1. Current, P.D. & Resistance

a. Charge and current

6091 14(a)–(c) · IGCSE 4.2.2

Lesson flow

Toggle conventional/electron flow early and keep the label visible. The counter integrates I over time at the dashed line so Q = It is seen, not just stated.

Predict-then-test

Prompt: The ammeter reads 0.50 A. Predict the charge passing in 20 s.
Outcome: Q = 0.50 A × 20 s = 10 C.

Challenge (auto-checked)

Task: Set the rheostat so that exactly 6.0 C of charge passes the ammeter every 12 s.

Answer: Rheostat at 4.0 Ω: I = 3.0 V ÷ (2.0 + 4.0) Ω = 0.50 A; 0.50 A × 12 s = 6.0 C.

Exam-style questions and answers

1. A current of 0.40 A flows in a lamp for 2.0 minutes. Calculate the charge that passes through the lamp. [2]
   48 C. Q = It = 0.40 × 120 = 48 C. Convert minutes to seconds first.
2. 90 C of charge flows through a heater in 30 s. Calculate the current. [1]
   3 A. I = Q/t = 90/30 = 3.0 A.
3. Which statement about current in a metal wire is correct? [1]
   A  Conventional current and electron flow are in the same direction
   B  Electrons flow from the negative terminal of the cell round the circuit to the positive terminal
   C  Current is the energy carried by each electron
   D  Current is measured with a voltmeter connected in parallel
   B (Electrons flow from the negative terminal of the cell round the circuit to the positive terminal).
4. What is one ampere? [1]
   A  One joule per coulomb
   B  One coulomb per second
   C  One volt per ohm per second
   D  One joule per second
   B (One coulomb per second). Current is the rate of flow of charge: 1 A = 1 C/s.
5. State what is meant by an electric current, and distinguish between conventional current and electron flow. [3]
   Current is the rate of flow of charge (Q/t). Conventional current is taken as flowing from the positive terminal of the supply, through the circuit, to the negative terminal. Electron flow is the actual movement of (negatively charged) electrons, from the negative terminal to the positive terminal, opposite to conventional current.

b. E.m.f. and potential difference

6091 14(d)–(f) · IGCSE 4.2.3

Lesson flow

Link volts to joules per coulomb with the energy readout. The reversed-cell prediction tests whether students treat cells as simply “adding up”.

Predict-then-test

Prompt: Four 1.5 V cells in series; the third is reversed. Predict V_cells.
Outcome: 3.0 V.

Challenge (auto-checked)

Task: Make the lamp current exactly 0.25 A (the lamp is 12 Ω). Use the cell settings.

Answer: e.m.f. must be 3.0 V: two cells forwards and two removed, or three forwards and one reversed. 3.0 V ÷ 12 Ω = 0.25 A.

Exam-style questions and answers

1. 36 J of energy is transferred when 3.0 C of charge passes through a lamp. Calculate the p.d. across the lamp. [2]
   12 V. V = W/Q = 36/3.0 = 12 V.
2. A 9.0 V battery drives 0.20 A round a circuit for 50 s. Calculate the work done by the battery. [2]
   90 J. Q = It = 10 C; W = QV = 10 × 9.0 = 90 J.
3. Three 1.5 V cells are connected in series, but one faces the opposite way to the other two. What is the total e.m.f.? [1]
   A  0 V
   B  1.5 V
   C  3.0 V
   D  4.5 V
   B (1.5 V).
4. How is a voltmeter connected to measure the p.d. across a resistor? [1]
   A  In series with the resistor
   B  In parallel with the resistor
   C  In series with the cell only
   D  Across the ammeter
   B (In parallel with the resistor).
5. Distinguish between the e.m.f. of a cell and the p.d. across a lamp. [2]
   The e.m.f. is the work done per unit charge by the cell in driving charge round the complete circuit (energy given to each coulomb). The p.d. across the lamp is the work done per unit charge in driving charge through the lamp (energy transferred by each coulomb in the lamp).

c. Resistance

6091 14(g), (h) · IGCSE 4.2.4

Lesson flow

Misconception focus. In Predict the voltmeter across the battery stays at 6.0 V while the current drops: put both readings in front of the class. The mystery X challenge practises R = V/I with real readings; its value is randomised per visit (15–39 Ω).

