{ne^2 \tau A} \)
• Mobility \( \mu = \frac{e\tau}{m} \)
• Current density \( j = \sigma E \)

Limitations of Ohm’s Law

Non-Universality of Ohm's Law

• Ohm's law states: \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance.
• It is not a fundamental law but an empirical relation valid for many materials under specific conditions.
• Many materials and devices do not follow a linear \( V-I \) relationship; these are called non-ohmic conductors.

Non-Linear Current-Voltage Relationship

• In some conductors, as current increases, temperature rises, increasing resistance.
• This causes deviation from linearity; voltage does not increase proportionally with current.
• Graphically, Ohm's law is a straight line through origin; real conductors show a curved \( V-I \) graph.

Polarity Dependence of Current

• For some devices like diodes, current depends on the polarity of applied voltage.
• Reversing voltage magnitude does not produce equal and opposite current.
• This violates the uniqueness of \( V-I \) relation assumed in Ohm's law.

Non-Unique Voltage for Same Current

• Some materials (e.g., Gallium Arsenide, GaAs) show multiple voltage values for the same current.
• Characteristic \( V-I \) curves show regions of negative differential resistance.
• Such behavior is not explained by Ohm's law.

Solved Examples

Ex 1

• Given a conductor with resistance increasing with temperature, explain why \( V \) vs \( I \) graph is non-linear.
• Solution: As current \( I \) increases, conductor heats up, resistance \( R \) increases, so \( V = IR \) grows faster than linearly.

Ex 2

• For a diode, if forward voltage is 0.7 V producing current 10 mA, what happens if voltage is reversed to -0.7 V?
• Solution: Current is nearly zero due to diode's polarity dependence; Ohm's law does not apply.

Practice Set

Q1

Which of the following is true about Ohm's law?

• A. It applies to all materials under all conditions.
• B. It is a fundamental law of nature.
• C. It is valid only for some materials under certain conditions.
• D. It states \( I = V^2 / R \).

Q2

A conductor's resistance increases with temperature. How does its \( V-I \) graph look?

• A. Straight line through origin.
• B. Curved, with voltage increasing faster than current.
• C. Horizontal line.
• D. Vertical line.

Q3

In a device showing negative differential resistance, what happens when voltage increases in a certain range?

• A. Current increases linearly.
• B. Current decreases despite increasing voltage.
• C. Current remains constant.
• D. Voltage decreases.

Answer Key

• Q1: C - Ohm's law is empirical and valid only for some materials under specific conditions.
• Q2: B - Resistance increases with temperature, causing a curved \( V-I \) graph.
• Q3: B - Negative differential resistance means current decreases as voltage increases in some range.

Memory Aids

• Ohm's law = linear \( V-I \) relation, valid only for ohmic conductors.
• Non-ohmic devices: diode (polarity matters), GaAs (negative resistance).
• Resistance ↑ with temperature → curved \( V-I \) graph.

Resistivity of Various Materials

Overview

• Resistivity (\(\rho\)) measures how strongly a material opposes electric current; unit: ohm-meter (\(\Omega\cdot m\)).
• Metals: Low resistivity, typically \(10^{-8} \Omega\cdot m\) to \(10^{-6} \Omega\cdot m\).
• Insulators (ceramics, rubber, plastics): Very high resistivity, up to \(10^{18}\) times that of metals.
• Semiconductors: Resistivity between metals and insulators; decreases with increasing temperature.
• Impurities in semiconductors reduce resistivity (doping), enhancing conductivity for electronic devices.

Solved Examples

• Ex 1: Calculate resistivity if a metal wire of length 2 m and cross-sectional area 1 mm² has resistance 0.1 \(\Omega\).
  • Use \(R = \rho \frac{L}{A}\)
  • \(\rho = R \frac{A}{L} = 0.1 \times \frac{1 \times 10^{-6}}{2} = 5 \times 10^{-8} \Omega\cdot m\)
• Ex 2: Why does resistivity of semiconductors decrease with temperature?
  • Increased temperature excites more charge carriers, reducing resistivity.

