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1 Electric current \((I)\)
2 Microscopic view - drift velocity
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.
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Current is charge flow per unit time
10 seconds
Charge flow, drift velocity, resistance, and power
100 seconds
The first current mini-drill and graph cue
10 minutes
AC r.m.s. values, diodes, and rectifiers
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.
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=tQ.
1.1 Mini-drill
A charge of 12mC passes a point in 4.0s.
I=3.0mA.
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 velocityv when an external field is applied. Equating the charge that crosses a cross-section per second gives
I=nAvq
where
Symbol
Meaning
n
number density of charge carriers
A
conductor cross-section area
q
charge on one carrier (1.60×10−19C for an electron)
Tip for WA practice: treat v as μE when mobility (μ) and field (E) are given.
3 Potential difference (V)
Potential difference is the electrical work done per unit charge:
V=QW.
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,P=I2R,P=RV2.
Timing hack: write P=VI at the top of data-handling questions; deriving the other two takes <15 s.
5 e.m.f. vs p.d.
e.m.f.
p.d.
Energy picture
energy supplied per coulomb by a source
energy converted to other forms per coulomb in a component
Circuit location
inside cells, generators
across resistors, lamps, etc.
Sign convention
raises potential
drops potential
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,
Irms=2I0,Vrms=2V0.
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=x0sinωt,
where ω=2πf.
7 Mean power in a resistive load
For R purely resistive and current I=I0sinωt:
Pmean=21I02R=Irms2R,
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.
9 Three WA timing rules (Currents edition)
Use syllabus pacing as a guide: Paper 2/3 average ~1.6 min/mark; Paper 4 ~3 min/mark.
Sketch peak and r.m.s. values before calculating - prevents factor-of-2 slips.
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=tQ.
(b) derive and use the equation I=nAvq for a current-carrying conductor, where n is the number density of charge carriers and v is the drift velocity.
(c) recall and solve problems using the equation for potential difference in terms of electrical work done per unit charge, V=QW.
(d) recall and solve problems using the equations for electrical power P=VI, P=I2R and P=RV2
(e) distinguish between electromotive force (e.m.f.) and potential difference (p.d.) using energy considerations.
(f) show an understanding of and use the terms period, frequency, peak value and root-mean-square (r.m.s.) value as applied to an alternating current or voltage.
(g) represent a sinusoidal alternating current or voltage by an equation of the form x=x0sinωt.
(h) deduce that the mean power in a resistive load is half the maximum (peak) power for a sinusoidal alternating current.
(i) distinguish between r.m.s. and peak values, and recall and use Irms=2I0
(j) explain the use of a single diode for the half-wave rectification of an alternating current.
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
Quantity / concept
Expression / highlight
Current definition
I=dtdQ
Drift velocity
I=nAvq
Charge carrier density
n = number of carriers per m−3
Resistive power
P=VI=I2R=RV2
r.m.s. values (sinusoid)
Irms=2I0,Vrms=2V0
Mean power (AC resistor)
Pmean=Irms2R
e.m.f. vs terminal voltage
E=Vterminal+Ir (internal resistance r)
Half-wave rectified average current
Iavg=πI0 (sinusoidal input)
Derivations & reasoning to master
Drift velocity formula: equate charge passing per second through cross-section to nAvq.
Power relationships: combine Ohm's law V=IR with P=VI.
r.m.s. derivation: integrate I02sin2ωt over one period to show the 21 factor.
Rectifier output: sketch the current waveform with diode removal of the negative half-cycle and compute average/mean power.
Worked example 1 - drift velocity
A copper wire (area 1.5mm2) carries 4.0A. Given free-electron density 8.5×1028m−3, find the drift velocity. Comment on why it is much smaller than electron random speeds.
Outline: convert area to m2, apply v=nAqI. Typical answer ≈0.20mm⋅s−1.
A=1.5mm2=1.5×10−6m2.
nAq=(8.5×1028)(1.5×10−6)(1.60×10−19)=2.04×104.
v=2.04×1044.0=1.96×10−4m⋅s−1.
So v≈0.196mm⋅s−1, consistent with the quoted ≈0.20mm⋅s−1.
Worked example 2 - internal resistance & power
A 12V cell with internal resistance 0.40ohm supplies a 4.0ohm 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=R+rE, compute terminal voltage E−Ir, compare powers I2R and I2r.
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.
