Basics of Electricity — AQA A-level Physics

SPEC 3.5.1.1, 3.5.1.2 • ELECTRICITY

Current, Charge and Resistance

Understanding the fundamental quantities of electric circuits and their I-V characteristics

Free Electrons in Metals

In a metal, there are many free electrons that are not bound to any particular atom and can move freely through the metal lattice.

Yellow: Metal ions (fixed, positive) | Blue: Free electrons (mobile, negative)

Key Concept

When a potential difference is applied, free electrons experience a force and drift towards the positive terminal. This flow of electrons is electric current.

Electric Current

Definition

Electric current is the rate of flow of electric charge.

I = \frac{Δ Q}{Δ t}

SymbolQuantityUnit

I

CurrentAmpere (A)

Δ Q

ChargeCoulomb (C)

Δ t

TimeSecond (s)

Key Concept

1 ampere = 1 coulomb per second

A current of 1 A means 1 C of charge flows past a point every second.

Charge and Electrons

Electric charge is carried by electrons. Each electron carries a tiny amount of charge:

Elementary Charge

e = 1.6 \times 1 0^{- 19}

CThe charge on one electron is −1.6 × 10⁻¹⁹ C (negative)

The total charge Q depends on the number of electrons n:

Charge Formula

Q = n \times e

Useful Conversion

1 coulomb = 6.25 × 10¹⁸ electrons

Worked Example01

A current of 2.5 A flows through a wire. Calculate the charge that passes a point in 3 minutes.

1

Known: I = 2.5 A, t = 3 min = 180 s

2

Use

Δ Q = I \times Δ t

3

Δ Q = 2.5 \times 180 = 450

C

Answer: 450 C

Worked Example02

A current of 25 mA flows for 10 ms. (a) What charge has passed? (b) How many electrons have flowed?

(a)

I = 25 mA = 0.025 A, t = 10 ms = 0.010 s

•

Q = I \times t = 0.025 \times 0.010 = 2.5 \times 1 0^{- 4}

C

(b)

n = \frac{Q}{e} = \frac{2.5 \times 1 0 ^{- 4}}{1.6 \times 1 0 ^{- 19}} = 1.56 \times 1 0^{15}

electrons

Answer: (a) 2.5 × 10⁻⁴ C (b) 1.56 × 10¹⁵ electrons

Current Direction

There are two ways to describe current flow:

The historical convention assumes positive charges flow. Used in all circuit analysis.

What actually happens in metals. Electrons are negative so move opposite to I.

Exam Rule

Always use conventional current unless the question specifically asks about electron flow. The two directions are opposite to each other.

Measuring Current

Connection Rule

An ammeter must be connected in series with the component.

It can be placed anywhere in the series circuit — the current is the same at all points.

Same current flows through ammeter and R

Ideal Ammeter

In exams, assume ammeters have zero resistance unless told otherwise.

Potential Difference

Definition

Potential difference (p.d.) is the work done per unit charge in moving charge between two points.

V = \frac{W}{Q}

SymbolQuantityUnit

V

Potential differenceVolt (V)

W

Work done / EnergyJoule (J)

Q

ChargeCoulomb (C)

Key Concept

1 volt = 1 joule per coulomb

Voltage can be thought of as the "push" that moves charge around a circuit.

Worked Example03

A battery transfers 36 J of energy when 8 C of charge passes through a lamp. Calculate the p.d. across the lamp.

1

Known: W = 36 J, Q = 8 C

2

Use

V = \frac{W}{Q}

3

V = \frac{36}{8} = 4.5

V

Answer: 4.5 V

Measuring Voltage

Connection Rule

A voltmeter must be connected in parallel across the component.

Voltmeter measures p.d. across R

Ideal Voltmeter

In exams, assume voltmeters have infinite resistance unless told otherwise.

Resistance

Definition

Resistance is defined as the ratio of potential difference to current.

R = \frac{V}{I}

SymbolQuantityUnit

R

ResistanceOhm (Ω)

V

Potential differenceVolt (V)

I

CurrentAmpere (A)

Physical Meaning

Resistance measures how hard it is for current to flow. 1 ohm = 1 volt per ampere

Measuring Resistance

To find the resistance of component X, measure both current through it and p.d. across it:

Method

Ammeter in series — measures current I through X

Voltmeter in parallel — measures p.d. V across X

Calculate:

R = \frac{V}{I}

Why This Works

The ammeter reads the exact current flowing through X. The voltmeter reads the exact p.d. across X. Using R = V/I gives the resistance.

Worked Example04

A 6 V battery is connected to an electromagnet and a current of 1.5 A flows. What is the resistance?

1

Known: V = 6 V, I = 1.5 A

2

Use

R = \frac{V}{I}

3

R = \frac{6}{1.5} = 4

Ω

Answer: 4 Ω

Ohm's Law

For an ohmic conductor, current is directly proportional to potential difference, provided physical conditions (especially temperature) remain constant.

I \propto V

(at constant temperature)

Key Distinction

R = V/I is the DEFINITION of resistance — applies to everything

Ohm's Law (I ∝ V) is a SPECIAL CASE — only applies to ohmic conductors

Common Misconception

A filament lamp has resistance (R = V/I works), but does NOT obey Ohm's Law because I is not proportional to V.