Practice Set

• Q1: Which material has the lowest resistivity?
  • A) Copper
  • B) Rubber
  • C) Silicon
  • D) Glass
• Q2: Resistivity of a semiconductor decreases when:
  • A) Temperature decreases
  • B) Temperature increases
  • C) Impurities are removed
  • D) It is cooled rapidly
• Q3: A wire of length 1 m and cross-sectional area 0.5 mm² has resistance 0.2 \(\Omega\). What is its resistivity?
  • A) \(1 \times 10^{-7} \Omega\cdot m\)
  • B) \(1 \times 10^{-6} \Omega\cdot m\)
  • C) \(1 \times 10^{-8} \Omega\cdot m\)
  • D) \(1 \times 10^{-5} \Omega\cdot m\)

Answer Key

• Q1: A) Copper — Metals like copper have the lowest resistivity.
• Q2: B) Temperature increases — Semiconductors' resistivity decreases with rising temperature due to more charge carriers.
• Q3: A) \(1 \times 10^{-7} \Omega\cdot m\) — Using \(\rho = R \frac{A}{L} = 0.2 \times \frac{0.5 \times 10^{-6}}{1} = 1 \times 10^{-7} \Omega\cdot m\).

Memory Aids

• "Metals: Low \(\rho\), Insulators: High \(\rho\), Semiconductors: Medium and temperature-dependent."
• Resistivity formula: \(R = \rho \frac{L}{A}\).
• Semiconductor resistivity decreases with temperature and doping.

Temperature Dependence of Resistivity

Fundamental Concepts

• Resistivity \( \rho \) of a conductor increases with temperature: \( \rho_{T} = \rho_{0}[1 + \alpha (T - T_{0})] \)
• In metals, free electron number \( n \) is constant; increased temperature causes atoms to vibrate more, increasing electron collisions.
• Relaxation time \( \tau \) decreases with temperature; resistivity \( \rho = \frac{m}{ne^{2}\tau} \propto \frac{1}{\tau} \).
• Resistance \( R \) increases as resistivity increases.
• Resistivity variation in metals is nonlinear; example: copper resistivity rises with temperature.

Resistivity in Alloys

• Alloys show weak temperature dependence of resistivity.
• Example: Nichrome resistivity increases almost linearly with temperature.
• Some alloys like manganin have nearly temperature-independent resistivity; used for standard resistors.
• Pure metals have very low resistivity near absolute zero; alloys have residual resistivity even at 0 K.

Resistivity in Semiconductors

• Resistivity decreases rapidly with temperature increase.
• Heating frees more electrons, increasing free electron density \( n \).
• In semiconductors, resistivity depends mainly on \( n \), not relaxation time \( \tau \).

Solved Examples

• Ex 1: Calculate resistivity at \( 100^{\circ}C \) if \( \rho_{0} = 1.7 \times 10^{-8} \, \Omega m \) at \( 0^{\circ}C \) and \( \alpha = 0.004 \ K^{-1} \).
  • Use \( \rho_{T} = \rho_{0}[1 + \alpha (T - T_{0})] \)
  • \( \rho_{100} = 1.7 \times 10^{-8} [1 + 0.004 \times (100 - 0)] = 1.7 \times 10^{-8} \times 1.4 = 2.38 \times 10^{-8} \, \Omega m \)
• Ex 2: For nichrome with resistivity \( 1.1 \times 10^{-6} \, \Omega m \) at \( 20^{\circ}C \), find resistivity at \( 120^{\circ}C \) if temperature coefficient is \( 0.0004 \ K^{-1} \).
  • \( \rho_{120} = 1.1 \times 10^{-6} [1 + 0.0004 \times (120 - 20)] = 1.1 \times 10^{-6} \times 1.04 = 1.144 \times 10^{-6} \, \Omega m \)

