2. D.C. Circuits

a. Series circuits

6091 15(b), (c), (f) · IGCSE 4.3.2

Lesson flow

Start with the three ammeters showing equal readings, then reveal that V₁ + V₂ equals the e.m.f. In Predict, A₃ is hidden; most students who hold the “used up” idea pick the first option. Use the Test sliders to show V₁ : V₂ = R₁ : R₂.

Predict-then-test

Prompt: A₁ reads 0.20 A. Predict the reading on A₃ (after both resistors).
Outcome: A₃ = 0.20 A (same as A₁).

Challenge (auto-checked)

Task: Make the ammeter read 0.25 A using the 6.0 V battery and 12 Ω lamps. Fill the gap in the top wire.

Answer: Two 12 Ω lamps in series with the battery and ammeter: 6.0 V ÷ 24 Ω = 0.25 A.

Exam-style questions and answers

1. Three identical lamps are connected in series to a 9.0 V battery. What is the p.d. across each lamp? [1]
   A  1.0 V
   B  3.0 V
   C  9.0 V
   D  27 V
   B (3.0 V). The 9.0 V is shared equally between three identical lamps.
2. A 4.0 Ω resistor and an 8.0 Ω resistor are connected in series to a 12 V supply. Calculate the current. [2]
   1 A. R = 4.0 + 8.0 = 12 Ω; I = V / R = 12 / 12 = 1.0 A.
3. For the same circuit, calculate the p.d. across the 8.0 Ω resistor. [1]
   8 V. V = IR = 1.0 × 8.0 = 8.0 V.
4. A lamp and a variable resistor are in series with a cell. The resistance of the variable resistor is increased. What happens? [1]
   A  Current increases; p.d. across lamp increases
   B  Current decreases; p.d. across lamp decreases
   C  Current decreases; p.d. across lamp increases
   D  Current and p.d. across lamp do not change
   B (Current decreases; p.d. across lamp decreases). Total resistance rises, so the current falls; the lamp’s p.d. = IR_lamp also falls.
5. A student says that the current after a lamp is smaller than the current before it, because the lamp uses up current. Explain why the student is wrong and state what the lamp does use. [2]
   Current is the rate of flow of charge, and charge is not used up: ammeters before and after the lamp read the same. The lamp transfers energy from the charges (to thermal energy and light); the p.d. across it is the energy transferred per unit charge.

b. Parallel circuits

6091 15(d), (e) · IGCSE 4.3.2

Lesson flow

Use the four meters to establish I = I₁ + I₂ and V₁ = V₂. The Predict question targets the idea that a battery pushes a fixed current that has to “go somewhere”. With an ideal cell the lamp current does not change.

Predict-then-test

Prompt: Switch S is opened. Predict what happens to A₁ (lamp branch) and to the main ammeter A.
Outcome: A₁ stays at 0.50 A; main ammeter falls from 1.00 A to 0.50 A.

Challenge (auto-checked)

Task: Make the main ammeter read 1.50 A by adding 12 Ω lamps. Drop each lamp so that its two ends sit on the two vertical wires.

Answer: Two more 12 Ω lamps between the rails: three branches × 0.50 A = 1.50 A.

Exam-style questions and answers

1. Two lamps of different resistance are connected in parallel to a 6.0 V supply. What is the p.d. across each lamp? [1]
   A  3.0 V across each
   B  6.0 V across each
   C  12 V across each
   D  It depends on each lamp’s resistance
   B (6.0 V across each). Branches in parallel have the same p.d. as the supply.
2. The current from the supply to two parallel branches is 0.90 A. The current in one branch is 0.35 A. Calculate the current in the other branch. [1]
   0.55 A. I₂ = 0.90 − 0.35 = 0.55 A.
3. A 3.0 Ω resistor and a 6.0 Ω resistor are connected in parallel to a 6.0 V battery. Calculate the current from the battery. [2]
   3 A. I = 6/3 + 6/6 = 2.0 + 1.0 = 3.0 A.
4. Why are the lamps in a home wired in parallel? [1]
   A  Each lamp gets the full mains p.d. and can be switched on and off independently
   B  The mains p.d. is shared equally between the lamps
   C  It makes the total resistance of the circuit larger
   D  The current in each lamp equals the current from the supply
   A (Each lamp gets the full mains p.d. and can be switched on and off independently).
5. In the circuit in this chapter, the lamp filament breaks. State and explain what happens to (a) the current in the resistor branch, (b) the reading on the main ammeter. [3]
   (a) It does not change: the resistor branch still has the full 6.0 V across it and its resistance is unchanged, so I = V/R is unchanged. (b) It falls, by the current that the lamp branch was carrying, because the main current is the sum of the branch currents.