Misconception focus. The Predict step locks in each student’s prediction and shows it next to the measured result.

Predict-then-test

Prompt: 6.0 V battery, 12 Ω resistor, 0.50 A. Predict the current and battery p.d. with a 30 Ω resistor instead.
Outcome: 0.20 A and 6.0 V.

Challenge (auto-checked)

Task: Component X is hidden. Use the meters (change the battery if you like) to find its resistance, then enter it.

Answer: X is randomly 15, 18, 22, 27, 33 or 39 Ω. Students divide the voltmeter reading by the ammeter reading.

Exam-style questions and answers

1. The p.d. across a resistor is 4.5 V and the current through it is 0.15 A. Calculate its resistance. [2]
   30 Ω.
2. A 6.0 V battery is connected to a 24 Ω resistor. Calculate the current. [1]
   0.25 A.
3. The same ideal 6.0 V battery is connected in turn to a 10 Ω resistor and to a 20 Ω resistor. Which statement is correct? [1]
   A  The current is the same both times
   B  The battery’s e.m.f. is the same; the current halves with 20 Ω
   C  The e.m.f. doubles with 20 Ω
   D  The current doubles with 20 Ω
   B (The battery’s e.m.f. is the same; the current halves with 20 Ω). A battery provides a fixed e.m.f.; the current depends on the resistance.
4. Which of these is equal to one ohm? [1]
   A  One ampere per volt
   B  One volt per ampere
   C  One volt × one ampere
   D  One joule per coulomb
   B (One volt per ampere). R = V/I, so 1 Ω = 1 V/A. (1 J/C is one volt.)
5. A student says “a battery pushes out the same current whatever is connected to it”. Using the idea of e.m.f., explain why this is wrong. [2]
   An (ideal) battery provides a fixed e.m.f.: it does the same work on each coulomb. The current depends on the resistance of the circuit, I = V/R, so a larger resistance gives a smaller current from the same battery.

d. I–V characteristics

6091 14(j), (k) · IGCSE 4.2.4

Lesson flow

Model note: the lamp uses R = max(R_cold, 12 Ω × (V/6 V)^0.45), so I ∝ V^0.55 above a small p.d.; the diode is piecewise linear (off below 0.7 V, then 0.5 Ω). Plot the resistor first as the reference line, then the lamp, then the diode with negative p.d.

Predict-then-test

Prompt: Sketch the I–V graph you expect for a filament lamp from −6 V to +6 V.
Outcome: A curve that becomes less steep.

Challenge (auto-checked)

Task: Component Y is hidden. Plot its I–V graph and decide what it is.

Answer: Y is chosen at random from resistor, filament lamp and diode. Resistor: straight line through origin. Lamp: symmetric S-shaped curve, less steep at higher |V|. Diode: zero in reverse, steep rise above 0.7 V forward.

Exam-style questions and answers

1. Which I–V graph is a straight line through the origin? [1]
   A  Filament lamp
   B  Metal wire at constant temperature
   C  Diode
   D  Thermistor as it warms up
   B (Metal wire at constant temperature).
2. Why does the I–V graph of a filament lamp curve? [1]
   A  The current is used up in the filament
   B  The filament’s temperature rises, so its resistance increases
   C  The lamp only conducts in one direction
   D  The p.d. across the lamp stays constant
   B (The filament’s temperature rises, so its resistance increases).
3. A filament lamp carries 0.20 A at 1.0 V and 0.50 A at 6.0 V. Calculate its resistance at 6.0 V. [1]
   12 Ω. R = V/I = 6.0/0.50 = 12 Ω (it was 5.0 Ω at 1.0 V).
4. A diode is connected so that it is reverse-biased. What is the current? [1]
   A  Very large
   B  Almost zero
   C  The same as when forward-biased
   D  Equal to V/R of the diode’s forward resistance
   B (Almost zero).
5. Sketch and describe the I–V characteristic of a semiconductor diode, for both positive and negative p.d. [3]
   For negative (reverse) p.d. the current is almost zero. For positive (forward) p.d. the current stays almost zero until about 0.7 V, then rises steeply, so the graph is close to vertical. The diode lets current through in one direction only.

e. Resistance of a wire

6091 14(i), (j) · IGCSE 4.2.4 (ρ is IGCSE / extension)

Lesson flow

Values used: ρ(nichrome) = 1.10 × 10⁻⁶ Ω m, ρ(constantan) = 4.9 × 10⁻⁷ Ω m, ρ(copper) = 1.7 × 10⁻⁸ Ω m, with R = ρ(1 + αΔθ)L/A. O-Level 6091 asks only for R ∝ L and R ∝ 1/A; ρ is shown for IGCSE and stretch. The ρ formula is labelled accordingly.