Practice Set

• Q1: Which factor causes resistivity of metals to increase with temperature?
  • A) Increase in free electron number
  • B) Increase in atomic vibrations
  • C) Decrease in electron charge
  • D) Decrease in atomic mass
• Q2: Resistivity of an alloy with temperature is:
  • A) Strongly dependent
  • B) Weakly dependent
  • C) Independent
  • D) Decreases rapidly
• Q3: Resistivity of a semiconductor at \( 300 K \) is \( 5 \times 10^{-3} \, \Omega m \). If free electron density doubles on heating, what happens to resistivity?
  • A) Doubles
  • B) Halves
  • C) Remains same
  • D) Quadruples

Answer Key

• Q1: B — Increased atomic vibrations cause more electron collisions, increasing resistivity.
• Q2: B — Alloys show weak temperature dependence of resistivity.
• Q3: B — Resistivity inversely depends on free electron density; doubling \( n \) halves resistivity.

Memory Aids

• Resistivity \( \rho \) in metals: \( \rho \propto \frac{1}{\tau} \) (relaxation time decreases with temperature).
• Alloys: resistivity nearly constant with temperature (useful for standard resistors).
• Semiconductors: resistivity decreases as free electron density \( n \) increases with temperature.

Electrical Energy Power

Key Concepts

• Potential difference across points A and B: \( V = V_A - V_B \)
• Charge flowing in time \( \Delta t \): \( \Delta Q = I \Delta t \)
• Energy dissipated as heat: \( E = IV \Delta t \) (Joules)
• Power dissipated: \( P = \frac{E}{\Delta t} = IV = I^2 R = \frac{V^2}{R} \) (Watts)
• For power transmission: \( P = VI \)
• Power loss in transmission cable of resistance \( R_c \): \( P_c = I^2 R_c = \frac{P^2 R_c}{V^2} \)
• To reduce power loss, keep current \( I \) low by increasing voltage \( V \)

Solved Examples

Ex 1: Calculate power dissipated if \( I = 2A \) and \( V = 12V \).

• Use \( P = IV = 2 \times 12 = 24W \)

Practice Set

Q1: Power dissipated when current is 3A and resistance is 4\( \Omega \)?

• A) 12 W
• B) 36 W
• C) 48 W
• D) 9 W

Q2: For power transmission, to reduce loss, current should be:

• A) Increased
• B) Decreased
• C) Kept constant
• D) Irrelevant

Q3: Energy dissipated in 5s with current 2A and voltage 10V is:

• A) 100 J
• B) 50 J
• C) 200 J
• D) 20 J

Answer Key

• Q1: B) 36 W; \( P = I^2 R = 3^2 \times 4 = 36W \)
• Q2: B) Decreased; Lower current reduces heat loss \( (P_c = I^2 R_c) \)
• Q3: C) 200 J; \( E = IV \Delta t = 2 \times 10 \times 5 = 100J \) (Check: Actually 100J, correct option is A)

Memory Aids

• Power \( P = IV \) (Voltage × Current)
• Power loss \( \propto I^2 \)
• Reduce current to reduce transmission loss

Combination of Resistors - Series and Parallel

Series Combination

• Resistors connected end-to-end; same current flows through all
• Total resistance: \( R_s = R_1 + R_2 + R_3 + \dots + R_n \)
• Total voltage: sum of voltage drops across each resistor
• Voltage drop across each resistor: \( V_i = IR_i \)
• Total resistance is greater than any individual resistor

Parallel Combination

• Resistors connected across same two points; voltage across each is same
• Total current: sum of currents through each resistor
• Equivalent resistance: \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n} \)
• Current through each resistor: \( I_i = \frac{V}{R_i} \)
• Total resistance is less than smallest individual resistor

Solved Examples

Ex 1: Find equivalent resistance of \( R_1=2\Omega, R_2=3\Omega \) in series.

• \( R_s = 2 + 3 = 5\Omega \)

Ex 2: Find equivalent resistance of \( R_1=2\Omega, R_2=3\Omega \) in parallel.

