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. 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)
- potential difference across the component (V)
- current through the component (A)
- resistance (Ω, ohms)
- resistance of the wire (Ω)
- 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)
- the individual resistances (Ω)
- electrical power — energy transferred per second (W, watts)
- current through the component (A)
- potential difference across the component (V)
- resistance of the component (Ω)
- electrical energy transferred (J; or kWh for bills)
- power (W; or kW for bills)
- time the component runs (s; or hours for bills)
- emf of the cell (V) — the energy it gives each coulomb
- current drawn from the cell (A)
- external (load) resistance (Ω)
- internal resistance of the cell (Ω)
- 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.
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.
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.
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.
🧠 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.