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GCSE / IGCSE Physics – Electricity

Circuits, resistance, series and parallel, I-V graphs, mains electricity.

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12 questions  ·  60 marks  ·  answers included  ·  GCSE/IGCSE
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This worksheet covers:

  • Charge
  • Current and potential difference
  • Resistance
  • Series and parallel circuits
  • I–V characteristics
  • Electrical power
  • Mains electricity

Calculator allowed; allow about 60 minutes. Use $Q = It$, $V = IR$, $P = IV$, $E = Pt$ and $E = QV$.

Questions

Section A – Skills

1

A current of $2.5\,\text{A}$ flows through a lamp for $4$ minutes. Calculate the charge that flows through the lamp.

Show answer
$Q = It = 2.5 \times 240 = 600\,\text{C}$
[2]
2

A resistor has a potential difference of $12\,\text{V}$ across it and a current of $3\,\text{A}$ through it.(a) Calculate its resistance.(b) State what happens to the current if the potential difference is doubled and the resistance stays constant.

Show answer
(a) $R = \frac{V}{I} = 4\,\Omega$(b) Current doubles to $6\,\text{A}$
[3]
3

A $4\,\Omega$ resistor and a $6\,\Omega$ resistor are connected in series to a $12\,\text{V}$ battery.(a) Calculate the total resistance.(b) Calculate the current in the circuit.(c) Calculate the potential difference across the $6\,\Omega$ resistor.

Show answer
(a) $10\,\Omega$(b) $I = \frac{12}{10} = 1.2\,\text{A}$(c) $V = 1.2 \times 6 = 7.2\,\text{V}$
[4]
4

The same $4\,\Omega$ and $6\,\Omega$ resistors are now connected in parallel across the $12\,\text{V}$ battery.(a) Calculate the current through each resistor.(b) Calculate the total current from the battery.(c) Hence find the total resistance of the parallel combination.

Show answer
(a) $3\,\text{A}$ through $4\,\Omega$, $2\,\text{A}$ through $6\,\Omega$(b) $5\,\text{A}$(c) $R = \frac{12}{5} = 2.4\,\Omega$ (less than the smallest resistor)
[4]
5

A kettle is rated $230\,\text{V}$, $2300\,\text{W}$.(a) Calculate the current in the kettle.(b) Calculate the resistance of the heating element.(c) Calculate the energy transferred when the kettle runs for $3$ minutes.

Show answer
(a) $I = \frac{P}{V} = 10\,\text{A}$(b) $R = \frac{V}{I} = 23\,\Omega$(c) $E = Pt = 2300 \times 180 = 414\,000\,\text{J}$ ($414\,\text{kJ}$)
[5]

Section B – Exam-style

6

A student measures the current through a filament lamp as the potential difference across it is increased from $0$ to $12\,\text{V}$.(a) Describe the shape of the I–V graph for the filament lamp.(b) Explain, in terms of the metal ions and electrons, why the resistance of the lamp changes as the current increases.(c) Describe how the I–V graph for a fixed resistor at constant temperature differs from that of the lamp.

Show answer
(a) Curve through the origin with gradient decreasing as V increases (flattens); symmetric for negative V(b) Higher current heats the filament; metal ions vibrate more; electrons collide more often with ions, so resistance increases(c) Straight line through the origin (ohmic conductor); current is directly proportional to p.d.
[6]
7

A thermistor is connected in series with a fixed resistor and a $6\,\text{V}$ battery. A voltmeter is connected across the fixed resistor.(a) State what happens to the resistance of the thermistor as its temperature rises.(b) Explain what happens to the voltmeter reading as the temperature rises.(c) Suggest one practical use of this circuit.

