FAD1022 L10 — EMF and Internal Resistance

This lecture covers the fundamental concepts of electromotive force (EMF), internal resistance in electrical sources, and the relationship between EMF and terminal voltage. It also includes a review of electrical conduction, resistivity, and Ohm's Law.

Overview

flowchart TB
    A[L10 EMF & Internal Resistance] --> B[10.1 Electrical Conduction]
    A --> C[10.2 Conductivity & Resistivity]
    A --> D[10.3 Ohm's Law]
    A --> E[10.4 Electromotive Force]
    A --> F[10.5 Internal Resistance]
    A --> G[10.6 Source of EMF & Internal Resistance]
    
    B --> B1[Current definition I=ΔQ/Δt]
    C --> C1[ρ = RA/l]
    C --> C2[σ = 1/ρ]
    D --> D1[V = IR]
    D --> D2[Ohmic vs Non-ohmic]
    E --> E1[ε = P/I]
    E --> E2[EMF vs Potential Difference]
    F --> F1[Ir voltage drop]
    G --> G1[V = ε - Ir]
    G --> G2[Power dissipation]

10.1 Electrical Conduction

Key Concepts

Electric current: Flow of electric charges
Current definition: Rate at which charges flow across any cross-sectional area

$$I = \frac{\Delta Q}{\Delta t}$$

where:

$I$ = current (Amperes, A)
$\Delta Q$ = charge (Coulombs, C)
$\Delta t$ = time (seconds, s)
SI unit: C s⁻¹ or Ampere (A)
Direction of current: Direction of positive charge movement
Electron movement: Electrons move in the opposite direction of conventional current

Worked Example #10.1

Problem: A wire carries 6.0 A current. Determine: (a) Charge flowing in 5.0 s (b) Number of electrons in 5.0 s

Solution:

(a) Using $I = \frac{\Delta Q}{\Delta t}$:

$$\Delta Q = I \Delta t = (6.0)(5.0) = 30 \text{ C}$$

(b) Using $Q = ne$ where $e = 1.6 \times 10^{-19}$ C:

$$n = \frac{Q}{e} = \frac{30}{1.6 \times 10^{-19}} = 1.88 \times 10^{20} \text{ electrons}$$

10.2 Electrical Conductivity and Resistivity

Resistivity ($\rho$)

Definition: Resistance per unit length per unit cross-sectional area
Resistivity is an intrinsic property of the material (does not depend on size or shape)

$$R = \rho \frac{l}{A}$$

where:

$R$ = resistance (Ω)
$\rho$ = resistivity (Ω·m)
$l$ = length (m)
$A$ = cross-sectional area (m²)

Alternatively:

$$\rho = \frac{RA}{l}$$

Conductivity ($\sigma$)

Definition: Reciprocal of resistivity; measures ability to conduct electricity

$$\sigma = \frac{1}{\rho} = \frac{l}{RA}$$

Unit: S/m (Siemens per meter)

Material Classification

Property	Good Conductors	Poor Conductors (Insulators)
Resistivity ($\rho$)	Low ($\sim 10^{-8}$ Ω·m)	High ($\sim 10^{10}$+ Ω·m)
Conductivity ($\sigma$)	High ($\sim 10^{7}$ S/m)	Low ($\sim 10^{-7}$ S/m)
Examples	Silver, Copper, Gold	Carbon steel, non-metals

Table of Resistivity and Conductivity at 20°C

Material	$\rho$ (Ω·m)	$\sigma$ (S/m)
Silver	$1.59 \times 10^{-8}$	$6.30 \times 10^{7}$
Copper	$1.68 \times 10^{-8}$	$5.96 \times 10^{7}$
Gold	$2.44 \times 10^{-8}$	$4.10 \times 10^{7}$
Aluminum	$2.82 \times 10^{-8}$	$3.5 \times 10^{7}$
Iron	$1.0 \times 10^{-7}$	$1.00 \times 10^{7}$
Carbon steel	$\sim 10^{-10}$	$1.43 \times 10^{-7}$

Worked Example #10.2

Problem: Calculate the resistance of gold (conductivity = 36 MS m⁻¹) with diameter 0.02 m and length 0.08 m.

