Q: What does IP Physics Notes (Upper Secondary, Year 3-4): 11) Current of Electricity cover? A: Relate charge flow, potential difference, resistance, and I-V characteristics to core IP circuit analysis.
Quick recap - Electric current measures how quickly charge moves. Define the energy supplied (emf) and used (p.d.), then apply Ohm's law, resistivity relations, and characteristic curves to decode circuit behaviour.
The core idea is simple: Current is charge flow per second.
Use it as a working check: Link charge, current, time, voltage, resistance, and energy carefully. Use Ohm's law only when the component behaves ohmically at constant temperature.
Then go one layer deeper: Use the resistivity and I-V sections to practise identifying variables, reading graph shape, and explaining why metals, lamps, and diodes behave differently.
Keep your practice loop tight via our IP Physics tuition hub-it links each topic here to quizzes, diagnostics, and WA-style problem sets.
For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.
charge flow, conventional current, electron flow, e.m.f., potential difference, resistance, wire dimensions, temperature effects, I-V characteristics, and resistance measurement follow Marcus Pang's current Chapter 11 route.
Eclat extension depth: the resistivity equation and experiment provide a more quantitative route than the proportional relationships alone.
2027 national comparison: K323 Topic 14 overlaps with the Eclat core and also names total e.m.f. for sources in series. That short case is retained below even though it is not a separate Marcus objective.
Check your school: follow the graph-axis convention and whether resistivity is assessed quantitatively in the current task.
Use this map before choosing a formula. It separates the flow question from the energy question.
cell or battery
supplies energy to each coulomb
emf = energy supplied per coulomb
|
v
charge moves around the circuit
current = charge per second
|
v
component uses the energy
p.d. = energy transferred per coulomb
|
v
resistance controls how much current flows for that p.d.
If the question gives time, start with charge flow: Q=It. If it gives energy per charge, start with voltage: V=W/Q. If it asks how easily current passes, use resistance: R=V/I.
Formula-choice checkpoint
Underline the phrase that tells you what is being counted before choosing a formula.
Question phrase
Quantity to find first
Formula lane
Unit check
"charge passing a point"
Charge moved
Q=It
Time must be in seconds.
"energy transferred per coulomb"
Voltage or p.d.
V=W/Q
Joules per coulomb gives volts.
"opposition to current"
Resistance
R=V/I
Volts per ampere gives ohms.
"energy transferred by a component"
Work done or energy
W=VQ
Charge and voltage together give joules.
Misconception check: do not use R=V/I just because a circuit question contains volts and amperes. If the question asks for charge over time, start from Q=It first.
Charge & Current
Charge Q measured in coulombs; 1 electron carries −1.6×10−19C.
Current is rate of charge flow: I=ΔtΔQ
Conventional current: positive to negative. Electron flow: negative to positive.
Total charge moved: Q=It.
Worked Example: Charge Flow
A lamp has a steady current of 0.40A for 3.0min. The time must be converted first:
t=3.0×60=180s
Q=It=0.40×180=72C
Common trap: amperes already mean coulombs per second, so do not multiply by minutes unless you have converted minutes to seconds.
Electromotive Force vs Potential Difference
EmfE: work done by source per coulomb round the entire circuit, E=QWsource.
Potential differenceV: work done per coulomb across a component, V=QWcomponent.
Both measured in volts; emf describes supply, p.d. describes energy drop in loads.
For cells connected in series in the same direction, total e.m.f. is the sum of their e.m.f. values. If one cell is reversed, its e.m.f. opposes the others and is subtracted.
Read the wording carefully:
If the phrase says...
It is asking about...
Usual symbol
"energy supplied by the source per unit charge"
emf
E
"energy converted in the component per unit charge"
p.d. across that component
V
"work done moving charge through the whole circuit"
source-side energy supply
EQ
"work done in a resistor, lamp, or motor"
load-side energy transfer
VQ
Resistance & Ohm's Law
Resistance opposes current flow: R=IV
Ohmic conductor: V∝I at constant temperature.
Resistivity relation for uniform wires: R=ρAL
ρ material property (Ω⋅m), L length, A cross-sectional area.
Temperature effects: metallic resistance increases with temperature; semiconductors often show opposite trend.
