Electricity and Magnetism

Electricity and Magnetism carries more marks than any other 0625 topic. Across recent Paper 4 sessions it has supplied 20-25% of the theory marks. It runs from simple bar magnets to transformers, and it links three equations students must use fluently: $Q = It$, $V = IR$ and $P = IV$.

What does this topic cover in IGCSE Physics 0625?

Seven areas: magnetism and magnetic fields, electric charge, current and resistance, series and parallel circuits, full circuit calculations, electrical safety and power, then electromagnetic effects. The last area (induction, the motor effect and transformers) is where Core and Extended differ most. Extended students must use the transformer equation $\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}$ and explain why high-voltage transmission cuts power loss.

Why do students lose marks here?

Three reasons appear in examiner reports every session. First, mixing up series and parallel rules. Current is the same everywhere in series; voltage is the same across each branch in parallel. Students swap these under pressure. Second, unit errors. A current of 250 mA must become 0.25 A before it goes into $V = IR$. Milliamps, kilowatt-hours and megaohms all trip students in Paper 2 and Paper 4. Third, direction rules. Fleming’s left-hand rule questions reward a clear stated method; vague answers about “the field pushing the wire” score zero.

Worried the topic is too abstract to self-study? It is the most pattern-based topic in 0625. The same five circuit setups (series resistors, parallel lamps, potential divider, diode, thermistor) cover almost every past-paper question since 2016.

How should you revise Electricity and Magnetism?

Work circuit-first, theory-second. Draw the circuit, label every known value with units, then choose the equation. Practise the resistor combination rules until automatic: add resistances in series; in parallel use $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$. Memorise the three definitions examiners ask verbatim: current as charge per unit time, e.m.f. as work done per unit charge by the source, p.d. as work done per unit charge between two points. Then drill 10 transformer and induction questions, because electromagnetic effects is the highest-weighted Extended section.

Stuck on circuits? Book a free 1-hour taught trial. It is 1-to-1, online, RM80/hr after that. We teach a real circuits lesson in the trial and start fixing the series/parallel confusion the same hour. Message us on WhatsApp.

How Electricity and Magnetism Is Assessed Across the Papers

This is the highest-weighted topic on the theory papers, so it appears in force every session. Papers 1 (Core) and 2 (Extended) carry a steady run of multiple-choice items: circuit symbols, $V = IR$, series and parallel rules, and field patterns around magnets and wires. Papers 3 (Core) and 4 (Extended) build multi-part structured questions, often the longest on the paper, combining a circuit calculation with a definition or an explanation. The Core and Extended split is widest in electromagnetic effects. The Supplement tier owns the transformer equation $\left(\dfrac{V_p}{V_s} = \dfrac{N_p}{N_s}\right)$, the reasoning behind high-voltage transmission, electromagnetic induction, and the motor effect with Fleming’s left-hand rule. Papers 5 (Practical) and 6 (Alternative to Practical) test circuit building: connecting an ammeter in series and a voltmeter in parallel, plotting current against voltage to find resistance, and reading a meter scale correctly. Because the topic is so heavily weighted, a single confident method, draw the circuit and label every value with its unit before choosing an equation, protects marks across all three paper pairs.

A Worked Example That Shows the Method

A $4.0\ \Omega$ resistor and an $8.0\ \Omega$ resistor are connected in series across a $12\ \text{V}$ supply. Calculate the current in the circuit and the power dissipated in the whole circuit. [4]

Worked solution:

$R_{\text{total}} = R_1 + R_2 = 4.0 + 8.0 = 12\ \Omega$

$I = \dfrac{V}{R} = \dfrac{12}{12} = 1.0\ \text{A}$

$P = IV = 1.0 \times 12 = 12\ \text{W}$

So the current is $1.0\ \text{A}$ and the power dissipated is $12\ \text{W}$ (2 significant figures). The method matters: in series the resistances add, the current is the same everywhere, and the total power equals the supply voltage times that current. Had the resistors been in parallel, you would combine them with the reciprocal rule first, so always identify the arrangement before reaching for an equation.

Mark scheme:

• M1: total resistance $= 4.0 + 8.0 = 12\ \Omega$ (series resistances add)
• A1: $I = \dfrac{V}{R} = 1.0\ \text{A}$ with the unit
• M1: $P = IV$ with correct substitution
• A1: $P = 12\ \text{W}$

Before you start the subtopics below, print the full equation set from our physics equations list and keep the calculation-question method guide open beside your practice papers.