Key Idea: This topic builds a complete picture of an electric circuit: charge flowing as a current, voltage as the energy each coulomb carries, resistance opposing the flow, and power as the energy delivered each second.
Measure it
- I = Δq/Δt, through: ammeter in series
- V = W/q, across: voltmeter in parallel
- R = V/I; ρ = RA/L
Combine it
- Series: same I, R adds
- Parallel: same V, 1/R adds
- Potential divider: pd shared in the ratio of R
Power it
- P = IV = I²R = V²/R
- ε = I(R + r); V = ε − Ir
- V–I graph: intercept ε, gradient −r
It finishes with real cells, which lose a little voltage inside themselves. It is examined on both papers — quick Paper 1A multiple-choice (rank resistor combinations, read a resistance off an I–V graph, why terminal p.d. drops) and longer Paper 2 structured questions ('show that' a current, an equivalent resistance, a power ratio, or an internal resistance).
Key formulas
Almost all of these are given in the data booklet (look for the booklet badge). The only one you must build yourself is the terminal-p.d. form, which drops straight out of ε = I(R + r).
- electric current (A, amperes)
- charge that flows past a point (C, coulombs)
- time for that charge to flow (s)
- potential difference / voltage (V, volts)
- energy given to the charge (J, joules)
- amount of charge moved (C, coulombs)
- resistance (Ω, ohms)
- resistivity of the material (Ω m) — a property of the substance
- length of the wire (m)
- cross-sectional area of the wire (m²)
- total (equivalent) resistance of resistors in series (Ω)
- the individual resistances (Ω)
- total (equivalent) resistance of resistors in parallel (Ω)
- electrical power — energy transferred per second (W, watts)
- electrical energy transferred (J; or kWh for bills)
- time the component runs (s; or hours for bills)
- emf of the cell (V) — the energy it gives each coulomb
- internal resistance of the cell (Ω)
Not in the booklet: write it out yourself.
- terminal p.d. — the voltage actually delivered to the circuit (V)
- emf of the cell (V)
- 'lost volts' used up inside the cell (V)
The five ideas side by side
| Idea | Key relationship | What to remember |
|---|---|---|
| Current, charge & voltage | I = Δq/Δt and V = W/q | Current is charge per second (through a component); voltage is energy per coulomb (across it). Don't divide voltage by time. |
| Resistance & resistivity | R = V/I and R = ρL/A | Read R off any point of a straight I–V line. Longer wire ⇒ more R; thicker (bigger A) ⇒ less R. |
| Series vs parallel | Rₛ = ΣR; 1/Rₚ = Σ(1/R) | Series: same current, voltages add, R goes up. Parallel: same voltage, currents add, R goes down (below the smallest). |
| Power & energy | P = IV = I²R = V²/R; E = Pt | Power is energy per second. Choose the P-form by what you know. Bills use kWh: power in kW × time in hours. |
| emf & internal resistance | ε = I(R + r); V = ε − Ir | A real cell loses Ir volts inside itself, so terminal p.d. V < ε. Only when r ≈ 0 does V ≈ ε. |
Series vs parallel — the rules in full
| Quantity | Series (one loop) | Parallel (branches) |
|---|---|---|
| Current | Same through every component | Splits — branch currents add to the total |
| Voltage (p.d.) | Adds up — the p.d.s share out the supply | Same across every branch |
| Total resistance | Rₛ = R₁ + R₂ + … (goes up) | 1/Rₚ = 1/R₁ + 1/R₂ + … (goes down, below the smallest) |
| Add a component | Total resistance increases | Total resistance decreases |
After adding the reciprocals you have 1/Rₚ, not Rₚ. You must flip (take the reciprocal) for the final answer. Sanity check: a parallel combination is always smaller than the smallest resistor in it. Two equal resistors in parallel give exactly half of one.
Worked exam-style questions
A copper wire has length 15 m and cross-sectional area 2.0 × 10⁻⁶ m². Copper has resistivity 1.7 × 10⁻⁸ Ω m. (a) Find the wire's resistance. (b) State the resistance of an identical wire whose cross-sectional area is doubled.
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
Next:
A 12 V supply (negligible internal resistance) drives a 30 Ω resistor in series with two 60 Ω resistors that are in parallel with each other. (a) Find the total resistance. (b) Find the current drawn from the supply. (c) Find the current in each 60 Ω branch.
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
Next:
A heater is a wire of resistance R connected across the mains, dissipating 1200 W at 230 V. A second heater uses the same wire material and the same cross-sectional area, but with three times the length, connected across the same 230 V. Determine the power of the second heater.
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
Next:
A cell of emf 1.5 V drives a current of 0.30 A through a 4.5 Ω resistor. (a) Find the internal resistance of the cell. (b) Find the terminal potential difference across the cell.
Model answer plan
See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.
Important: At a fixed pd more resistance means LESS power (P = V²/R); at a fixed current it means MORE power (P = I²R). Decide what is fixed before you choose the form.
Quick self-check
Tap each card to reveal the answer.
Through or across — how is current measured, and how is voltage? Current is measured through a component (ammeter in series, in the line). Voltage is measured across it (voltmeter in parallel).
Double a wire's length — what happens to its resistance? It doubles. R = ρL/A, so R is proportional to length. Doubling the cross-sectional area instead would halve R.
Add another resistor in parallel — does the total resistance rise or fall? It falls. Extra parallel paths make it easier for current to flow, so 1/Rₚ grows and Rₚ drops below the smallest resistor.
You know the voltage and the resistance — which power formula is quickest? P = V²/R. Pick the P-form that uses the two quantities you already have (P = IV, P = I²R, or P = V²/R).
Why is a cell's terminal p.d. less than its emf when current flows? Some voltage — the lost volts Ir — is used up driving the current through the cell's own internal resistance r. So V = ε − Ir < ε.
In a single series loop, where is the current largest? Nowhere — it is the same everywhere. Charge is conserved, so the same current passes through every component in the loop.
Exam tips
Exam tips
- Keep the two definitions straight: current = charge ÷ time (I = Δq/Δt, measured THROUGH, ammeter in series); voltage = energy ÷ charge (V = W/q, measured ACROSS, voltmeter in parallel).
- Read resistance off an I–V graph as V ÷ I at a point. A straight line through the origin is ohmic (constant R); a curve (e.g. a filament lamp) is non-ohmic.
- In R = ρL/A: longer ⇒ more resistance (R ∝ L), thicker ⇒ less (R ∝ 1/A). Resistivity ρ is a property of the material, not the shape.
- Series adds resistances and shares the voltage at one current; parallel adds reciprocals and shares the current at one voltage. After 1/Rₚ, always FLIP — a parallel total is smaller than the smallest resistor.
- Choose the power form by the quantities given: P = IV, P = I²R, or P = V²/R. For ratio questions, write the formula for each case and divide so the common factors cancel.
- Electricity bills use kilowatt-hours: power in kW × time in HOURS = energy in kWh; cost = kWh × unit price. Convert minutes/seconds before using hours.
- A real cell obeys ε = I(R + r). Don't forget r: the terminal p.d. is V = ε − Ir, always a little below the emf. Only when r ≈ 0 does V ≈ ε.