Solution:

Given:

$\sigma = 36 \times 10^{6}$ S/m
$d = 0.02$ m → $r = 0.01$ m
$l = 0.08$ m

Step 1: Find resistivity $$\rho = \frac{1}{\sigma} = \frac{1}{36 \times 10^{6}} = 2.78 \times 10^{-8} \text{ Ω·m}$$

Step 2: Find cross-sectional area $$A = \pi r^{2} = \pi (0.01)^{2} = 3.142 \times 10^{-4} \text{ m}^{2}$$

Step 3: Calculate resistance $$R = \frac{\rho l}{A} = \frac{(2.78 \times 10^{-8})(0.08)}{3.142 \times 10^{-4}} = 7.08 \times 10^{-6} \text{ Ω}$$

10.3 Ohm's Law

Statement

Ohm's Law gives the relationship between potential difference across a conductor and the current flowing through it:

$$V \propto I \quad \text{or} \quad V = IR$$

where $R$ is the resistance (constant for ohmic conductors).

Ohmic Conductors

Resistance remains unchanged when current and voltage change
Current and voltage are directly proportional (linear relationship)
V vs I graph is a straight line
Examples: Pure metals (copper, aluminum)

$$R = \frac{1}{\text{slope of V-I graph}}$$

Non-Ohmic Conductors

Resistance changes with current, voltage, or temperature
Current and voltage are NOT directly proportional
V vs I graph is curved
Examples: Semiconductors, carbon, p-n junction diodes

Worked Examples

Problem #10.3: A bulb draws 0.1 A at 3.5 V. Find its resistance.

$$R = \frac{V}{I} = \frac{3.5}{0.1} = 35 \text{ Ω}$$

Problem #10.4: A toaster ($R = 126.25$ Ω) operates on 240 V. Find the current.

$$I = \frac{V}{R} = \frac{240}{126.25} = 1.90 \text{ A}$$

Problem #10.5: A wire ($R = 15$ Ω) has 850 C passing per minute. Find the potential difference.

First, find current: $$I = \frac{\Delta Q}{\Delta t} = \frac{850}{60} = 14.17 \text{ A}$$

Then: $$V = IR = (14.17)(15) = 212.55 \text{ V}$$

10.4 Electromotive Force (EMF)

Definition

EMF ($\varepsilon$) is the electrical energy per unit charge generated by a source, converted from chemical or other forms of energy.

$$\varepsilon = \frac{P}{I}$$

where:

$P$ = power (Watts)
$I$ = current (Amperes)

Key Characteristics

EMF is generated by a source so that charge can flow from one terminal to another
EMF is the potential difference of a source when NO current is being drawn
When current flows, the terminal voltage is always less than the EMF

Sources of EMF

Batteries
Generators
Solar cells
Fuel cells

EMF vs Potential Difference

EMF	Potential Difference
Energy supplied per unit charge by the source	Energy dissipated per unit charge in the circuit
Exists even when no current flows	Exists only when current flows
Measured across the source terminals (open circuit)	Measured across any component or the source (closed circuit)
Always greater than terminal voltage when delivering current	Equal to EMF only when $I = 0$

10.5 Internal Resistance

Concept

Internal Resistance ($r$) is the resistance to current flow within the source itself.

Every real source (battery, generator) has internal resistance
It acts as if it were a resistor in series with the battery
When current $I$ flows, there is a potential drop $Ir$ across the internal resistance

Why Terminal Voltage < EMF

When there is current in a circuit:

$$V_{\text{terminal}} = \varepsilon - Ir$$

The potential difference across the terminals is always less than the EMF of the source due to the voltage drop across the internal resistance.

Circuit Representation

flowchart LR
    A[+] --- B[ε]
    B --- C[r]
    C --- D[-]
    D --- E[External R]
    E --- A
    
    style B fill:#90EE90
    style C fill:#FFB6C1

EMF source ($\varepsilon$) in series with internal resistance ($r$)
External load resistance ($R$) connected across the terminals

10.6 Source of EMF and its Internal Resistance

Fundamental Equations

The EMF of a source can be expressed as:

$$\varepsilon = IR + Ir = V + Ir$$

where:

$IR$ = potential drop across external resistance (terminal voltage $V$)
$Ir$ = potential drop across internal resistance

The terminal voltage is:

$$V = \varepsilon - Ir$$

Three Conditions

Condition	Formula	Explanation
Delivering current (discharging)	$V = \varepsilon - Ir$	Battery supplies current to load; terminal voltage < EMF
Receiving current (charging)	$V = \varepsilon + Ir$	Battery is being charged; terminal voltage > EMF
No current (open circuit)	$V = \varepsilon$	No internal voltage drop; terminal voltage equals EMF

Terminal Voltage

Terminal Voltage ($V$) is the operating potential difference of a source when connected to a circuit.

