Edexcel A-level Physics (9PH0) · Topic 3: Electric Circuits
Mini-Lesson
Electric Circuits
Topic 3 takes GCSE circuits and gives them a microscopic engine: I = nAvq, resistivity, internal resistance and the potential divider.
I = ΔQ/Δt · V = W/Q · R = V/I · ρ = RA/Lcurrent is charge flow; pd is energy transferred per coulomb
Definitions worth memorising word for word: the volt is one joule per coulomb. Potential difference is the energy transferred from the charge to the component; EMF is the energy transferred to the charge by the source. Same unit, opposite direction of energy flow.
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
Topic 3 · charge carriers
Current, charge and drift velocity
I = ΔQ/Δt · I = nAvqn = number density of carriers (m⁻³) · A = cross-sectional area · v = drift velocity · q = charge per carrier
Current is the rate of flow of charge. In a metal the carriers are free electrons, each carrying −1.60 × 10⁻¹⁹ C.
The drift velocity is astonishingly slow — typically 10⁻⁴ m s⁻¹ in a copper wire. The lamp lights instantly because the electric field propagates at close to the speed of light, nudging every electron at once.
Metals have huge n (≈10²⁹ m⁻³) → good conductors. Semiconductors have much smaller n. Insulators have almost none.
Squeeze the same current through a narrower wire (smaller A) and v must increase.
Conventional current flows from + to −, which is the direction positive charge would move. The electrons actually drift the other way. Every exam answer uses conventional current.
Calculate
Your turn — charge flow
1A current of 0.25 A flows for 2.0 minutes. Calculate the charge that passes. Give your answer in C.
C
Hint: Convert first: 2.0 minutes = 120 s. Then Q = It = 0.25 × 120.
Topic 3 · resistivity
Resistance and resistivity
R = ρL / Aρ = resistivity in Ω m — a property of the MATERIAL, not of the sample
Resistance depends on the object: double the length and R doubles; double the cross-sectional area and R halves. Resistivity strips that away and describes the material alone.
Copper: ρ ≈ 1.7 × 10⁻⁸ Ω m. Glass: ρ ≈ 10¹² Ω m — twenty orders of magnitude apart.
In a metal, heating increases lattice vibration → more collisions → resistivity rises.
In a semiconductor (e.g. an NTC thermistor), heating releases many more charge carriers → n rises sharply → resistivity falls.
Watch the area
For a wire of diameter d, A = πd²/4 = πr². A diameter of 0.50 mm gives r = 0.25 mm = 2.5 × 10⁻⁴ m.
A = π × (2.5 × 10⁻⁴)² = 1.96 × 10⁻⁷ m² — a classic place to lose a mark by using d instead of r.
Calculate
Your turn — resistivity
2A copper wire is 2.0 m long with a diameter of 0.50 mm. Copper has resistivity 1.7 × 10⁻⁸ Ω m. Calculate its resistance. Give your answer in milliohms (mΩ) to 3 significant figures.
mΩ
Hint: r = 0.25 mm = 2.5 × 10⁻⁴ m, so A = πr² = 1.963 × 10⁻⁷ m². R = ρL/A = (1.7 × 10⁻⁸ × 2.0) ÷ 1.963 × 10⁻⁷ = 0.173 Ω. Now convert to mΩ.
Topic 3 · I–V characteristics
I–V characteristics
Plot current against potential difference and the shape tells you what the component is doing.
Straight line through the origin = constant resistance. Curving towards the V-axis = resistance increasing.
Ohmic conductor (metal at constant temperature): straight line through the origin — R is constant. That is Ohm's law, and it only holds at constant temperature.
Filament lamp: as current rises the filament heats up, lattice vibrations increase, resistance rises, so the graph bends towards the V-axis.
Semiconductor diode: conducts only in forward bias, and only above a threshold pd of about 0.6 V. Enormous resistance in reverse.
NTC thermistor: resistance falls as temperature rises.
Sort it
Sort the components by their I–V behaviour
Tap a component, then tap the box describing its I–V characteristic.
📏 Ohmic (straight line)
🔥 Temperature-dependent
➡️ Conducts one way only
Quick check
Why does the lamp curve?
?The I–V graph for a filament lamp curves towards the V-axis at higher currents. Why?
Topic 3 · combinations
Series and parallel
series: R = R₁ + R₂ + … · parallel: 1/R = 1/R₁ + 1/R₂ + …current is the same in series; potential difference is the same in parallel
Series: the same current passes through every component; the pds add up to the supply pd (conservation of energy).
Parallel: each branch has the full supply pd; the branch currents add up to the total (conservation of charge).
Adding a resistor in parallel always lowers the total resistance below the smallest branch — you have opened another route for the charge.
Worked example
6.0 Ω and 3.0 Ω in parallel: 1/R = 1/6 + 1/3 = 1/2, so R = 2.0 Ω.