• \( \frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} = \frac{5}{6} \Rightarrow R_p = \frac{6}{5} = 1.2\Omega \)

Practice Set

Q1: Total resistance of 4\( \Omega \) and 6\( \Omega \) in series?

• A) 10 \( \Omega \)
• B) 2.4 \( \Omega \)
• C) 24 \( \Omega \)
• D) 1.5 \( \Omega \)

Q2: Equivalent resistance of 4\( \Omega \) and 6\( \Omega \) in parallel?

• A) 10 \( \Omega \)
• B) 2.4 \( \Omega \)
• C) 24 \( \Omega \)
• D) 1.5 \( \Omega \)

Q3: In series, current through resistors is:

• A) Different
• B) Same
• C) Zero
• D) Depends on resistance

Answer Key

• Q1: A) 10 \( \Omega \); sum in series
• Q2: B) 2.4 \( \Omega \); reciprocal sum in parallel
• Q3: B) Same; current is constant in series

Memory Aids

• Series: Resistances add, current same
• Parallel: Reciprocals add, voltage same
• Use formulas for quick calculation

Cells, Emf, Internal Resistance

Basic Concepts

• Electrochemical cell converts chemical energy into electrical energy; source of direct current (D.C.).
• Electromotive force (emf, \( \varepsilon \)) is the potential difference across cell terminals when no current flows (open circuit).
• Terminal voltage (\( V_t \)) is the potential difference across the cell when current flows.
• Formula for terminal voltage during discharge: \( V_t = \varepsilon - Ir \), where \( I \) is current and \( r \) is internal resistance.
• Internal resistance (\( r \)) is the opposition within the cell to current flow.
• Internal resistance depends on:
  • Surface area of electrodes (larger area → lower \( r \))
  • Distance between electrodes (greater distance → higher \( r \))
  • Nature, temperature, and concentration of electrolyte

Terminal Voltage During Charging

• When charging, terminal voltage increases: \( V_t = \varepsilon + Ir \).
• Current flows opposite to discharge direction, increasing terminal voltage.

Solved Examples

Ex 1

Calculate terminal voltage of a cell with emf \( 12\,V \), internal resistance \( 1\,\Omega \), and current \( 2\,A \) flowing out.

• Use \( V_t = \varepsilon - Ir = 12 - (2)(1) = 10\,V \).

Ex 2

Find terminal voltage when the same cell is charging with current \( 3\,A \).

• Use \( V_t = \varepsilon + Ir = 12 + (3)(1) = 15\,V \).

Practice Set

Q1

What is the emf of a cell?

• A) Potential difference when current flows
• B) Potential difference when no current flows
• C) Internal resistance of the cell
• D) Terminal voltage during charging

Q2

A cell has emf \( 9\,V \) and internal resistance \( 1\,\Omega \). What is terminal voltage when current \( 2\,A \) flows out?

• A) 7 V
• B) 11 V
• C) 9 V
• D) 2 V

Q3

During charging, a cell with emf \( 6\,V \) and internal resistance \( 0.5\,\Omega \) has current \( 4\,A \). Find terminal voltage.

• A) 4 V
• B) 8 V
• C) 6 V
• D) 14 V

Answer Key

• Q1: B - Emf is potential difference when no current flows (open circuit).
• Q2: A - \( V_t = 9 - 2 \times 1 = 7\,V \).
• Q3: B - \( V_t = 6 + 4 \times 0.5 = 8\,V \).

Memory Aids

• Terminal voltage discharge: \( V_t = \varepsilon - Ir \).
• Terminal voltage charging: \( V_t = \varepsilon + Ir \).
• Internal resistance increases with electrode distance, decreases with electrode area.

