Dr Desouky physics tutoring logo
IGCSEA-LevelResultsFree resourcesContactالعربية
Join now
IGCSEA-LevelResultsFree resourcesContactالعربيةJoin now
Free resources/Edexcel IAL Physics revision/Electricity and d.c. circuits

Edexcel IAL Physics notes · AS — Electricity

Electricity and d.c. circuits

Current measures charge flow; potential difference measures energy transferred per charge. A circuit calculation must satisfy charge conservation at junctions and energy conservation around loops.

By Dr Desouky Physics AcademyUpdated 25 September 202611 min reading + practice

Your review checklist is saved on this browser. No sign-up needed.

In this lesson
  1. 01The key idea
  2. 02Separate what flows from what is transferred
  3. 03Model the energy transfer inside a cell
  4. 04Check temperature and loading assumptions
  5. 05Key definitions
  6. 06Equations & units
  7. 07Worked examples
  8. 08Practice & solutions
  9. 09Common mistakes
  10. 10Questions, explained
  11. 11Learning checklist
  12. 12Sources & citation

The key idea

IGCSE circuits with three upgrades that carry most of the AS electricity marks: resistivity, internal resistance, and the potential divider. Master those three and Paper 2's circuit section becomes routine.

Understand it · 01

Separate what flows from what is transferred↗

Current is the rate of flow of charge through a cross-section. In a metal, mobile electrons drift opposite to conventional current, while the circuit can transfer energy rapidly even though the drift speed is small. A potential difference of one volt means one joule is transferred per coulomb of charge between two points. Charge is not used up by a resistor; the moving charges transfer energy as they interact with the material.

A junction equation expresses charge conservation: total current entering equals total current leaving in a steady circuit. A loop equation expresses energy conservation: energy supplied per coulomb by sources equals energy transferred per coulomb to components around a closed loop. These rules explain why series components share current and parallel branches share potential difference. They do not require identical components, so use the actual resistance values rather than dividing current equally by default.

Understand it · 02

Model the energy transfer inside a cell↗

An emf is the energy supplied per unit charge by a source. A real cell can be represented by an ideal emf ε in series with internal resistance r. When it supplies current I, the terminal potential difference is V = ε − Ir. The term Ir is the energy transferred internally per unit charge. The terminal voltage is therefore usually smaller under load than when measured with a high-resistance voltmeter and negligible current.

Combining the model with an external resistor R gives I = ε/(R + r). A plot of terminal voltage against current has intercept ε and gradient −r if the source parameters remain constant. Take readings promptly or allow recovery between readings to limit heating and changes in the cell. Do not interpret the decreasing terminal voltage as charge disappearing; the distinction is between energy transfers inside and outside the source.

A cell model places ideal emf and internal resistance in series with an external load, with terminal voltage across the load.
Figure 2. The source supplies ε joules per coulomb; Ir is transferred internally and V is available to the load.Open full-size SVG diagram ↗

Pause & explain

Why does a cell’s terminal voltage fall when the current it supplies increases?

Check your reasoning

A larger current causes a larger internal voltage drop Ir, so V = ε − Ir decreases.

Understand it · 03

Check temperature and loading assumptions↗

Resistance is V/I at an operating point. Ohm’s law describes a component whose current is proportional to potential difference while relevant physical conditions, particularly temperature, stay constant. A filament lamp heats as current grows, changing its resistance and giving a curved characteristic. Resistivity separates the material from specimen dimensions through R = ρL/A; temperature still matters. Use a low current or short measurement time when determining the resistivity of a wire.

A potential divider splits a supply voltage across series resistances. For an unloaded output across R₂, Vout = VinR₂/(R₁ + R₂). Connecting a low-resistance load across the output makes it parallel with R₂ and changes the divider ratio. Light-dependent resistors and thermistors can provide a changing resistance for sensing, but predict the output direction from their position in the circuit. Memorising that a sensor makes voltage rise is unreliable without knowing which component the output spans.

Pause & explain

Are currents equal in two parallel branches with different resistances?

Check your reasoning

No. Their potential differences are equal; the lower-resistance branch carries the larger current.

