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Edexcel International GCSE Physics · Section 2 Electricity · 2 (c) Energy and voltage in circuits
Electricity Formulas: Charge, Voltage, Power and Energy
In this lesson you'll learn to: use the five electricity formulae, P = I × V, E = I × V × t,
V = I × R, Q = I × t and E = Q × V, explain that current is the rate of flow of charge and
voltage is the energy transferred per unit charge, pick the right current and voltage in series and
parallel circuits, and convert kΩ, mA and minutes before you substitute.
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Prefer to watch? Scroll to the bottom for the video on this topic
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A video walkthrough is coming soon — it will appear at the bottom of this page
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Start here — have a play
Don't read anything yet. Pick a circuit: one component, two in series or two in parallel. Read the ammeters and voltmeters (hide them if they get in the way), then use Calculate it to find the power, energy, voltage or charge for the cell or for one component. The gold buttons load the questions from pages 12, 16 and 17 of the workbook: try each one yourself first, then check.
▲ The moving dots are charge. In a parallel circuit, watch the dots split at the junction.
Five formulae, one circuit
Every question on this topic uses one of five formulae. The trick is choosing the right current and the right voltage, and converting the units first. The notes show you how.
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Revision notes
The five formulae
| Find | Formula | In words | Units |
|---|---|---|---|
| Power | P = I × V | power = current × voltage | W = A × V |
| Energy transferred | E = I × V × t | energy = current × voltage × time | J = A × V × s |
| Voltage | V = I × R | voltage = current × resistance | V = A × Ω |
| Charge | Q = I × t | charge = current × time | C = A × s |
| Energy (by charge) | E = Q × V | energy = charge × voltage | J = C × V |
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Current = flow of charge
Current is the rate of flow of charge. 1 A means 1 coulomb passes a point every second.
That is why Q = I × t: more current, or longer, means more charge passes.🔋
Voltage = energy per coulomb
Voltage is the energy transferred per unit charge. 1 V = 1 J/C.
A 12 V cell gives 12 J to every coulomb that passes through it. That is why E = Q × V.Which current? Which voltage?
Most lost marks come from using the cell's numbers for a single component. Before you substitute, ask where you are in the circuit:
| Series | Parallel | |
|---|---|---|
| Current | Same everywhere: the cell's current goes through every component | Splits: the cell's current = the branch currents added up |
| Voltage | Shared: each component gets part; they add up to the cell's voltage | Same: every branch gets the full voltage of the cell |
| Resistance | Adds up: R = R₁ + R₂ | Less than the smallest branch |
⚠ Convert before you substitute
k (kilo) = × 1000: 4.0 kΩ = 4000 Ω · m (milli) = ÷ 1000: 30 mA = 0.030 A · minutes × 60: 3.5 min = 210 s.
Put 4.0 instead of 4000, or 30 instead of 0.030, and your answer is 1000 times too big or too small.
How to write it in the exam
Every calculation, in the same order
Which I and V? (series or parallel) → equation
→ convert (kΩ, mA, min) → substitute
→ answer + unit, to the significant figures asked.
think first show the working the answer mark
These are the questions from pages 12, 16 and 17 of the workbook. Load each one in the animation and try it first.
Page 12 · the five formulae, one circuit · a 12 V supply drives 2.0 A through a lamp for 30 s
- Power: P = I × V = 2.0 × 12 = 24 W
- Energy: E = I × V × t = 2.0 × 12 × 30 = 720 J
- Voltage & resistance: 0.50 A through a 24 Ω resistor: V = I × R = 0.50 × 24 = 12 V
- Charge: Q = I × t = 2.0 × 30 = 60 C · Key definition: current is the flow of negatively charged particles (electrons)
- Energy & charge: E = Q × V = 60 × 12 = 720 J, the same as the energy row
Page 16 · Question 1 · series · a 6.0 V cell, two identical bulbs, 0.50 A, 2 minutes · 6 marks
- (a) t = 2 × 60 = 120 s (1); Q = I × t = 0.50 × 120 = 60 C (1)
- (b) E = Q × V = 60 × 6.0 (1) = 360 J (1)
- (c) Identical bulbs in series share the voltage: 6.0 ÷ 2 = 3.0 V each (1); E = 60 × 3.0 = 180 J (1)
Page 16 · Question 2 · parallel · a 12 V cell, two identical bulbs, 0.30 A through each, 5 minutes · 6 marks
- (a) The branch currents add: I = 0.30 + 0.30 = 0.60 A (1)
- (b) t = 5 × 60 = 300 s (1); E = I × V × t = 0.60 × 12 × 300 (1) = 2160 J (1)
- (c) Each bulb gets the full 12 V and 0.30 A: E = 0.30 × 12 × 300 (1) = 1080 J (1)
Page 17 · Question 1 · a 2.0 kΩ and a 4.0 kΩ resistor in series with 12 V, 3.5 minutes · energy in the 4.0 kΩ resistor, 2 s.f. · 5 marks
- Total resistance = 2.0 + 4.0 = 6.0 kΩ = 6000 Ω (1)
- Current = V ÷ R = 12 ÷ 6000 = 0.0020 A (1)
- Voltage across the 4.0 kΩ resistor = 0.0020 × 4000 = 8.0 V (1)
- t = 3.5 × 60 = 210 s; E = I × V × t = 0.0020 × 8.0 × 210 = 3.36 J (1)
- = 3.4 J to 2 s.f. (1)
Where marks go missing: using 12 V. The 12 V is shared, so only 8.0 V is across the 4.0 kΩ resistor.
Page 17 · Question 2 · lamp A (250 Ω) and lamp B (30 mA) in parallel with 6.0 V, 2.0 minutes · total charge, 2 s.f. · 5 marks
- In parallel each lamp has the full 6.0 V (1)
- Lamp A: I = V ÷ R = 6.0 ÷ 250 = 0.024 A (24 mA) (1)
- Total current = 24 + 30 = 54 mA = 0.054 A (1)
- t = 2.0 × 60 = 120 s; Q = I × t = 0.054 × 120 = 6.48 C (1)
- = 6.5 C to 2 s.f. (1)
Shortcut: you never need lamp B's resistance: its current is given.
✅ Check your answer
Did you use the current and voltage for that part of the circuit? Is everything in A, V, Ω and s? Have you given the unit and rounded to the significant figures asked? The last mark is often just for the rounding.
📘 Now do it in your workbook
Write each formula, what it means and your own example; work out the energy and charge in a series and a parallel circuit; then two 5-mark questions with unit prefixes and no steps to follow.
Edexcel iGCSE Physics Workbook · ELECTRICITY · pages 12, 16 & 17Free sample = the pages for this lesson. The full workbook covers the whole of Section 2 Electricity, with exam-style questions and worked answers.
✅ Can you do it? Tick as you go
0% complete🎉 Nice work! You've ticked off every objective for this spec point.
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Prefer to watch? Here's the whole thing
Video walkthrough — coming soon
Everything above, explained out loud — useful for a last-minute recap, or if you'd rather hear it than read it.
Everything above, explained out loud — useful for a last-minute recap, or if you'd rather hear it than read it. The video is on its way.
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Video coming soon
The walkthrough for this lesson is being filmed and will appear right here. Until then, the animation and the notes above cover every mark.
Now put it into practice
Pages 12, 16 and 17 of the Electricity workbook are free: the five formulae, energy and charge in series and parallel circuits, and two 5-mark challenge questions.
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