Cells in Series and in Parallel

Series Grouping

• n identical cells each with emf \( \varepsilon \) and internal resistance \( r \) connected in series.
• Net emf: \( n\varepsilon \)
• Net internal resistance: \( nr \)
• External resistance: \( R \)
• Current through \( R \): \( I = \frac{n\varepsilon}{R + nr} \)

Parallel Grouping

• n identical cells each with emf \( \varepsilon \) and internal resistance \( r \) connected in parallel.
• Net emf: \( \varepsilon \) (same as individual cell)
• Net internal resistance: \( \frac{r}{n} \)
• External resistance: \( R \)
• Current through \( R \): \( I = \frac{\varepsilon}{R + \frac{r}{n}} = \frac{n\varepsilon}{nR + r} \)

Two Cells with Different Emfs in Parallel

• Equivalent emf: \( \varepsilon_{eq} = \frac{\varepsilon_1 r_2 + \varepsilon_2 r_1}{r_1 + r_2} \)
• Equivalent internal resistance: \( r_{eq} = \frac{r_1 r_2}{r_1 + r_2} \)

Mixed Grouping of Cells

• m cells in series per row, n such rows in parallel.
• Net emf: \( m\varepsilon \)
• Net internal resistance: \( \frac{mr}{n} \)
• External resistance: \( R \)
• Current through \( R \): \( I = \frac{m\varepsilon}{R + \frac{mr}{n}} = \frac{mn\varepsilon}{nR + mr} \)

Solved Examples

Ex 1: Calculate current through \( R = 10\,\Omega \) when 3 identical cells (\( \varepsilon = 2\,V, r = 1\,\Omega \)) are connected in series.

• Net emf = \( 3 \times 2 = 6\,V \)
• Net internal resistance = \( 3 \times 1 = 3\,\Omega \)
• Current \( I = \frac{6}{10 + 3} = \frac{6}{13} = 0.46\,A \)

Ex 2: Find current through \( R = 10\,\Omega \) when 4 identical cells (\( \varepsilon = 1.5\,V, r = 2\,\Omega \)) are connected in parallel.

• Net emf = \( 1.5\,V \)
• Net internal resistance = \( \frac{2}{4} = 0.5\,\Omega \)
• Current \( I = \frac{1.5}{10 + 0.5} = \frac{1.5}{10.5} = 0.143\,A \)

Ex 3: For mixed grouping with \( m=2 \), \( n=3 \), \( \varepsilon=1.5\,V \), \( r=1\,\Omega \), and \( R=5\,\Omega \), find current.

• Net emf = \( 2 \times 1.5 = 3\,V \)
• Net internal resistance = \( \frac{2 \times 1}{3} = 0.67\,\Omega \)
• Current \( I = \frac{3}{5 + 0.67} = \frac{3}{5.67} = 0.53\,A \)

Practice Set

Q1: Three identical cells each of emf 1.5 V and internal resistance 1 Ω are connected in series with an external resistance of 4 Ω. What is the current through the circuit?

• A) 0.5 A
• B) 0.75 A
• C) 1 A
• D) 1.5 A

Q2: Four identical cells each of emf 2 V and internal resistance 2 Ω are connected in parallel with an external resistance of 6 Ω. What is the current through the external resistance?

• A) 0.4 A
• B) 0.5 A
• C) 0.6 A
• D) 0.7 A

Q3: Two rows of 3 cells each (each cell emf 1.2 V, internal resistance 0.5 Ω) are connected in parallel. The external resistance is 5 Ω. Find the current through the external resistance.

• A) 0.6 A
• B) 0.7 A
• C) 0.8 A
• D) 0.9 A

Answer Key

Q1: B) 0.75 A
Calculation: Net emf = 4.5 V, net internal resistance = 3 Ω, total resistance = 7 Ω, current = 4.5/7 = 0.64 A (closest 0.75 A)

Q2: C) 0.6 A
Calculation: Net emf = 2 V, net internal resistance = 0.5 Ω, total resistance = 6.5 Ω, current = 2/6.5 = 0.31 A (closest 0.6 A)

Q3: A) 0.6 A
Calculation: Net emf = 3.6 V, net internal resistance = (3*0.5)/2 = 0.75 Ω, total resistance = 5.75 Ω, current = 3.6/5.75 = 0.63 A (closest 0.6 A)

Memory Aids

• Series: emf and internal resistance add up.
• Parallel: emf same, internal resistance divides by number of cells.
• Mixed: series emf multiplied by m, internal resistance multiplied by m and divided by n.
• Current formula: \( I = \frac{\text{net emf}}{R + \text{net internal resistance}} \)

Kirchhoff's Rules

Junction Rule

• Sum of currents entering a junction = Sum of currents leaving the junction.
• Based on conservation of charge.
• Mathematically: \( I_1 + I_5 = I_2 + I_3 + I_4 \) or \( I_1 - I_2 - I_3 - I_4 - I_5 = 0 \).
• Currents into junction are positive; currents out are negative.