The value of internal resistance of a battery is typically small
Terminal voltage is only slightly less than EMF under normal operation

Power Dissipation

Power dissipated by the internal resistance:

$$P_{r} = I^{2}r = \frac{V_{r}^{2}}{r}$$

where $V_{r} = Ir$ is the voltage drop across internal resistance.

Worked Examples

Problem #10.6: An AA battery ($\varepsilon = 1.50$ V, $r = 0.25$ Ω) supplies 50 mA. Calculate terminal voltage.

$$V = \varepsilon - Ir = 1.50 - (50 \times 10^{-3})(0.25) = 1.50 - 0.0125 = 1.4875 \text{ V}$$

Problem #10.7: A dry cell delivers 1.50 A with terminal voltage 1.25 V. If EMF is 1.50 V, find internal resistance.

From $V = \varepsilon - Ir$:

$$r = \frac{\varepsilon - V}{I} = \frac{1.50 - 1.25}{1.50} = \frac{0.25}{1.50} = 0.17 \text{ Ω}$$

Problem #10.8: A car battery ($\varepsilon = 12$ V, $r = 0.050$ Ω) is being charged with 60 A.

(a) Terminal voltage (charging): $$V = \varepsilon + Ir = 12 + (60)(0.050) = 12 + 3 = 15 \text{ V}$$

(b) Rate of thermal energy dissipated (power in internal resistance): $$P = I^{2}r = (60)^{2}(0.050) = 180 \text{ W}$$

Problem #10.9: Ammeter-voltmeter method measures unknown resistance $R$. Voltmeter reads 3.0 V, ammeter reads 0.5 A. Find $R$.

$$R = \frac{V}{I} = \frac{3.0}{0.5} = 6 \text{ Ω}$$

Key Formulas Summary

Concept	Formula	Variables
Current	$I = \frac{\Delta Q}{\Delta t}$	$I$=current, $Q$=charge, $t$=time
Resistance	$R = \rho \frac{l}{A}$	$\rho$=resistivity, $l$=length, $A$=area
Conductivity	$\sigma = \frac{1}{\rho}$	$\sigma$=conductivity, $\rho$=resistivity
Ohm's Law	$V = IR$	$V$=voltage, $I$=current, $R$=resistance
EMF	$\varepsilon = \frac{P}{I}$	$\varepsilon$=EMF, $P$=power, $I$=current
Terminal Voltage (discharging)	$V = \varepsilon - Ir$	$V$=terminal voltage, $r$=internal resistance
Terminal Voltage (charging)	$V = \varepsilon + Ir$
Internal Resistance	$r = \frac{\varepsilon - V}{I}$
Power Dissipated	$P = I^{2}r$	Power lost in internal resistance

Related Concepts

Electromotive Force (EMF) — Definition, sources, and EMF vs potential difference
Internal Resistance — Concept, circuit analysis, and power dissipation
Terminal Voltage — Relationship to EMF and internal resistance
Ohm's Law — Linear relationship between voltage and current
Electrical Resistivity — Material property affecting resistance
Electrical Conductivity — Reciprocal of resistivity

Practice Problems

The lecture includes 9 worked problems (#10.1 through #10.9):

#10.1: Charge and electron count from current
#10.2: Resistance calculation from conductivity and geometry
#10.3: Resistance from Ohm's Law
#10.4: Current calculation from Ohm's Law
#10.5: Potential difference with time-varying charge
#10.6: Terminal voltage calculation (battery discharging)
#10.7: Internal resistance calculation
#10.8: Charging battery with power dissipation (3 parts)
#10.9: Ammeter-voltmeter method for resistance measurement

Source: FAD1022 Lecture 10 — Electromotive Force and Internal Resistance (38 slides) Institution: Universiti Malaya, Pusat Asasi Sains

FAD1022 L10 — EMF and Internal Resistance

Overview

10.1 Electrical Conduction

Key Concepts

Worked Example #10.1

10.2 Electrical Conductivity and Resistivity

Resistivity ($\rho$)

Conductivity ($\sigma$)

Material Classification

Table of Resistivity and Conductivity at 20°C

Worked Example #10.2

10.3 Ohm's Law

Statement

Ohmic Conductors

Non-Ohmic Conductors

Worked Examples

10.4 Electromotive Force (EMF)

Definition

Key Characteristics

Sources of EMF

EMF vs Potential Difference

10.5 Internal Resistance

Concept

Why Terminal Voltage < EMF

Circuit Representation

10.6 Source of EMF and its Internal Resistance

Fundamental Equations

Three Conditions

Terminal Voltage

Power Dissipation

Worked Examples

Key Formulas Summary

Related Concepts

Related Course Links

Practice Problems