Key definitions

Resistivity (ρ)
A property of the material defined by ρ = RA ÷ L — numerically the resistance of a cube of the material of unit dimensions. Units: Ω m.
Internal resistance (r)
The resistance within a source of e.m.f., causing the terminal potential difference to drop below the e.m.f. when current flows: V = E − Ir.
Kirchhoff's first law
The sum of currents into a junction equals the sum of currents out — a consequence of conservation of charge.
Kirchhoff's second law
Around any closed loop, the sum of e.m.f.s equals the sum of potential differences — a consequence of conservation of energy.

Equations, units & conditions

Charge carriers

I = n A v q

n = number density, v = drift velocity

Resistivity

ρ = R A ÷ L

Units: Ω m

E.m.f. and internal resistance

E = I(R + r)

Potential divider

Vout = Vin × R₂ ÷ (R₁ + R₂)

Unloaded output across R₂. A connected load changes its effective resistance.

Open the Edexcel IAL Physics formula sheet →

Worked examples, step by step

Example 1

A battery of e.m.f. 12 V and internal resistance 0.50 Ω drives a current through an external resistor of 5.5 Ω. Find the current and the terminal p.d.

Follow the method

  1. E = I(R + r): 12 = I × (5.5 + 0.50) = 6.0I, so I = 2.0 A.
  2. Terminal p.d. V = E − Ir = 12 − (2.0 × 0.50).

I = 2.0 A, V = 11 V

Practice questions with worked solutions

Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.

  1. Build · 3 suggested marks

    1. A loaded cell

    A cell of emf 6.0 V and internal resistance 0.50 Ω supplies a 2.5 Ω resistor. Find the current and terminal voltage.

    Show solution and self-check: question 1
    1. Total resistance = 2.5 + 0.50 = 3.0 Ω.
    2. I = 6.0/3.0 = 2.0 A.
    3. Terminal voltage = IR = 2.0 × 2.5 = 5.0 V.

    2.0 A and 5.0 V.

    Self-check · one mark per point

    • Include internal resistance in the total.
    • Find 2.0 A.
    • Calculate the terminal voltage across the external resistor.
  2. Build · 3 suggested marks

    2. An unloaded divider

    A 12 V supply is connected across 2.0 kΩ and 4.0 kΩ resistors in series. Find the unloaded output across the 4.0 kΩ resistor. Explain what a finite-resistance load in parallel with it would do.

    Show solution and self-check: question 2
    1. Vout = 12 × 4.0/(2.0 + 4.0) = 8.0 V.
    2. A parallel load reduces the effective lower resistance, reducing its fraction of the total supply voltage.

    8.0 V unloaded; a finite-resistance load reduces the output.

    Self-check · one mark per point

    • Apply the correct divider fraction.
    • Obtain 8.0 V.
    • Explain loading using the reduced parallel resistance.

Common mistakes & how to avoid them

  • Confusing resistivity with resistance. For the same material and temperature, resistivity is independent of the specimen dimensions; resistance also depends on length and cross-sectional area. 'The resistivity doubles when the length doubles' scores zero.
  • For a source supplying current, V = ε − Ir. With negligible current, a high-resistance voltmeter estimates the emf. A rechargeable cell being charged has the opposite current direction, so its terminal voltage need not be below its emf.
  • Potential divider logic inverted: the output is taken across one resistor, and increases when that resistor's share of the total resistance increases. For a thermistor circuit, decide what happens to its resistance first (for an NTC thermistor, temperature up → resistance down), then follow the ratio.
  • Filament lamp I-V curve explained without physics: current heats the filament, the lattice ions vibrate more, collisions with electrons increase, so resistance rises — that is why the graph bends.

Understand the why

Questions, explained

Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.

  1. What is internal resistance and why does it matter?
  2. How does a potential divider with a thermistor work?
  3. What do Kirchhoff's laws conserve?
  4. How do I find emf and internal resistance from a V–I graph?
  5. Why does connecting a load change a potential divider’s voltage?
  6. What is the difference between resistance and resistivity, and why does temperature matter?
Browse all physics questions →

What is internal resistance and why does it matter?

Internal resistance represents energy transferred inside a source when it supplies current. In the simple cell model, terminal voltage is V = ε − Ir, so a greater supplied current leaves less energy per coulomb available to the external circuit.

