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.
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.
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.
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
- E = I(R + r): 12 = I × (5.5 + 0.50) = 6.0I, so I = 2.0 A.
- 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.
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
- Total resistance = 2.5 + 0.50 = 3.0 Ω.
- I = 6.0/3.0 = 2.0 A.
- 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.
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
- Vout = 12 × 4.0/(2.0 + 4.0) = 8.0 V.
- 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.
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.
- Internal voltage drop = 1.60 − 1.40 = 0.20 V.
- r = (ε − V)/I = 0.20/0.50.
0.40 Ω.
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.
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.
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 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.
- The voltage intercept is 3.0 V.
- Gradient = (2.2 − 3.0)/(2.0 − 0) = −0.40 V A⁻¹.
- Internal resistance is the negative of the gradient.
Emf 3.0 V; internal resistance 0.40 Ω.
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.
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.
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
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.
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
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