That pair is then in series with 4.0 Ω: total = 2.0 + 4.0 = 6.0 Ω.
Calculate
Your turn — combining resistors
3A 6.0 Ω and a 3.0 Ω resistor are connected in parallel. This combination is then connected in series with a 4.0 Ω resistor. Calculate the total resistance. Give your answer in Ω.
Ω
Hint: Parallel pair: 1/R = 1/6 + 1/3 = 1/2, so R = 2.0 Ω. Then add the series resistor: 2.0 + 4.0.
Topic 3 · internal resistance
EMF and internal resistance
ε = I(R + r) → V = ε − Irε = EMF · r = internal resistance · V = terminal potential difference
Every real source has resistance inside it. Some of the energy given to each coulomb is therefore dissipated before the charge even leaves the battery — the pd across the terminals is always less than the EMF whenever current flows.
The term Ir is called the lost volts.
Plot terminal pd V (y) against current I (x): the gradient = −r and the y-intercept = ε.
On open circuit (I = 0) the terminal pd equals the EMF. That is how a high-resistance voltmeter measures ε directly.
Worked example
ε = 12.0 V, r = 0.50 Ω, external R = 5.5 Ω.
I = ε ÷ (R + r) = 12.0 ÷ 6.0 = 2.0 A
Terminal pd V = ε − Ir = 12.0 − (2.0 × 0.50) = 11.0 V
Calculate
Your turn — terminal pd
4A cell of EMF 12.0 V and internal resistance 0.50 Ω is connected to a 5.5 Ω resistor. Calculate the terminal potential difference. Give your answer in V.
V
Hint: I = ε ÷ (R + r) = 12.0 ÷ 6.0 = 2.0 A. Then V = ε − Ir = 12.0 − 2.0 × 0.50.
Quick check
Reading the graph
?Terminal pd V is plotted against current I for a cell. What do the gradient and the y-intercept represent?
Topic 3 · potential dividers
The potential divider
Vout = Vin × R₂ / (R₁ + R₂)the supply pd is shared between the resistors in the ratio of their resistances
Two resistors in series across a supply split the pd in proportion to their resistance. Swap one for a sensor and you have an automatic control circuit:
Use an LDR: in the dark its resistance is huge, so it takes most of the pd → a light-activated switch.
Use a thermistor: as it warms, its resistance falls, so its share of the pd drops → a temperature-controlled circuit.
A potentiometer (sliding contact) gives a continuously variable Vout from 0 up to Vin.
Warning: connecting a load across the output changes the resistance of that arm (they are now in parallel), so Vout falls. That is why examiners ask for a high-resistance voltmeter.
Calculate
Your turn — potential divider
5A 12 V supply is connected across a 4.0 kΩ resistor in series with an 8.0 kΩ resistor. Calculate the potential difference across the 8.0 kΩ resistor. Give your answer in V.
V
Hint: V_out = 12 × 8.0 ÷ (4.0 + 8.0) = 12 × 8/12.
Topic 3 · power
Electrical power and energy
P = VI = I²R = V²/R · W = VItchoose the version that uses the two quantities you actually know
All three power expressions come straight from P = VI combined with V = IR.
P = I²R shows why power lines transmit at very high voltage: the same power delivered at higher V means lower I, and heat loss goes as I², so it collapses.
Energy in kWh: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.
Maximum power transfer to the load happens when the external resistance equals the internal resistance, R = r.
Match it
Match the quantity to its defining equation
Tap an item on the left, then its partner on the right.
Quantity
Equation
Quick check
Does resistivity change?
?A copper wire is replaced by another copper wire of twice the length at the same temperature. What happens to its resistance and its resistivity?
Quick check
Drift velocity
?The same current flows through a thick wire and then through a thin wire joined to it. What happens to the drift velocity of the electrons in the thin wire?
Recap
The big ideas to know
Current: I = ΔQ/Δt and I = nAvq — drift velocity is tiny; the field sets electrons moving almost instantly
Resistivity: R = ρL/A — ρ is a material property; a metal's ρ rises with temperature, a semiconductor's falls
I–V: ohmic = straight line (constant T) · lamp curves (R rises) · diode conducts one way above ≈0.6 V
Combinations: series R = R₁ + R₂ (same current) · parallel 1/R = 1/R₁ + 1/R₂ (same pd)
Internal resistance: ε = I(R + r), V = ε − Ir; graph of V against I gives gradient −r and intercept ε
Potential divider: V_out = V_in R₂/(R₁ + R₂); swap in an LDR or thermistor for a sensing circuit
Power: P = VI = I²R = V²/R; 1 kWh = 3.6 × 10⁶ J
That is the whole of Edexcel Topic 3 — and the platform for capacitors in Topic 7. Press Finish to see your score.
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