c. Effective resistance

6091 15(f), (g) · IGCSE 4.3.2

Lesson flow

Misconception focus. Run Predict before any teaching of the parallel formula. The side-by-side box keeps each student’s prediction visible next to the measured result. Then use Test to check 1/R = 1/R₁ + 1/R₂ against measured values.

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

Predict-then-test

Prompt: One 12 Ω resistor: ammeter reads 0.50 A. A second 12 Ω resistor is added in parallel. Predict the new reading.
Outcome: 0.50 A → 1.00 A; effective resistance 12 Ω → 6.0 Ω.

Challenge (auto-checked)

Task: Using only 10 Ω resistors, complete the circuit so that the ammeter reads 0.40 A. Join the open end of the top wire to the open end on the right.

Answer: One 10 Ω resistor in series with two 10 Ω resistors in parallel: 10 + 5 = 15 Ω, so I = 6.0 / 15 = 0.40 A. (Other networks that total 15 Ω are also accepted, since the check uses the ammeter reading.)

Exam-style questions and answers

1. Calculate the effective resistance of a 6.0 Ω resistor and a 3.0 Ω resistor connected in parallel. [2]
   2 Ω. 1/R = 1/6 + 1/3 = 1/2, so R = 2.0 Ω.
2. Two identical lamps are connected in parallel across a battery. A third identical lamp is connected in parallel with them. What happens to the effective resistance of the lamps and to the current from the battery? [1]
   A  Resistance increases; current decreases
   B  Resistance decreases; current increases
   C  Resistance decreases; current decreases
   D  Resistance stays the same; current increases
   B (Resistance decreases; current increases). Another parallel path lowers the effective resistance, so more current flows from the same e.m.f.
3. Two 6.0 Ω resistors are connected in parallel, and this pair is connected in series with a third 6.0 Ω resistor. Calculate the effective resistance. [2]
   9 Ω. 6.0 ∥ 6.0 = 3.0 Ω; 3.0 + 6.0 = 9.0 Ω.
4. Which arrangement has the smallest effective resistance? [1]
   A  One 10 Ω resistor
   B  Two 10 Ω resistors in series
   C  Two 10 Ω resistors in parallel
   D  A 10 Ω and a 20 Ω resistor in parallel
   C (Two 10 Ω resistors in parallel). 10 ∥ 10 = 5.0 Ω; 10 ∥ 20 = 6.7 Ω.
5. Explain, in terms of the paths available to charge, why connecting a resistor in parallel with another resistor decreases the effective resistance. [2]
   The parallel resistor gives charge an additional path. With the same p.d., the original branch carries the same current as before and the new branch carries extra current, so the total current is larger. Since R = V/I, the effective resistance is smaller.

d. Potential dividers

6091 15(h) · IGCSE 4.3.2

Lesson flow

Begin with equal resistors (4.5 V each), predict the effect of doubling R₂, then use Test to verify the ratio formula over several rows. Switch to the potentiometer to show continuous control. The voltmeter is ideal so there is no loading effect; mention that a real load in parallel with R₂ lowers Vout.

Predict-then-test

Prompt: R₁ = R₂ = 1.0 kΩ across 9.0 V. Predict Vout when R₂ becomes 2.0 kΩ.
Outcome: Vout rises from 4.50 V to 6.00 V.

Challenge (auto-checked)

Task: Set R₁ and R₂ so that V_out = 3.0 V and the battery supplies a current of 1.0 mA.

Answer: Total resistance 9.0 V ÷ 1.0 mA = 9.0 kΩ; Vout/9.0 = R₂/9.0 kΩ → R₂ = 3.0 kΩ, R₁ = 6.0 kΩ.