Show answer
(a) Resistance decreases(b) Total resistance falls, so current increases; p.d. across the fixed resistor ($V = IR$) increases, and the thermistor takes a smaller share of the $6\,\text{V}$(c) Thermostat / temperature sensor for a heating system or fire alarm
[5]
8

Mains electricity in the UK is supplied at $230\,\text{V}$, $50\,\text{Hz}$.(a) Explain what is meant by $50\,\text{Hz}$.(b) State the potential of the live wire and the neutral wire with respect to earth.(c) A hair dryer with a metal casing is rated $920\,\text{W}$. Calculate the current and choose a suitable fuse from $3\,\text{A}$, $5\,\text{A}$ and $13\,\text{A}$.(d) Explain how the earth wire and fuse together protect the user if the live wire touches the metal casing.

Show answer
(a) The alternating current changes direction $50$ times per second (50 complete cycles per second)(b) Live about $230\,\text{V}$; neutral about $0\,\text{V}$(c) $I = \frac{920}{230} = 4\,\text{A}$; use the $5\,\text{A}$ fuse(d) A large current flows from live through the casing to earth via the low-resistance earth wire; this current exceeds the fuse rating, the fuse wire melts and disconnects the live supply, so the casing cannot give a shock
[8]
9

A battery transfers $1500\,\text{C}$ of charge at a potential difference of $230\,\text{V}$ through a heater.(a) Calculate the energy transferred.(b) The heater runs for $50\,\text{s}$. Calculate the power of the heater.

Show answer
(a) $E = QV = 1500 \times 230 = 345\,000\,\text{J}$(b) $P = \frac{345\,000}{50} = 6900\,\text{W}$
[4]

Section C – Challenge

10

A $2\,\Omega$ resistor is connected in series with a parallel combination of a $3\,\Omega$ resistor and a $6\,\Omega$ resistor. The circuit is connected to a $12\,\text{V}$ supply.(a) Show that the total resistance of the circuit is $4\,\Omega$.(b) Calculate the current from the supply.(c) Calculate the potential difference across the parallel combination.(d) Calculate the current in the $3\,\Omega$ resistor.

Show answer
(a) Parallel pair: $\frac{1}{R} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$ so $2\,\Omega$; total $2 + 2 = 4\,\Omega$(b) $3\,\text{A}$(c) $V = 3 \times 2 = 6\,\text{V}$(d) $I = \frac{6}{3} = 2\,\text{A}$ (and $1\,\text{A}$ in the $6\,\Omega$)
[7]
11

A student investigates how the resistance of a wire depends on its length using an ammeter, a voltmeter, a metre rule and a low-voltage supply.(a) Describe how the student should carry out the investigation, including two ways to keep the results accurate.(b) A $0.50\,\text{m}$ length of the wire has resistance $2.0\,\Omega$. Predict the resistance of a $1.25\,\text{m}$ length of the same wire and explain your reasoning.

Show answer
(a) Connect the wire in series with the ammeter and supply, voltmeter in parallel across the chosen length; use a crocodile clip to vary the length in steps (e.g. every $10\,\text{cm}$); record V and I and calculate $R = V/I$; keep current low or switch off between readings to prevent heating; repeat and average; check for zero errors; keep the wire straight and taut(b) Resistance is directly proportional to length, so $R = 2.0 \times \frac{1.25}{0.50} = 5.0\,\Omega$
[6]
12

(a) A $3\,\text{kW}$ electric heater is used for $5$ hours. Electricity costs $15\,\text{p}$ per kWh. Calculate the cost.(b) Electricity is transmitted around the National Grid at very high voltage. Explain why step-up transformers are used at the power station and why this reduces energy losses.

Show answer
(a) $3 \times 5 = 15\,\text{kWh}$; cost $15 \times 15 = 225\,\text{p} = \pounds 2.25$(b) Step-up transformers increase the p.d.; for the same power $P = IV$, the current is smaller; heating losses in the cables depend on $I^2R$, so a lower current greatly reduces energy wasted as heat; step-down transformers reduce the p.d. to $230\,\text{V}$ for safe domestic use
[6]

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