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
Reviewed by
Chee Wei Jie·Academic Advisor (Physics)
, and energy over time only after the current path is clear.
Current-electricity decision map
Question clue
First check
Main relation
Trap to avoid
Charge passes a point in a stated time
Convert charge to coulombs and time to seconds
I=Q/t
Leaving mC, μC, or nC unconverted.
Wire area, carrier density, and carrier charge are given
Area must be in m2
I=nAvq
Treating drift speed as the random electron speed.
Energy per charge or cell behavior is described
Decide whether the source supplies energy or a component uses it
V=W/Q and e.m.f. versus p.d. language
Saying p.d. is stored in a resistor.
Resistor heating or energy use is asked
Identify which of P, V, I, and R are known
P=VI
Sinusoidal AC is involved
Check whether the value is peak or r.m.s.
Irms=I0/2
A single diode is in series with AC
Sketch which half-cycle is blocked
Half-wave rectification
Calling the output steady d.c.; it is pulsating before smoothing.
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=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.
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.
Quantity in the question
First setup move
Common trap
Wire radius or diameter
Convert to metres, then calculate A=πr2.
Using diameter as radius or leaving mm2 as if it were m2.
Carrier density n
Keep it as carriers per m3.
Treating n as number of moles or number of electrons in the whole wire.
Charge per carrier q
Use the magnitude 1.60×10−19C for electrons unless direction is being discussed.
Making the current negative just because electrons are negative.
Required drift speed
Rearrange to v=nAqI.
Expecting a large speed because the lamp turns on quickly.
Worked check: for I=2.0A, n=8.5×1028m−3, radius r=0.50mm, and q=1.60×10−19C, first convert r=5.0×10−4m, so A=π(5.0×10−4)2=7.85×10−7m2. Then v=(8.5×1028)(7.85×10−7)(1.60×10−19)2.0=1.9×10−4m⋅s−1.
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=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.
Power formula checkpoint
Choose the power formula from the component values you actually know. The voltage and resistance in P=RV2 must belong to the same component.
Question gives
Use first
Check before substituting
Current through a resistor and its resistance
P=I2R
The same current passes through that resistor.
P.d. across a resistor and its resistance
P=RV2
The voltage is across that resistor, not across the whole supply plus internal resistance.
Current and p.d. for the same component
P=VI
V is the p.d. across the component using that current.
Cell with internal resistance
Split the power into load loss and internal loss
Use I2R for the external load and I2r for the internal resistance.
Worked check: a 12V cell with r=1.0ohm supplies a 5.0ohm resistor. The current is I=12/(5.0+1.0)=2.0A. The load power is I2R=(2.0)2(5.0)=20W, while the internal heating is I2r=(2.0)2(1.0)=4.0W. Do not use 122/5.0 for the load, because 12V is the e.m.f. of the cell, not the p.d. across the 5.0ohm resistor.
Misconception check: P=VI, P=I2R, and P=V2/R are not three unrelated formulas. They are equivalent only when V, I, and R describe the same ohmic component.
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.
6.2.1 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.
What the question asks for
Use this value first
Why
Common wrong move
Maximum current or voltage on the graph
Peak value, I0 or V0
The graph reaches this at the crest.
Dividing by 2 too early.
Heating effect or power in a resistor
R.m.s. value, Irms or Vrms
R.m.s. gives the equivalent d.c. heating effect.
Putting peak current into P=I2R
Instantaneous value at a stated time
Substitute into x=x0sinωt
The sign and magnitude depend on the phase.
Reporting the peak value just because amplitude is given.
Average over a full sinusoidal cycle
Zero for current or voltage; non-zero for power
Positive and negative halves cancel for I or V, but power depends on a square.
Saying mean current and mean power both vanish.
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 I2R, 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=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.
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.
Feature to mark
What happens
Exam trap
Conducting half-cycle
The load current follows the allowed half of the input waveform.
Drawing a constant output just because the current has one direction.
Blocked half-cycle
The load current is zero because the diode is reverse-biased.
Reflecting the negative half upward as if it were a full-wave rectifier.
Average current
For a positive half-wave sinusoid, use Iavg=I0/π.
Using the r.m.s. value when the question asks for average current.
Smoothing capacitor
The output ripple is reduced, but it is not perfectly steady in the simple model.
Calling the unsmoothed half-wave output smooth d.c.
Worked check: if the input current would be I=I0sinωt, a positive half-wave rectifier gives the positive sine hump from 0 to π, then zero from π to 2π. The average over the full cycle is I0/π, not I0/2, because the sine hump is curved rather than rectangular.
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)
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.