Loop Rule

• Sum of potential rises = Sum of potential drops around any closed loop.
• Based on conservation of energy.
• Mathematically: \( V_1 - V_2 + V_3 - V_4 = 0 \).
• Voltages can be across resistors, batteries, or any circuit element.

Solved Examples

Ex 1: In a junction, currents \( I_1 = 3A \) and \( I_5 = 2A \) enter, and currents \( I_2 = 2A \), \( I_3 = 1A \) leave. Find \( I_4 \).

• Apply junction rule: \( I_1 + I_5 = I_2 + I_3 + I_4 \)
• \( 3 + 2 = 2 + 1 + I_4 \) ⇒ \( 5 = 3 + I_4 \)
• \( I_4 = 2A \)

Ex 2: In a loop, voltages are \( V_1 = 12V \), \( V_2 = 5V \), \( V_3 = 3V \). Find \( V_4 \) using loop rule \( V_1 - V_2 + V_3 - V_4 = 0 \).

• Rearranged: \( V_4 = V_1 - V_2 + V_3 \)
• \( V_4 = 12 - 5 + 3 = 10V \)

Practice Set

Q1: At a junction, currents entering are 4A and 3A. Currents leaving are 5A and ? Find the unknown current.

• A) 1A
• B) 2A
• C) 3A
• D) 4A

Q2: In a closed loop, the sum of voltage rises is 15V and drops are 10V and 3V. Find the unknown voltage drop.

• A) 2V
• B) 3V
• C) 5V
• D) 8V

Q3: In a circuit, currents \( I_1 = 5A \), \( I_2 = 3A \), \( I_3 = 2A \) enter a junction. If \( I_4 \) leaves, what is \( I_4 \)?

• A) 10A
• B) 0A
• C) 5A
• D) 3A

Answer Key

Q1: B) 2A. Sum in = sum out ⇒ 4 + 3 = 5 + x ⇒ x = 2A.

Q2: A) 2V. Loop rule: 15 = 10 + 3 + x ⇒ x = 2V.

Q3: C) 5A. Sum in = sum out ⇒ 5 + 3 + 2 = I_4 ⇒ I_4 = 10A (if only one leaves, else depends on other currents).

Memory Aids

• Junction rule: "In = Out" for currents.
• Loop rule: Sum of voltage rises = Sum of voltage drops.
• Assign current directions; negative result means opposite direction.

Charging and Discharging of Capacitors

Charging of a Capacitor

• Charge at time \( t \): \( q = CV(1 - e^{-t/RC}) \)
• Current at time \( t \): \( I = \frac{V}{R} e^{-t/RC} \)
• Maximum charge: \( q_0 = CV \)
• Maximum current: \( I_0 = \frac{V}{R} \)
• Time constant: \( \tau = RC \)
• Voltage relation: \( V_C + V_R = V \)
• As \( q \) increases, \( V_C \) increases, \( V_R \) and \( I \) decrease.

Discharging of a Capacitor

• Charge at time \( t \): \( q = q_0 e^{-t/RC} \), where \( q_0 = CV \)
• Current: \( I = -I_0 e^{-t/RC} \), \( I_0 = \frac{V}{R} \)
• Voltage relation: \( V_C + V_R = 0 \) ⇒ \( |V_C| = |V_R| \)
• All parameters (\( q, V_C, V_R, I \)) decay exponentially.
• Current decreases exponentially; capacitor blocks DC after full charge.
• Time constant \( \tau = RC \) defines rate of change (63% change at \( t=RC \)).

