📘 GCSE Physics · Notes & worksheet
Electricity.
Notes
Key equations
| Quantity | Equation |
|---|---|
| Charge | Q = It |
| Potential difference | V = IR |
| Energy transferred | E = QV |
| Power | P = IV = I²R |
| Energy | E = Pt |
Series and parallel circuits
| Series | Parallel | |
|---|---|---|
| Current | same everywhere | splits between branches; total = sum of branches |
| Potential difference | shared between components | same across each branch |
| Total resistance | R = R₁ + R₂ | 1/R = 1/R₁ + 1/R₂ (always smaller than the smallest) |
Current–voltage graphs
- Resistor at constant temperature: straight line through the origin (ohmic).
- Filament lamp: curve that flattens. Resistance increases as the filament heats up.
- Diode: conducts in one direction only. Very high resistance in reverse.
- Thermistor: resistance falls as temperature rises. LDR: resistance falls as light intensity rises.
Potential divider
For two resistors in series, the p.d. across each is proportional to its resistance. With 9 V across 2 kΩ and 1 kΩ, the 1 kΩ resistor gets 3 V.
Mains electricity
Mains is a.c. (alternating current) at about 230 V and 50 Hz. Cables: live (brown), neutral (blue), earth (green and yellow). The fuse and switch must be on the live wire. A fuse melts if the current is too high. The earth wire prevents a metal case becoming live.
Cost of electricity
Energy in kWh = power (kW) × time (h). Cost = kWh × price per kWh.
Common mistakes
- Using minutes instead of seconds in Q = It and E = Pt (unless working in kWh).
- Adding currents in series or resistances in parallel.
- Forgetting that the fuse rating should be just above the normal working current.
Exam tips
- For circuit problems, draw the circuit, label the known values and work step by step.
- If asked to "explain" a graph, link the shape to the physical cause (for example heating of the filament).
Worksheet
Prints the questions only, without answers. Choose “Save as PDF” in the print window to download.
Diagram placeholderThe diagrams for Q1, Q2 and Q7 are described in words in the questions. The drawn diagrams are being added to this page.
Total: 31 marks.
- A 12 V battery is connected across a 3.0 Ω resistor and a 6.0 Ω resistor in series. Calculate the current and the p.d. across each resistor. (4)
- The same two resistors are now connected in parallel across the 12 V battery. Calculate the total resistance, the current in each resistor, and the total current. (5)
- A current of 2.5 A flows for 3.0 minutes through a 12 V lamp.
(a) Calculate the charge that flows. (2)
(b) Calculate the energy transferred. (2) - A kettle is rated 2.3 kW at 230 V.
(a) Calculate the current. (2)
(b) Calculate the resistance of the heating element. (2)
(c) Choose a suitable fuse from 3 A, 5 A, 13 A and explain your choice. (2) - Calculate the cost of using the 2.3 kW kettle for 5 minutes if electricity costs 28p per kWh. Give your answer in pence to 1 decimal place. (3)
- Explain the shape of the I–V graph of a filament lamp. (3)
- 9 V is applied across a 2 kΩ resistor and a 1 kΩ resistor in series. Calculate the p.d. across the 1 kΩ resistor. (3)
- Explain why the fuse and switch in a plug are placed in the live wire. (2)
- Explain how a thermistor in a circuit could be used to switch on a heater when the temperature falls. (1) (Challenge: state the key idea.)
Show mark scheme and answers (tutor copy)
Try the worksheet first, then check your work here.
- R = 3 + 6 = 9 Ω (M1). I = V ÷ R = 12 ÷ 9 = 1.33 A (A1). V₃ = 1.33 × 3 = 4.0 V and V₆ = 1.33 × 6 = 8.0 V (A1, A1). The p.d.s add up to 12 V ✓.
- 1/R = 1/3 + 1/6 = 1/2, so R = 2.0 Ω (M1, A1). I₃ = 12 ÷ 3 = 4.0 A (A1). I₆ = 12 ÷ 6 = 2.0 A (A1). Total current = 6.0 A (A1), which equals 12 ÷ 2.0 ✓.
- (a) t = 180 s. Q = It = 2.5 × 180 (M1) = 450 C (A1).
(b) E = QV = 450 × 12 (M1) = 5400 J (A1). - (a) I = P ÷ V = 2300 ÷ 230 (M1) = 10 A (A1).
(b) R = V ÷ I = 230 ÷ 10 = 23 Ω (M1, A1).
(c) 13 A. (B1) A 3 A or 5 A fuse would blow at the normal 10 A working current; 13 A is the next rating above 10 A. (B1) - Energy = 2.3 kW × (5 ÷ 60) h = 0.1917 kWh (M1). Cost = 0.1917 × 28 (M1) = 5.4p (A1).
- The graph is a curve that becomes less steep as p.d. increases (B1). As current increases the filament gets hotter (B1), so its resistance increases because the ions vibrate more and collide more with the moving electrons, giving less current per volt. (B1)
- Total R = 3 kΩ (M1). I = 9 ÷ 3000 = 0.003 A (M1). V = 0.003 × 1000 = 3.0 V (A1). (Alternatively V = 9 × 1/3.)
- If a fault occurs and the fuse is in the live wire, the appliance is disconnected from the high voltage (B1). A switch in the neutral wire would leave the appliance live even when switched off. (B1)
- A thermistor's resistance rises as temperature falls; in a potential divider this changes the p.d. across the thermistor, which can switch on a relay or transistor to power the heater. (B1)