The emf ε describes the source’s total energy supply per unit charge, while Ir describes the internal part of that transfer. A high-resistance voltmeter draws little current and can therefore estimate the emf. Under a heavier load the current is larger and the voltage drop becomes more noticeable. This model assumes the emf and internal resistance remain approximately constant during the measurement; heating and chemical changes can alter them. If a rechargeable cell is being charged, current direction reverses relative to discharge, so terminal voltage need not be below its emf.

Infer internal resistance

A cell has emf 1.60 V and terminal voltage 1.40 V while supplying 0.50 A. Find its internal resistance.

  1. Internal voltage drop = 1.60 − 1.40 = 0.20 V.
  2. r = (ε − V)/I = 0.20/0.50.

0.40 Ω.

  • Read the supporting explanation ↑

How does a potential divider with a thermistor work?

For an NTC thermistor, resistance falls as temperature rises. In an unloaded series divider, the voltage across the thermistor then falls while the voltage across the other resistor rises. The output direction depends on where it is measured.

Draw the supply, both components and the output terminals before predicting the response. The divider formula uses the resistance across which the output is taken in the numerator. A connected measuring circuit may load that branch, so its parallel resistance must be included if the loading is significant. The usual school thermistor is NTC, but not every temperature-dependent resistor follows the same trend. Calibration is needed if the circuit is to measure temperature rather than simply switch at a threshold; the thermistor’s resistance–temperature relation is generally nonlinear.

  • Read the supporting explanation ↑

What do Kirchhoff's laws conserve?

Kirchhoff’s junction rule expresses conservation of charge, and the loop rule expresses conservation of energy per charge. In a steady circuit, current entering a junction equals current leaving, while signed potential changes around a closed loop sum to zero.

A resistor transfers energy from the circuit but does not consume charge, so the current does not progressively shrink along one series branch. At a junction, different branch currents can occur because different resistances share the same node-to-node potential difference. For a loop equation, choose a direction and apply one consistent sign convention to emf rises and potential drops. If the algebra gives a negative assumed current, the physical current flows opposite to the arrow initially chosen; the negative result is information rather than evidence that the equation must be discarded.

  • Read the supporting explanation ↑

How do I find emf and internal resistance from a V–I graph?

For the model V = ε − Ir, a terminal-voltage–current graph has voltage intercept ε and gradient −r. A falling straight line therefore allows the source emf and internal resistance to be estimated from several measurements.

The axes matter: plotting current vertically against voltage reverses the gradient relationship. Use a suitable range of loads without overheating the source, and take repeated readings to assess scatter. A curve may indicate changing source conditions or an imperfect model rather than a simple arithmetic mistake. Extrapolating to zero current estimates the open-circuit voltage, but the fitted line should be supported by a reasonable measurement range. The magnitude of the gradient has units V/A, equivalent to ohms, providing a dimensional check on the internal-resistance calculation.

A terminal-voltage–current graph falls linearly from a 3.0 volt intercept to 2.2 volts at two amperes.
The intercept gives emf and the negative gradient gives internal resistance.Open full-size SVG diagram ↗

A cell graph

A straight V–I graph passes through (0 A, 3.0 V) and (2.0 A, 2.2 V). Find emf and internal resistance.

  1. The voltage intercept is 3.0 V.
  2. Gradient = (2.2 − 3.0)/(2.0 − 0) = −0.40 V A⁻¹.
  3. Internal resistance is the negative of the gradient.

Emf 3.0 V; internal resistance 0.40 Ω.

  • Read the supporting explanation ↑

Why does connecting a load change a potential divider’s voltage?

A load connected across a divider output is in parallel with that divider resistor. It reduces the effective resistance of that branch and usually lowers the output voltage, so the unloaded divider formula cannot be used with the original resistance alone.

First combine the output resistor and load using the parallel-resistance rule. Then use that equivalent resistance in the divider ratio. A voltmeter with resistance much larger than the output branch causes little change, which is why high input resistance is desirable in measurement circuits. A low-resistance device can change the voltage substantially and may make the divider unsuitable as a power supply. The same idea explains why connecting a new sensor or meter can alter the quantity being measured; the measuring instrument becomes part of the circuit.