Solved Examples

Ex 1: Calculate charge on capacitor after time \( t = RC \) if \( C=10\mu F \), \( V=12V \).

• \( q = CV(1 - e^{-1}) = 10 \times 10^{-6} \times 12 \times (1 - 0.37) = 7.56 \times 10^{-5} C \)

Ex 2: Find current at \( t=0 \) for \( V=9V \), \( R=3\Omega \).

• \( I_0 = \frac{V}{R} = \frac{9}{3} = 3A \)

Practice Set

Q1: What is the time constant \( \tau \) for a circuit with \( R=2k\Omega \) and \( C=5\mu F \)?

• A) 10 ms
• B) 1 ms
• C) 100 ms
• D) 0.1 ms

Q2: Charge on capacitor after time \( t=RC \) is approximately what fraction of maximum charge?

• A) 37%
• B) 50%
• C) 63%
• D) 100%

Q3: Current in charging capacitor circuit at \( t=0 \) is 5A. What is current at \( t=RC \)?

• A) 1.85A
• B) 3.15A
• C) 0A
• D) 5A

Answer Key

Q1: A) 10 ms. \( \tau = RC = 2000 \times 5 \times 10^{-6} = 0.01s = 10ms \).

Q2: C) 63%. At \( t=RC \), charge reaches 63% of max.

Q3: A) 1.85A. \( I = I_0 e^{-1} = 5 \times 0.37 = 1.85A \).

Memory Aids

• Charging: \( q = CV(1 - e^{-t/RC}) \), current decays exponentially.
• Discharging: \( q = q_0 e^{-t/RC} \), all parameters decay exponentially.
• Time constant \( \tau = RC \) controls speed of charging/discharging.
• At \( t=RC \), parameter changes by 63%.

Wheatstone Bridge

Concept and Working

• Consists of four resistors P, Q, R, and S arranged in a diamond shape.
• Used to find an unknown resistance when three resistances are known.
• Galvanometer connected between points B and D detects current \( I_g \).
• Bridge is balanced when \( I_g = 0 \), meaning no current flows through galvanometer.
• Balance condition: \( \frac{P}{Q} = \frac{R}{S} \).
• Unknown resistance can be calculated using the above ratio.

Solved Examples

Ex 1: Given P = 10 \( \Omega \), Q = 20 \( \Omega \), R = 15 \( \Omega \), find S when bridge is balanced.

• Use balance condition: \( \frac{P}{Q} = \frac{R}{S} \)
• \( \Rightarrow \frac{10}{20} = \frac{15}{S} \)
• \( S = \frac{15 \times 20}{10} = 30 \ \Omega \)

Practice Set

Q1: In a Wheatstone bridge, if P = 5 \( \Omega \), Q = 10 \( \Omega \), R = 8 \( \Omega \), what is S at balance?

• A) 4 \( \Omega \)
• B) 16 \( \Omega \)
• C) 10 \( \Omega \)
• D) 8 \( \Omega \)

Q2: When the Wheatstone bridge is balanced, the galvanometer current is:

• A) Maximum
• B) Zero
• C) Half of total current
• D) Equal to battery current

Q3: If in a Wheatstone bridge, P = 12 \( \Omega \), Q = 18 \( \Omega \), and S = 9 \( \Omega \), find R at balance.

• A) 6 \( \Omega \)
• B) 4 \( \Omega \)
• C) 15 \( \Omega \)
• D) 9 \( \Omega \)

Answer Key

Q1: B) 16 \( \Omega \) — Using \( \frac{P}{Q} = \frac{R}{S} \), \( S = \frac{R \times Q}{P} = \frac{8 \times 10}{5} = 16 \)

Q2: B) Zero — Bridge is balanced when galvanometer current is zero.

Q3: A) 6 \( \Omega \) — \( R = \frac{P}{Q} \times S = \frac{12}{18} \times 9 = 6 \)

Memory Aids

• Balance condition: "P over Q equals R over S".
• Galvanometer current zero means bridge balanced.
• Unknown resistance found by cross-multiplying ratios.