Exam-style questions and answers

1. A 2.0 kΩ resistor and a 4.0 kΩ resistor are connected in series across a 12 V supply. Calculate the p.d. across the 4.0 kΩ resistor. [2]
   8 V. V = 4.0/6.0 × 12 = 8.0 V.
2. In a potential divider, V_out is taken across R₂. Which change increases V_out? [1]
   A  Increasing R₁
   B  Increasing R₂
   C  Decreasing the supply p.d.
   D  Decreasing R₂
   B (Increasing R₂). R₂ takes a larger share of the supply p.d.
3. A 6.0 V supply is connected across R₁ = 3.0 kΩ in series with R₂. What value of R₂ gives 2.0 V across R₂? Give your answer in kΩ. [2]
   1.5 kΩ. V₁ = 4.0 V, so I = 4.0/3000 = 1.33 mA; R₂ = 2.0 / 1.33 mA = 1.5 kΩ (or R₂/R₁ = 2/4).
4. A potentiometer track is connected across a 9.0 V supply. The slider is one third of the way along the track from end X. What is the p.d. between the slider and X? [1]
   A  1.5 V
   B  3.0 V
   C  6.0 V
   D  9.0 V
   B (3.0 V). One third of the track resistance, so one third of 9.0 V.
5. Describe how a potentiometer can provide a p.d. that can be varied continuously from zero to the supply p.d. [3]
   The supply is connected across the whole resistance track. The output is taken between the sliding contact and one end of the track. Moving the slider changes the fraction of the track resistance between the output terminals, so the output p.d. is that fraction of the supply p.d.: zero with the slider at that end, the full supply p.d. at the other end.

e. Sensing circuits

6091 15(i), (j) · IGCSE 4.3.2

Lesson flow

Establish that the sensor’s resistance falls with temperature (or light), then predict the effect on Vout with the sensor at the bottom. Use Swap to show that the position decides the direction of the response. The challenge needs both decisions: position and value.

Predict-then-test

Prompt: Thermistor at the bottom of the divider. Predict what happens to Vout as it is heated from 20 °C to 80 °C.
Outcome: Vout falls (about 3.3 V → 0.8 V with the 1.0 kΩ fixed resistor).

Challenge (auto-checked)

Task: Design a fire alarm: the output must be ON at 60 °C and OFF at 30 °C. Choose where the thermistor goes and the value of the fixed resistor.

Answer: Thermistor as the TOP resistor, fixed resistor (output) at the bottom, with R between about 300 Ω and 820 Ω (e.g. 500 Ω). Then Vout = 6R/(R + R_th): below 3 V at 30 °C (R_th ≈ 820 Ω) and above 3 V at 60 °C (R_th ≈ 290 Ω).

Exam-style questions and answers

1. The light shining on an LDR becomes brighter. What happens to its resistance? [1]
   A  It increases
   B  It decreases
   C  It stays the same
   D  It increases, then decreases
   B (It decreases).
2. An NTC thermistor is the lower resistor of a potential divider and V_out is taken across it. The temperature rises. What happens to V_out? [1]
   A  It rises
   B  It falls
   C  It stays the same
   D  It becomes equal to the supply p.d.
   B (It falls). The thermistor’s resistance falls, so its share of the supply p.d. falls.
3. An LDR is connected in series with a 2.0 kΩ resistor across a 6.0 V supply. In the dark the LDR has a resistance of 10 kΩ. Calculate the p.d. across the LDR. [2]
   5 V. V = 10/12 × 6.0 = 5.0 V.
4. In bright light the same LDR has a resistance of 400 Ω. Calculate the new p.d. across it. [2]
   1 V. V = 0.4/2.4 × 6.0 = 1.0 V.
5. A lamp must switch on automatically when it gets dark. The switching unit turns on when its input p.d. rises above 3 V. State whether the LDR should be the top or bottom resistor of the potential divider, and explain how the circuit works. [3]
   The LDR should be the bottom resistor (across the output). When it gets dark the LDR’s resistance rises, so it takes a larger share of the supply p.d.; Vout rises above 3 V and the lamp switches on. In bright light its resistance falls, Vout falls below 3 V and the lamp switches off.