H2 Physics Current Electricity Notes | A-Level 9478

Q: What does A-Level Physics: 15) Currents Guide cover?
A: From drift velocity to diode rectifiers, this post unpacks Sub-topic 15 Currents of the 2026 H2 Physics syllabus for IP students and parents.

TL;DR
Treat current electricity as the “traffic system” of Paper 2. Mastering charge flow, r.m.s. AC values and rectifiers turns once-scary graph questions into free marks - and locks in concepts needed for Magnetism, Quantum and Practical.

Concrete example: how to use this page

If a question gives charge passing a point over time, use $I=Q/t$. If it gives power or energy in a resistor, connect $P=IV$, $V=IR$, and energy over time only after the current path is clear.

Current-electricity decision map

Misconception check: Conventional current direction and electron flow direction are opposite in a metal. Use conventional current for circuit analysis unless the question explicitly asks about electron motion.

Keep the electromagnetism arc tight by revisiting the H2 Physics notes hub; it threads this chapter together with Electric Fields, Circuits, and electromagnetic force topics.

1 Electric current $(I)$

Definition
Electric current is the rate of flow of charge through a surface:

$$I = \dfrac{Q}{t}.$$

1.1 Mini-drill

A charge of $\pu{12 mC}$ passes a point in $\pu{4.0 s}$.

$I = \pu{3.0 mA}$.

Exam cue: always convert milli-, micro- and nano-coulombs to coulombs before substituting.

2 Microscopic view - drift velocity

In a metal, free electrons move randomly but acquire a drift velocity $v$ when an external field is applied. Equating the charge that crosses a cross-section per second gives

$$I = n A v q$$

where

Tip for WA practice: treat $v$ as $\mu E$ when mobility $( \mu )$ and field $( E )$ are given.

Drift velocity setup checkpoint

Before using $I=nAvq$, make the microscopic quantities consistent. Most wrong answers come from area conversion or from treating electron charge as a negative current.

Worked check: for $I=\pu{2.0 A}$, $n=\pu{8.5\times10^{28} m-3}$, radius $r=\pu{0.50 mm}$

Misconception check: a small drift velocity does not mean the circuit signal is slow. Electrons already fill the conductor, and the electric field establishes the drift throughout the circuit.

3 Potential difference $(V)$

Potential difference is the electrical work done per unit charge:

$$V = \dfrac{W}{Q}.$$

Parents: remind your child to track units - joule per coulomb is the volt.

4 Electrical power

Combine charge-flow and Ohm's law ideas to get the “power trio”

$$P = VI, \qquad P = I^2 R, \qquad P = \dfrac{V^2}{R}.$$

Timing hack: write $P = VI$ at the top of data-handling questions; deriving the other two takes <15 s.

Power formula checkpoint

Choose the power formula from the component values you actually know. The voltage and resistance in $P=\dfrac{V^2}{R}$ must belong to the same component.

Worked check: a $\pu{12 V}$ cell with $r=\pu{1.0 ohm}$ supplies a $\pu{5.0 ohm}$ resistor. The current is $I=12/(5.0+1.0)=\pu{2.0 A}$

Misconception check: $P=VI$, $P=I^2R$, and $P=V^2/R$

5 e.m.f. vs p.d.

Remember: a cell's internal resistance turns some of its e.m.f. into heat inside the cell - that lost voltage never reaches the external circuit.

6 Alternating current essentials

6.1 Period & frequency

• Period $(T)$: time for one full cycle.
• Frequency $(f)$: $f = 1/T$.

6.2 Peak & r.m.s. values

For a sinusoid,

$$I_{\text{rms}} = \dfrac{I_0}{\sqrt{2}} , \qquad V_{\text{rms}} = \dfrac{V_0}{\sqrt{2}}.$$

6.2.1 AC quantity checkpoint

Before substituting, name the quantity the question is asking for. Most errors in this chapter come from using the right waveform but the wrong representative value.

Fast check: if the final answer is a power or heating comparison, your working should contain an r.m.s. value or an explicit average of $I^2R$, not just the peak value copied from the graph.

6.3 Why r.m.s.?

R.m.s. current produces the same heating effect in a resistor as a d.c. current of the same magnitude.

6.4 Equation of a sine wave

$$x = x_0 \sin \omega t,$$ where $\omega = 2\pi f$.

7 Mean power in a resistive load

For $R$ purely resistive and current $I = I_0 \sin \omega t$:

$$P_{\text{mean}} = \tfrac{1}{2} I_0^2 R = I_{\text{rms}}^2 R,$$

hence mean power is half the peak power.

8 Half-wave rectification

A single diode placed in series with a load blocks one half-cycle of the a.c. supply, allowing only positive (or negative) halves to pass. The output is a “pulsating d.c.” that still requires smoothing if a steady voltage is needed.

Rectifier waveform checkpoint

For half-wave rectifier questions, decide which half-cycle conducts before drawing or calculating. The diode does not change a sinusoid into a flat d.c. line; it removes one side of the waveform.

Worked check: if the input current would be $I=I_0\sin\omega t$, a positive half-wave rectifier gives the positive sine hump from $0$ to $\pi$, then zero from $\pi$

Misconception check: half-wave rectification and smoothing are separate ideas. The diode selects half-cycles; a capacitor is needed if the question wants reduced ripple.

