What is internal resistance and why does it matter?

Every real source resists the current it drives; energy is dissipated inside the source itself. The terminal voltage falls as current increases (V = E − Ir), which is why batteries deliver less voltage under heavy load and why 'lost volts' questions are AS favourites.

How does a potential divider with a thermistor work?

As temperature rises, the thermistor's resistance falls, so its share of the input voltage falls — and the voltage across the other resistor rises. Which output you take across decides whether the circuit turns things on when hot or when cold.

What do Kirchhoff's laws conserve?

The first law conserves charge: current into a junction equals current out. The second law conserves energy: around any loop, the energy given per coulomb (e.m.f.) equals the energy transferred per coulomb (sum of p.d.s).

← Edexcel IAL Physics revision

Edexcel IAL Physics revision · AS — Electricity

Electricity and d.c. circuits

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.

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What the syllabus demands

—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

Definitions that earn marks

Clear definitions to practise — check your course mark scheme

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.

The equations

Charge carriersI = n A v qn = number density, v = drift velocity

Resistivityρ = R A ÷ L · Ω m

E.m.f. and internal resistanceE = I(R + r)

Potential dividerVout = Vin × R₂ ÷ (R₁ + R₂)

More equations to practise: the Edexcel IAL formula sheet.

Where the marks die

Common mistakes to check

01
Confusing resistivity with resistance. Resistivity belongs to the material and never changes with the wire's shape; resistance belongs to the component and does. 'The resistivity doubles when the length doubles' scores zero.
02
Forgetting that terminal p.d. < e.m.f. only when current flows. With no current (open circuit), the voltmeter across the terminals reads the full e.m.f. — that is how e.m.f. is measured.
03
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 (temperature up → resistance down), then follow the ratio.
04
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.

One worked example, done properly

Question

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.

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

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