Predict-then-test

Prompt: A wire twice as long and twice the area (same material). Predict its resistance compared with 2.20 Ω.
Outcome: Unchanged, 2.20 Ω.

Challenge (auto-checked)

Task: Using nichrome wire at 20 °C, choose a length and area so that the ammeter reads 0.75 A from the 1.5 V cell.

Answer: R = 2.0 Ω: L/A = 2.0/1.10 = 1.82 m per mm², e.g. L = 1.00 m with A = 0.55 mm², or L = 0.91 m with A = 0.50 mm² (within 1.6%).

Exam-style questions and answers

1. A wire of length 0.80 m has a resistance of 4.0 Ω. What is the resistance of 2.0 m of the same wire? [2]
   10 Ω. R ∝ L: 4.0 × 2.0/0.80 = 10 Ω.
2. A wire has a resistance of 6.0 Ω. Another wire of the same material and length has three times the cross-sectional area. Calculate its resistance. [1]
   2 Ω.
3. Wire P has length L and diameter d. Wire Q, of the same material, has length L and diameter 2d. What is the ratio resistance of Q : resistance of P? [1]
   A  1 : 4
   B  1 : 2
   C  2 : 1
   D  4 : 1
   A (1 : 4). Doubling the diameter makes the area four times larger, so R is a quarter.
4. What happens to the resistance of a metal wire as its temperature increases? [1]
   A  It increases
   B  It decreases
   C  It stays the same
   D  It becomes zero
   A (It increases).
5. Describe an experiment to show how the resistance of a wire depends on its length. State the readings you would take and how you would use them. [4]
   Connect a length of the wire (held on a metre rule with crocodile clips) in series with a cell, a switch and an ammeter, with a voltmeter across the wire. For several lengths, measure V and I and calculate R = V/I. Keep the material, area and temperature constant (switch off between readings so the wire does not heat). Plot R against L: a straight line through the origin shows R ∝ L.

f. Thermistor and LDR

6091 15(i) · IGCSE 4.2.4

Lesson flow

Model: thermistor R = 1.0 kΩ × exp[3500 (1/T − 1/298.15 K)]; LDR R = 100 kΩ × (200/100 000)^(light/100), i.e. 100 kΩ dark to 200 Ω bright. The sweep draws the characteristic curves; link forward to D.C. Circuits chapter e.

Predict-then-test

Prompt: Thermistor heated from 20 °C to 80 °C. Predict the change in the ammeter reading.
Outcome: The current increases.

Challenge (auto-checked)

Task: Set the temperature of the thermistor so that the ammeter reads 15 mA (±0.3 mA).

Answer: About 50 °C, where R ≈ 403 Ω and I = 14.9 mA.

Exam-style questions and answers

1. How does the resistance of an NTC thermistor change as it is heated? [1]
   A  It increases steadily
   B  It decreases
   C  It stays the same
   D  It increases, then decreases
   B (It decreases).
2. An LDR is covered so that no light reaches it. What happens to its resistance and to the current through it from a fixed supply? [1]
   A  Resistance rises; current falls
   B  Resistance falls; current rises
   C  Both rise
   D  Both fall
   A (Resistance rises; current falls).
3. A thermistor connected to a 6.0 V supply carries 2.0 mA. Calculate its resistance in kΩ. [2]
   3 kΩ. R = 6.0 / 0.0020 = 3000 Ω = 3.0 kΩ.
4. Which device would be most suitable for detecting a rise in temperature in a circuit? [1]
   A  Fixed resistor
   B  LDR
   C  NTC thermistor
   D  Diode
   C (NTC thermistor).
5. Contrast how the resistance of a metal wire and the resistance of an NTC thermistor change with temperature. [2]
   The resistance of a metal wire increases (slightly) as its temperature rises. The resistance of an NTC thermistor decreases (sharply) as its temperature rises.