Current, Potential Difference, e.m.f. and Resistance

Current, p.d., e.m.f. and resistance are the four quantities every IGCSE electricity question runs on, linked by $V = IR$, which has appeared on every 0625 Paper 4 since at least 2015. Examiners test the definitions as strictly as the calculations, so you need both word-perfect statements and a reliable substitution method.

What do current, p.d., e.m.f. and resistance actually mean?

Current is the charge passing a point per unit time. In words: current equals charge divided by time. In symbols: $I = \dfrac{Q}{t}$. Measure it with an ammeter connected in series.

Potential difference (p.d.) across a component is the work done by a unit charge passing through it. Electromotive force (e.m.f.) of a source is the work done by the source per unit charge driven round a complete circuit. Both are measured in volts with a voltmeter connected in parallel. The distinction matters: e.m.f. belongs to the supply; p.d. belongs to components.

Resistance is the ratio of p.d. to current. In words: resistance equals potential difference divided by current. In symbols: $R = \dfrac{V}{I}$, usually rearranged as $V = IR$.

Quantity	Symbol	Unit	Meter
Current	$I$	ampere (A)	ammeter, in series
Potential difference	$V$	volt (V)	voltmeter, in parallel
e.m.f.	$E$	volt (V)	voltmeter across the source
Resistance	$R$	ohm (Ω)	calculated from $V/I$
Charge	$Q$	coulomb (C)	(calculated)

Extended candidates also learn the wire rules: resistance is directly proportional to length and inversely proportional to cross-sectional area. Doubling the length doubles R; doubling the area halves R. Conventional current flows from positive to negative terminal outside the supply, opposite to electron flow.

How do you find resistance by experiment?

Connect the component in series with a cell, an ammeter and (ideally) a variable resistor. Connect a voltmeter across the component. Record V and I, then calculate $R = \dfrac{V}{I}$. The variable resistor lets you take several pairs of readings and plot a V-I graph; for a fixed resistor at constant temperature, the graph is a straight line through the origin. A filament lamp curves, because its resistance rises as the filament heats up.

Worked Exam Question

A filament lamp carries a current of 250 mA when the p.d. across it is 6.0 V. (a) Calculate the resistance of the lamp. (3 marks) (b) State and explain what happens to this resistance when the p.d. increases. (2 marks)

Solution. (a) Convert: $250\ \text{mA} = 0.25\ \text{A}$. Equation: $V = IR$, rearranged to $R = \dfrac{V}{I}$. Substitute: $R = \dfrac{6.0}{0.25}$. Answer: $R = 24\ \Omega$. (b) The resistance increases. A larger p.d. drives more current, the filament temperature rises, and a hotter filament has greater resistance.

Mark scheme:

• M1: converts 250 mA to 0.25 A.
• M1: $R = \dfrac{V}{I}$ with correct substitution.
• A1: 24 Ω with unit.
• B1: resistance increases.
• B1: link to temperature of filament rising.

Common Mistakes

• Leaving current in milliamps. Fix: convert mA to A before substituting. 250 mA into $V = IR$ uncorrected gives an answer 1000× too small or large.
• Defining e.m.f. as “the force of the battery”. Fix: e.m.f. is energy (work done) per unit charge, not a force, despite the name.
• Swapping ammeter and voltmeter placement. Fix: ammeter in series, voltmeter in parallel, and an ideal voltmeter has very high resistance.
• Writing $R = VI$. Fix: triangle-check the rearrangement. V sits on top; $R = \dfrac{V}{I}$.
• Calling the lamp graph “non-ohmic” with no reason. Fix: always add the temperature explanation for the curve.

Exam Technique Tip

Use the four-line layout on every calculation: equation in symbols, substitution with units converted, rearrangement if needed, answer with unit to 2 or 3 significant figures. Cambridge awards the method mark for a correct substitution even when the arithmetic slips, so never jump straight to a number. Our calculation-questions guide drills this layout across all 0625 equations.

How This Is Examined

This CS subtopic appears everywhere. Papers 1 and 2 test definitions, meter placement and one-step $V = IR$ multiple-choice. Papers 3 and 4 set structured calculations; Extended adds e.m.f. versus p.d. distinctions and the length/area resistance proportionality, sometimes as a ratio question. Papers 5 and 6 use the V-I experiment as a full practical: circuit drawing, tabulating readings and plotting the graph. Learn the experiment as a sequence of steps. Malaysian centres sitting Paper 6 (alternative to practical) get asked to describe it without ever touching the apparatus.