  • Read the supporting explanation ↑

What is the difference between resistance and resistivity, and why does temperature matter?

Resistance depends on the specimen’s length, cross-sectional area and material resistivity through R = ρL/A. Resistivity characterises the material at stated conditions, particularly temperature; it is not generally a temperature-independent constant.

A longer wire of unchanged cross-section offers more resistance, while a larger cross-section provides more paths for charge flow. These geometry changes do not by themselves change the material’s resistivity if its condition remains the same. Stretching a real wire, however, can alter several properties simultaneously, so read the assumptions carefully. In a metal, raising temperature usually increases resistivity as lattice vibrations increase scattering of charge carriers. During a practical resistivity measurement, a large current can heat the wire and undermine a comparison intended to vary only its length.

  • Read the supporting explanation ↑

Your learning checklist

Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.

  • Use I = nAvq for charge carriers in a conductor
  • Define resistivity: ρ = RA ÷ L
  • Apply Kirchhoff's first law (charge conservation) and second law (energy conservation)
  • Solve circuits with e.m.f. and internal resistance: E = I(R + r)
  • Design and analyse potential dividers, including with thermistors and LDRs
  • Sketch I-V characteristics: ohmic conductor, filament lamp, diode
Check which exam board you study

Sources & how to cite these notes

Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.

  • Pearson Edexcel International AS/A Level Physics — specification

Reference this page

Dr Desouky Physics Academy. Edexcel IAL Physics: Electricity and d.c. circuits. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/electricity-and-dc-circuits

Use the arrow beside an explanation heading to link straight to that section. Spotted a problem? Contact the academy with the topic and the step you want us to check.

Connect this to your next topic

AS / A-LevelPhysical quantities and units →AS / A-LevelCapacitance →AS / A-LevelElectric fields →

Practise the supporting skills

  • Practise graph gradients, areas and units

Make this topic part of your revision routine

Choose one idea from these notes, attempt a fresh question and use the result to plan your next session. The Physics revision system connects your course and current difficulty to a guide, a practical study task and teaching support when you need it.

Previous topicWaves and superpositionNext topicFluids, upthrust and viscosity →

Find your physics class

Your next chapter starts with one message.

Tell Dr Desouky your level, exam board and city. Choose in-person physics lessons in Abu Dhabi or live online lessons across the UAE and Gulf.

Abu Dhabi in-person tutoringOnline tutoring for DubaiOnline tutoring for Sharjah
Ask about physics lessons

AED 150 · 90-minute live online session

Contact Dr Desouky for in-person availability, location and fees.

Dr Desouky physics tutoring logo

IGCSE, AS and A-Level Physics with Dr Mohamed Desouky. In-person lessons in Khalidiya, near Abu Dhabi Corniche; live online tutoring for Dubai, Sharjah, Qatar, Saudi Arabia and Kuwait.

Message Dr DesoukyDr Desouky Academy on FacebookDr Desouky Physics Academy on Google Maps+971 50 445 7621

Explore

  • Contact & lesson location
  • About Dr Desouky
  • The method
  • The difference
  • Student results
  • The Journal
  • 2025 results

Tutoring

  • IGCSE Physics tutor
  • AS Physics tutor
  • A-Level Physics tutor
  • Cambridge IGCSE
  • Edexcel IGCSE
  • Edexcel modular
  • Edexcel A-Level Physics
  • Edexcel M1

Locations

  • UAE
  • Dubai
  • Abu Dhabi
  • Al Ain
  • Sharjah
  • Ajman
  • Ras Al Khaimah
  • Saudi Arabia
  • Riyadh
  • Jeddah
  • Dammam & Khobar
  • Qatar
  • Doha
  • Kuwait
  • Kuwait City

Free resources

  • How to revise Physics
  • All free resources
  • Unit converter
  • Graph skills
  • Revision planner
  • IGCSE revision
  • A-Level Physics revision
  • Formula sheet
  • Physics questions
  • Definitions A–Z
  • Grade calculator
  • Physics marks report

Dr Desouky / Physics. Simplified.

© 2026 Dr Desouky. All rights reserved.