4.20625 · Section 4: Electricity and Magnetism

Electrical quantities

Key learnings

(E) NTC thermistor and LDR resistance
(E) An NTC thermistor's resistance decreases as temperature rises (heat frees more charge carriers). An LDR (light-dependent resistor)'s resistance decreases as light intensity rises (more light frees more carriers). Both are the opposite of a metal wire, whose resistance increases with temperature (more lattice vibration → more collisions). Conductors: low $R$ ; insulators: very high $R$ ; semiconductors (e.g. silicon): in between.
Current, charge, p.d. and resistance
Current $I$ = rate of flow of charge; $I = Q/t$ ; unit ampere (A) (1 A = 1 C/s) — say "of charge", not "of electrons". Charge $Q$ in coulombs (C); $Q = It$ . Potential difference $V$ = energy transferred per unit charge; $V = E/Q$ ; unit volt (V) (1 V = 1 J/C). Resistance $R$ = p.d. ÷ current; $R = V/I$ ; unit ohm (Ω) (1 Ω = 1 V/A). e.g. 9.0 V and 0.30 A give $R = 9.0/0.30 = 30$ Ω.
(E) E.m.f.: work/charge round the circuit
(E) E.m.f. ( $\varepsilon$ ) = work done by the source per unit charge in driving it around a complete circuit; unit volt (V) (both "work done" and "per unit charge" are required). P.d. is work per unit charge between two external points; e.m.f. covers the whole circuit, including inside the source. With current flowing, terminal p.d. < e.m.f. as energy is lost to internal resistance $r$ : $\varepsilon = V + Ir$ .

Common mistakes

Drawn from real examiner reports.

Resistance is V ÷ I, not V × I

The top calculation slip is writing $R = V \times I$ instead of $R = V/I$ . With $V = 4.8$ V and $I = 0.4$ A: wrong $R = 4.8 \times 0.4 = 1.92$ Ω; correct $R = 4.8 \div 0.4 = 12$ Ω. Sense-check: 12 Ω with 4.8 V gives 0.4 A, but 1.92 Ω would give 2.5 A. Write $R = V/I$ before substituting to secure the method mark.

June 2024 Paper 31 Q7(b): "The most common error was using an incorrectly rearranged form of V = I × R. The most common of these was R = V × I = 4.8 × 0.4 = 1.92."

(E) E.m.f. is not the battery voltage

(E) "The voltage of the battery" scores zero, and e.m.f. is not a force (despite the name, it is measured in volts, not newtons). The mark scheme needs: (1) work done / energy transferred, (2) per unit charge, (3) around a complete circuit. That "complete circuit" element is what separates e.m.f. from p.d. (which is between two external points).

June 2024 Papers 31 and 41 Q7(a): "The expected definition that e.m.f. is the work done in moving unit charge around a complete circuit was seen in few responses. A common error was to define e.m.f. as a force."

Voltmeter in parallel, ammeter in series

Draw the ammeter in series (the same current flows through it and the component) and the voltmeter in parallel (across the component). A voltmeter placed in series adds high resistance and blocks the current; an ammeter placed in parallel short-circuits the component. Each meter's position is a separate mark, so a misplaced voltmeter loses marks on its own.

June 2024 Paper 31/33 Q8(a): "Many candidates placed the voltmeter in series with the ammeter." November 2024 Paper 41/42 alternative to practical: "Some candidates inserted a thermistor or a variable resistor instead of a voltmeter, or positioned the voltmeter incorrectly — either between the terminals of the power supply or in series with some part of the circuit."

Conventional current vs electron flow

Conventional current flows from + to − in the external circuit; the electrons actually flow the opposite way, from − to +. Keep the two directions straight. Also define current as the rate of flow of charge, not "of electrons" — the definition mark needs the word "charge", and the flow could be ions in some conductors.

Convert time to seconds in Q = It

In $Q = It$ (and energy $E = IVt$ ) the time must be in seconds. Convert first: 5 minutes $= 5 \times 60 = 300$ s. Leaving the time in minutes makes the answer 60× too small. Always convert the time to seconds before you substitute, and check the unit of your final answer (C for charge, J for energy).

(E) Terminal p.d. is below e.m.f. under load

(E) The terminal p.d. equals the e.m.f. only at zero current (open circuit). As soon as current flows, some energy is transferred to the source's internal resistance $r$ , so the terminal p.d. falls below the e.m.f.: $\varepsilon = V + Ir$ . Assuming terminal p.d. = e.m.f. while current flows is a common error.

Exam tips

Write the equation before substituting
Write the equation as the first line — $R = V/I$ , $I = Q/t$ , or $V = IR$ — then substitute on the next line. Cambridge gives a compensatory (C) method mark for the correct formula, so you keep marks even if the arithmetic slips.
Always give the unit (Ω, A, V, C)
Every final answer needs its unit: resistance in Ω, current in A, p.d. in V, charge in C. The answer mark needs both the correct number and the correct unit — if the value is wrong or the unit missing, only working marks are given.
Sense-check the size of your answer
Sense-check the magnitude. If $R = V/I$ gives a tiny value where you expected tens of ohms, you have probably multiplied instead of divided. A quick "does this current or voltage look sensible?" catches most rearrangement slips before they cost the answer mark.
State a named direction for R change
When asked how a thermistor or LDR responds, give a named direction: an NTC thermistor's resistance decreases as it gets hotter; an LDR's resistance decreases in brighter light. "It changes" scores zero — you must say increases or decreases.

Revision notes