Worked Example05

A current of 0.5 mA flows through a 100 kΩ resistor. What is the p.d. across it?

1

I = 0.5 mA = 0.0005 A, R = 100 kΩ = 100000 Ω

2

V = IR = 0.0005 × 100000 = 50 V

Answer: 50 V

I-V Characteristics

An I-V characteristic is a graph showing how current varies with potential difference.

Gradient = ΔI/ΔV = 1/R

Gradient Interpretation

For an I-V graph (I on y-axis, V on x-axis):

Gradient = $\frac{1}{R}$

A steeper gradient means lower resistance.

Check Your Axes!

If V is on y-axis and I on x-axis, then gradient = R (not 1/R)

Components to Know

You need to know the I-V characteristics for: Ohmic conductor, Filament lamp, and Semiconductor diode

Ohmic Conductor (Resistor)

Straight line through origin

Key Features

Current proportional to voltage (I ∝ V)

Constant gradient = constant R

Obeys Ohm's law

Symmetric for ±V

What It Is

A component with constant resistance. Examples: metal wires, carbon resistors (at constant temperature).

Exam Tip

Gradient = 1/R. Steeper line = lower resistance.

Filament Lamp

Curved line through origin

Key Features

Gradient decreases as |V| increases

I is NOT proportional to V

Resistance increases with current

Does NOT obey Ohm's law

What It Is

A thin tungsten wire that glows when hot. Higher current → higher temperature → more ion vibrations → more collisions → R increases.

Why It Curves

More current → hotter → higher R → less current increase per volt.

Worked Example06

Explain why the I-V characteristic of a filament lamp curves as p.d. increases.

1

As p.d. increases, current increases

2

Increased current causes filament temperature to rise

3

Higher temperature → metal ions vibrate more

4

Electrons collide more frequently → resistance increases

5

Higher resistance means gradient decreases

Answer: Temperature increases, causing more collisions and higher resistance.

Semiconductor Diode

Arrow shows current direction

Forward: conducts | Reverse: blocks

Key Features

Forward bias: Conducts above ~0.7V

Reverse bias: No current (very high R)

Acts as one-way valve for current

Does NOT obey Ohm's law

What It Is

Made from semiconductor (e.g. silicon). Only allows current in one direction. Threshold voltage (~0.7V) must be exceeded to conduct.

Breakdown

In reverse bias, exceeding breakdown voltage causes sudden conduction — can damage the diode.

Worked Example07

A diode has threshold 0.7 V. Describe what happens with: (a) 0.5 V forward, (b) 2.0 V forward, (c) 5.0 V reverse.

(a)

0.5 V < 0.7 V threshold, so virtually no current flows (high resistance)

(b)

2.0 V > 0.7 V threshold, so diode conducts freely (low resistance)

(c)

Reverse bias blocks current regardless of magnitude (no current)

Answer: (a) No current (b) Current flows (c) No current

Practice

Question 1: A charge of 1800 C passes through a heater in 10 minutes at 240 V. Calculate: (a) the current, (b) the energy transferred.

(a)

t = 10 min = 600 s;

I = \frac{Q}{t} = \frac{1800}{600} = 3

A

(b)

W = VQ = 240 × 1800 = 432000 J = 432 kJ

Answer: (a) 3 A; (b) 432 kJ

Practice

Question 2: A component has I = 50 mA at V = 6.0 V, and I = 80 mA at V = 12.0 V. Calculate R at each p.d. Does it obey Ohm's law?

1

At 6.0 V: R = 6.0/0.050 = 120 Ω

2

At 12.0 V: R = 12.0/0.080 = 150 Ω

3

R changed (120 → 150 Ω), so I is NOT proportional to V

Answer: R₁ = 120 Ω, R₂ = 150 Ω; Does NOT obey Ohm's law

Practice

Question 3: 20 C passes a point in 4 s. (a) What is the current? (b) How many electrons pass per second?

(a)

I = \frac{Q}{t} = \frac{20}{4} = 5

A

(b)

Charge per second = 5 C;

n = \frac{5}{1.6 \times 1 0 ^{- 19}} = 3.125 \times 1 0^{19}

electrons/s

Answer: (a) 5 A; (b) 3.125 × 10¹⁹ electrons per second

Summary

Key Formulas

FormulaMeaningUnits

I = \frac{Δ Q}{Δ t}

Current = charge ÷ timeA, C, s

V = \frac{W}{Q}

P.D. = energy ÷ chargeV, J, C

R = \frac{V}{I}

Resistance = p.d. ÷ currentΩ, V, A

Q = n e

Charge = electrons × eC, −, C

I-V Characteristics

Ohmic: Straight line through origin (constant R)

Filament lamp: Curve, R increases with temperature

Diode: Conducts above ~0.7 V forward only

Exam Tips

R = V/I is DEFINITION. Ohm's law (I ∝ V) is SPECIAL CASE.

Ammeter: series, zero R. Voltmeter: parallel, infinite R.

Gradient of I-V graph = 1/R