Worked Example: Calculating Resistivity
A copper wire (ρ=1.7×10−8Ω⋅m) is 0.80m long with diameter 0.90mm. Area A=π(0.90×10−3/2)2 so R=ρL/A≈2.7×10−2Ω.
Common trap: the diameter is not the radius. Halve the diameter before finding area, and convert millimetres to metres before substituting.
Resistivity variable-change checkpoint
When two wires are compared, keep the variables separate before substituting numbers. The relation R=ρL/A says resistance increases with ρ and L, but decreases when cross-sectional area increases.
Change made
What happens to resistance
Why
Same material, length doubled
Resistance doubles.
R∝L when ρ and A are unchanged.
Same material, cross-sectional area doubled
Resistance halves.
R∝1/A when ρ and L are unchanged.
Same material, diameter doubled
Resistance becomes one quarter as large.
Doubling diameter doubles radius, so area becomes four times as large.
Higher-resistivity material, same dimensions
Resistance increases in the same ratio as resistivity.
ρ is the material factor in R=ρL/A.
Worked check: wire B is made of the same material and has the same length as wire A, but B has twice the diameter. The radius is also twice as large, so AB=4AA. Since R∝1/A, RB=RA/4.
Misconception check: doubling diameter does not halve the resistance. Area depends on radius squared, so a diameter change is also an area-squared change.
I-V Characteristics to Memorise
Component
Graph
Key behaviour
Metal resistor
Straight line through origin
Constant R; obeys Ohm's law
Filament lamp
Curve flattening at higher V
Heating raises R; gradient decreases
Semiconductor diode
Nearly zero current reverse bias; sharp rise beyond threshold (~0.6V)
Conducts mainly in one direction
Gradient on V-against-I graph equals resistance; on I-vs-V, resistance is reciprocal of gradient.
I-V graph shape checkpoint
Before calculating a gradient, decide what the shape is telling you about the component.
Graph cue
What it means physically
How to use resistance
Common trap
Straight line through the origin
Current is directly proportional to p.d. at constant temperature.
One constant value of R=V/I works for the whole line.
Calling any straight line ohmic if it does not pass through the origin.
Filament lamp curve flattens as p.d. increases
The filament heats up, so its resistance increases.
Use R=V/I at the stated point, or compare tangent steepness if the question asks about gradient.
Quoting one fixed resistance for the whole curve.
Diode current rises sharply in one direction only
The diode conducts mainly under forward bias after the threshold region.
Treat it as non-ohmic; use the graph reading at the stated voltage or current.
Applying Ohm's law as if the diode were a metal resistor.
Axes are swapped
The gradient may be ΔV/ΔI or ΔI/ΔV.
For a V-against-I graph, gradient is resistance. For an I
Worked check: a filament lamp has V=6.0V and I=0.50A at one marked point, so its resistance there is R=6.0/0.50=12Ω. That does not mean the lamp is 12Ω at every point on the curve, because the filament temperature changes as the current changes.
Misconception check: a curved I-V graph is not automatically a faulty result. It can be the expected signature of a lamp heating up or a diode switching from very little current to strong forward conduction.
Worked Example: Reading a Graph Gradient
Suppose a V-against-I graph for a resistor passes through (0,0) and (0.50,3.0), where current is in amperes and voltage is in volts.
R=ΔIΔV=0.50−03.0−0=6.0Ω
If the axes are swapped and the graph is I-against-V, the gradient would be ΔI/ΔV, so resistance is the reciprocal. Always check the axis labels before using the gradient.
Practical Measurement Tips
Ammeter in series (assume negligible resistance); voltmeter in parallel (assume infinite resistance).
For accurate resistivity experiments, keep wire at constant temperature, measure length precisely, and average diameter readings with a micrometer.
Practice Quiz
Reinforce charge-time calculations, Ohm's law, and I-V curve interpretation with this currents checkpoint.
Key Takeaways
Use Q=It to connect time, current, and charge.
Differentiate supply emf from component p.d.; both share volt units but describe distinct energy transfers.
Resistivity links microscopic material properties to macroscopic resistance; temperature shifts can explain non-linear graphs.
Recognise signature I-V curves under exam pressure: straight line (ohmic), curved (filament), threshold (diode).
-against-
V
graph, resistance is the reciprocal of gradient.
Reading the graph shape correctly but using the wrong gradient ratio.