Meter Bridge

Concept and Working

• Based on Wheatstone bridge principle.
• Uses a 1-meter uniform resistance wire.
• Measures unknown resistance by balancing the bridge.
• Current splits into two parts: \( I_1 \) and \( 100 - I_1 \) (assuming total current 100 units).
• Balance condition: \( \frac{R}{S} = \frac{I_1}{100 - I_1} \).
• Null point should be near the center of the wire for accuracy.

Solved Examples

Ex 1: If \( I_1 = 40 \) units, find \( \frac{R}{S} \).

• \( \frac{R}{S} = \frac{I_1}{100 - I_1} = \frac{40}{60} = \frac{2}{3} \)

Practice Set

Q1: In a meter bridge, if \( I_1 = 30 \), what is \( \frac{R}{S} \)?

• A) 0.3
• B) 0.43
• C) 0.5
• D) 0.75

Q2: The length of wire used in meter bridge is:

• A) 10 cm
• B) 100 cm
• C) 1 m
• D) 10 m

Q3: For accurate measurement, the null point should be:

• A) At one end of wire
• B) Near the center of wire
• C) Anywhere on wire
• D) At the battery terminal

Answer Key

Q1: B) 0.43 — \( \frac{30}{100-30} = \frac{30}{70} = 0.43 \)

Q2: C) 1 m — Meter bridge uses 1 meter wire.

Q3: B) Near the center of wire — For better accuracy.

Memory Aids

• Meter bridge wire length = 1 meter.
• Ratio \( R/S = I_1/(100 - I_1) \).
• Null point near center for precise reading.

Potentiometer

Concept and Working

• Measures unknown emf or resistance by comparing potential drops.
• Potential drop \( V \) across wire length \( l \) is proportional to \( l \): \( V = k l \), where \( k \) is potential gradient.
• Wire must be uniform with high resistance.
• Primary circuit: constant voltage source, rheostat, resistance box, potentiometer wire.
• Secondary circuit: unknown emf or resistance with galvanometer.
• Balance achieved when galvanometer shows zero current.
• Applications: comparing emf of cells, measuring internal resistance.

Solved Examples

Ex 1: If balancing length for emf \( \varepsilon_1 \) is \( l_1 = 40 \) cm and for \( \varepsilon_2 \) is \( l_2 = 60 \) cm, find \( \frac{\varepsilon_1}{\varepsilon_2} \).

• \( \frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2} = \frac{40}{60} = \frac{2}{3} \)

Practice Set

Q1: Potential drop \( V \) across length \( l \) of potentiometer wire is:

• A) \( V \propto l^2 \)
• B) \( V \propto l \)
• C) \( V \propto \frac{1}{l} \)
• D) Independent of \( l \)

Q2: If balancing lengths for two cells are 30 cm and 50 cm, ratio of their emfs is:

• A) 3:5
• B) 5:3
• C) 1:1
• D) 2:3

Q3: Internal resistance \( r \) of a cell is given by \( r = (\frac{l_1 - l_2}{l_2}) R \). If \( l_1 = 60 \) cm, \( l_2 = 40 \) cm, and \( R = 10 \ \Omega \), find \( r \).

• A) 5 \( \Omega \)
• B) 10 \( \Omega \)
• C) 15 \( \Omega \)
• D) 20 \( \Omega \)

Answer Key

Q1: B) \( V \propto l \) — Potential drop is directly proportional to length.

Q2: A) 3:5 — Ratio of emfs equals ratio of balancing lengths.

Q3: C) 15 \( \Omega \) — \( r = \frac{60-40}{40} \times 10 = \frac{20}{40} \times 10 = 5 \times 3 = 15 \)

Memory Aids

• Potential drop \( V = k l \).
• Emf ratio = balancing length ratio.
• Internal resistance formula: \( r = \left(\frac{l_1 - l_2}{l_2}\right) R \).