9 Three WA timing rules (Currents edition)

1. Use syllabus pacing as a guide: Paper 2/3 average ~1.6 min/mark; Paper 4 ~3 min/mark.
2. Sketch peak and r.m.s. values before calculating - prevents factor-of-$\sqrt{2}$ slips.
3. For rectifier graphs, label axes with units first, then plot.

Need structured practice on Currents? Our H2 Physics tuition programme covers this topic with weekly problem sets and Paper 4 practical drills.

Comprehensive revision pack

9478 Section V, Topic 15 Syllabus outcomes

Candidates should be able to:

• (a) show an understanding that electric current is the rate of flow of charge and solve problems using $I = \dfrac{Q}{t}$.
• (b) derive and use the equation $I = nAvq$ for a current-carrying conductor, where $n$

Concept map (in words)

Charge carriers move with drift velocity when electric fields act. Potential difference measures energy per charge; power relations follow directly. e.m.f. supplies energy; p.d. dissipates it. AC descriptions require r.m.s. quantities, and rectifiers convert AC to pulsating DC.

Key relations

Derivations & reasoning to master

1. Drift velocity formula: equate charge passing per second through cross-section to $n A v q$.
2. Power relationships: combine Ohm's law $V = IR$ with $P = VI$.
3. r.m.s. derivation: integrate $I_0^2 \sin^2 \omega t$

Worked example 1 - drift velocity

A copper wire (area $\pu{1.5 mm^2}$) carries $\pu{4.0 A}$. Given free-electron density $8.5 \times 10^{28} \ \pu{m-3}$

Outline: convert area to $\pu{m^2}$, apply $v = \dfrac{I}{n A q}$. Typical answer $\approx \pu{0.20 mm.s-1}$

$$A = 1.5\,\pu{mm^2} = 1.5\times 10^{-6}\,\pu{m^2}.$$

$$nAq = (8.5\times 10^{28})(1.5\times 10^{-6})(1.60\times 10^{-19}) = 2.04\times 10^{4}.$$

$$v = \dfrac{4.0}{2.04\times 10^{4}} = 1.96\times 10^{-4}\,\pu{m.s-1}.$$

So $v \approx \pu{0.196 mm.s-1}$, consistent with the quoted $\approx \pu{0.20 mm.s-1}$

Worked example 2 - internal resistance & power

A $\pu{12 V}$ cell with internal resistance $\pu{0.40 ohm}$ supplies a $\pu{4.0 ohm}$ resistor. Find terminal p.d., current, and power dissipated in cell vs load. Suggest how results change if a second identical resistor is added in parallel.

Method: current $I = \dfrac{\mathcal{E}}{R + r}$, compute terminal voltage $\mathcal{E} - I r$, compare powers $I^2 R$

$$I = \dfrac{12}{4.0 + 0.40} = \pu{2.73 A},\qquad V_{\text{terminal}} = 12 - Ir = 12 - (2.73)(0.40) = \pu{10.9 V}.$$

\[ P\{\text{load}} = I^2R = (2.73^2)(4.0)=\pu{29.8 W},\qquad P\{\text{internal}} = I^2r = (2.73^2)(0.40)=\pu{3.0 W}. \]

With a second $\pu{4.0\,\Omega}$ resistor in parallel, the external resistance becomes $\pu{2.0\,\Omega}$, so current increases and terminal voltage drops:

$$I = \dfrac{12}{2.0 + 0.40} = \pu{5.0 A},\qquad V_{\text{terminal}}=12-(5.0)(0.40)=\pu{10.0 V}.$$

Practical & data tasks

• Measure I-V characteristics of a diode using datalogger; observe forward conduction and reverse blocking.
• Use oscilloscope to compare peak and r.m.s. values of mains waveform; annotate timebase.
• Investigate internal resistance by plotting terminal voltage vs current for a cell and finding gradient/intercept.

Common misconceptions & exam traps

• Confusing conventional current direction with electron flow.
• Forgetting to convert $\pu{mm2}$ to $\pu{m2}$ in drift velocity calculations.
• Using peak rather than r.m.s. values when working with power.
• Ignoring internal resistance when analysing supply voltage drops.

Quick self-check quiz

1. Define current in terms of charge flow. - $I = \dfrac{\mathrm{d}Q}{\mathrm{d}t}$.
2. What physical quantity does potential difference represent? - Work done per unit charge.
3. For $I = 0.5 \sin 100 t$ A, find

Revision workflow

1. Re-derive drift velocity and r.m.s. relations weekly to keep intuition fresh.
2. Practise I-V graph interpretation for resistors, bulbs, diodes, and identify regions of operation.
3. Work through past exam problems involving internal resistance and power distribution.
4. Summarise AC vs DC terminology (peak, r.m.s., average) on a cheat sheet and review before quizzes.

Practice Quiz

Test yourself on the key concepts from this guide.

10 Further reading

• SEAB H2 Physics 9478 syllabus (2026)
• Wikipedia - Electric current
• Britannica - Drift velocity
• Khan Academy - Electric potential & potential difference

11 Call-to-action

Parents: book a focused Currents clinic 1 week before WA 2 - it shores up both Electricity and upcoming EM induction. Students: condense Sections 6-8 onto an A5 “AC cheat sheet” and quiz yourself on every bus ride.

Last updated 14 Jul 2025. Next review when SEAB releases the 2027 draft syllabus.