Electricity: revision notes and practice

What you will learn

• Distinguish current, charge, potential difference and resistance.
• Apply charge, resistance, power and energy equations with correct units.
• Explain series and parallel circuits and measure them safely using low-voltage equipment.

Key rules

Potential difference

V = IR. Potential difference V is in volts, current I in amperes (A), and resistance R in ohms (Ω).

Electrical power

P = VI, so I = P/V. Power is in watts (W). Convert kilowatts to watts by multiplying by 1000.

Charge flow

Q = It for a steady current. Charge is in coulombs (C). Convert minutes to seconds and milliamperes to amperes: 1000 mA = 1 A.

Worked examples

Example 1: A 20 Ω resistor carries 250 mA. Find the potential difference across it.

1. Convert current: 250 ÷ 1000 = 0.25 A.
2. V = IR = 0.25 × 20 = 5 V.

5 V

Example 2: A steady current of 0.4 A flows for 3 minutes. Find the charge transferred.

1. Convert time: 3 × 60 = 180 s.
2. Q = It = 0.4 × 180 = 72 C.

72 C

A practice question

A resistor of 11 Ω carries a current of 0.5 A. Calculate the potential difference across it.

1. Current is moving charge, not moving energy

A current is the rate at which charge passes a point. One ampere means one coulomb each second. A closed conducting path and a source of potential difference are needed for a steady current. Metals contain mobile electrons; conventional current is drawn from positive to negative, opposite to the electrons' overall movement.

Components transfer energy, but they do not use up the charge. In one unbranched loop, the current is the same before and after a lamp. Opening a switch breaks the complete path, so the current stops throughout that loop.

I = Q/t; I in A, Q in C, t in s

2. Use charge flow and convert small currents

For steady current, Q = It. At 0.30 A for 40 s, 12 C passes a point. Rearrange to I = Q/t or t = Q/I. If current changes, one fixed current value only works when it represents the average over the interval.

Convert milliamperes to amperes by dividing by 1000: 250 mA = 0.250 A. Convert minutes by multiplying by 60. A current of 200 mA for 3 minutes transfers 0.200 × 180 = 36 C, rather than 600 C or 36 000 C.

Q = It; 1000 mA = 1 A

3. Potential difference and resistance have different jobs

Potential difference describes the energy transferred per unit charge between two points. Across 6 V, each coulomb transfers 6 J. Resistance describes how strongly a component opposes current. For the same potential difference, a greater resistance gives a smaller current.

Use V = IR, or rearrange to I = V/R and R = V/I. A 12 Ω resistor across 6 V carries 0.5 A. V is also the symbol for the volt, so read the context: in the equation it is a quantity; after a number it is a unit. Resistance is measured in ohms, Ω.

V = IR; E = QV

4. Recognise when resistance changes

An ohmic conductor has constant resistance when its temperature stays constant. Its current–potential difference graph is a straight line through the origin. A filament lamp heats as current increases, so its resistance rises and its graph curves. V = IR still gives resistance at a particular operating point; constant R is an extra assumption.

A typical GCSE thermistor's resistance falls as temperature rises. An LDR's resistance falls as light intensity increases. A diode conducts readily in one direction and very little in reverse. Explain the condition changing the resistance rather than assuming every component behaves like a fixed resistor.

5. Series means one route for current

In series, the current is the same through every component, the supply potential difference is shared and the resistances add. Two resistors of 2 Ω and 4 Ω have total resistance 6 Ω. Across a 12 V supply, current is 2 A, giving potential differences of 4 V and 8 V.

Adding a resistor in series increases total resistance and reduces current if the supply voltage stays fixed. One broken component interrupts the whole loop. These rules apply to the ideal circuit model; a real supply may also have internal resistance, which this introductory calculation ignores.

Rtotal = R₁ + R₂; supply V = V₁ + V₂

6. Parallel means branches across the supply

Parallel branches connect across the same two points, so each branch has the same potential difference. Current splits at a junction and rejoins; the total entering equals the total leaving. A 6 Ω branch across 12 V takes 2 A, while a 12 Ω branch takes 1 A. The supply current is 3 A.

That combination has an effective resistance of 12/3 = 4 Ω, smaller than either branch resistance. Adding another conducting branch lowers the total resistance. Parallel lamps can operate independently: breaking one branch does not normally break the other complete routes.

7. Connect power, energy and time

Electrical power is P = VI. A device carrying 2 A at 9 V transfers 18 J each second, so its power is 18 W. Combining this with V = IR gives P = I²R. Energy transferred is E = Pt or E = QV; choose the form matching the information given.

An 18 W device operating for 120 s transfers 2160 J. A kilowatt-hour is an energy unit, not a power: 1 kWh = 3.6 million J. A 2 kW appliance used for half an hour uses 1 kWh. Keep hours and kilowatts together only when calculating kWh.

P = VI = I²R; E = Pt = QV

8. Measure a circuit without changing the question

Connect an ammeter in series and a voltmeter in parallel across the component. In a teacher-supervised low-voltage resistance investigation, measure V and I and calculate R. When varying wire length, keep the material, thickness and temperature as constant as possible so you investigate one factor.

Switch off between readings to reduce wire heating, avoid short circuits and repeat measurements. Use the appropriate meter range and check units. Never copy the experiment using household mains electricity. Mains wiring, protective devices, the National Grid and static charge are further course content beyond this circuit-focused guide.

Check your understanding

How to earn the marks

• State whether a quantity is through a component or across it: current through, potential difference across.
• Check series and parallel connections from the junctions, not from how the drawing looks.
• Only assume constant resistance when the component and conditions justify it.
• For practical questions, name the controlled variable and explain how to control it.

A common mistake

Putting 250 mA into a formula as 250 A makes the result 1000 times too large.

Course context and official specifications

This guide covers selected skills. Some extensions are Higher content; choose your official tier separately from practice difficulty in the app. These are original questions and explanations.

• AQA GCSE Physics — 4.2.1.2–4.2.1.4; 4.2.4

  The core calculations in this bank are selected separate Physics skills. Foundation and Higher content differ within these larger sections; this is not complete topic coverage.

• Edexcel GCSE Physics — 10.9, 10.13; 10.17; 10.29–10.31

  The core calculations in this bank are selected separate Physics skills. Foundation and Higher content differ within these larger sections; this is not complete topic coverage.

• OCR Gateway GCSE Physics A — P3.2; PM3.2iii–PM3.2iv

  The core calculations in this bank are selected separate Physics skills. Foundation and Higher content differ within these larger sections; this is not complete topic coverage.

• OCR Twenty First Century GCSE Physics B — P3.2–P3.4

  The core calculations in this bank are selected separate Physics skills. Foundation and Higher content differ within these larger sections; this is not complete topic coverage.

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