Electric circuits — Physical Sciences | Learn on Stanmorephysics

4 exam-style question sets on Electric circuits (Paper 1), each with a hint and a fully worked answer. The app holds 20 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Rated bulbs in parallel branches & internal resistanceElectrical energy, power & unitsemf and internal resistance from two operating conditionsSeries–parallel circuits & switching

Practise this section in the app →

Question 1

Rated bulbs in parallel branches & internal resistanceApplication21 marksPaper 1

Two light bulbs, X and Y, have the following ratings:

Bulb	Rating
X	40 W ; 20 V
Y	18 W ; 12 V

The bulbs are connected in the circuit shown. One branch contains ammeter A₂, resistor R₁ and bulb Y in series; the other branch contains resistor R₂ = 2 Ω and bulb X in series. Ammeter A₁ is in the main circuit. The battery has an internal resistance of 0,4 Ω, and the ammeters and connecting wires have negligible resistance.

Both bulbs operate at their RATED values.

aDefine the term power. (2)
bCalculate the reading on ammeter A₂. (3)
cCalculate the reading on ammeter A₁. (3)
dCalculate the resistance of resistor R₁. (4)
eCalculate the emf of the battery. (4)
fBulb Y suddenly burns out. Assume that the resistance of bulb X remains constant. Will bulb X now operate ABOVE or BELOW its rated power? Support the answer with a suitable calculation. (5)

This question has a diagram, shown in the app.

Hint

Get each branch current from I = P/V; the two parallel branches must have equal total potential differences.

Worked answer

a) Power is the rate at which work is done (or the rate at which energy is transferred). ✓✓

b) A₂ reads the current in bulb Y's branch:
$P = V I \Rightarrow I = \frac{P}{V} = \frac{18}{12}$ ✓✓ $= 1, 50$ A ✓

c) Current in bulb X's branch: $I = \frac{40}{20} = 2, 00$ A ✓
$I_{A 1} = 2, 00 + 1, 50$ ✓ $= 3, 50$ A ✓

d) P.d. across the X-branch: $V = 20 + (2, 00) (2) = 24$ V ✓ (equal to the p.d. across the Y-branch, since the branches are in parallel)
P.d. across R₁: $V_{R 1} = 24 - 12 = 12$ V ✓
$R_{1} = \frac{V}{I} = \frac{12}{1 , 50}$ ✓ $= 8, 00$ Ω ✓

e) $ε = V_{e x t} + I r$ ✓ $= 24 + (3, 50) (0, 4)$ ✓✓ $= 25, 40$ V ✓

f) ABOVE. ✓ With the Y-branch open, only the X-branch carries current.
$R_{X} = \frac{V ^{2}}{P} = \frac{( 20 ) ^{2}}{40} = 10$ Ω ✓
$I = \frac{ε}{R _{X} + R _{2} + r} = \frac{25 , 40}{10 + 2 + 0 , 4} = 2, 05$ A ✓
$P = I^{2} R_{X} = (2, 05)^{2} (10)$ ✓ $= 41, 96$ W $> 40$ W, so bulb X operates above its rated power ✓

Modelled on Nov 2024 Paper 1, Question 8 — same skills, new scenario.

Question 2

Electrical energy, power & unitsRecall2 marksPaper 1

The joule per second (J·s⁻¹) is equivalent to the unit of …

Hint

Which quantity measures energy transferred per second?

Worked answer

A. Power is the rate at which energy is transferred: $P = \frac{W}{Δ t}$ , so 1 W = 1 J·s⁻¹. B is measured in joules alone. C is measured in coulombs. D is measured in volts, where 1 V = 1 J·C⁻¹ (joule per coulomb, not per second).

Modelled on Nov 2024 Paper 1, Question 1.8 — same skill, new scenario.

Question 3

emf and internal resistance from two operating conditionsApplication20 marksPaper 1

The battery of a game-viewing vehicle is connected to two identical spotlights, L₁ and L₂ (6 Ω each), which are in PARALLEL with each other, and to a winch motor M of resistance 0,15 Ω in a third parallel branch controlled by switch S. The emf (ε) and internal resistance (r) of the battery are unknown. Ammeter A₁ is in the main circuit and a voltmeter of very high resistance is connected across the battery. The connecting wires and the ammeter have negligible resistance.

Condition 1: The spotlights are removed from their holders and switch S is closed. Ammeter A₁ reads 60 A.

Condition 2: The spotlights are replaced and switch S is opened. Each spotlight now dissipates 24 W.

aDefine the term emf of a battery. (2)
bFor Condition 1, calculate the reading on the voltmeter. (3)
cFor Condition 2, calculate the current through spotlight L₁. (3)
dWrite down the reading on ammeter A₁ for Condition 2. (1)
eUse BOTH conditions to calculate the emf of the battery. (6)
fThe spotlights remain in place and switch S is now ALSO closed. How will (i) the reading on the voltmeter and (ii) the brightness of L₁ be affected? Choose from INCREASES, DECREASES or REMAINS THE SAME for each. Explain the answers WITHOUT any calculation. (5)

This question has a diagram, shown in the app.

Hint

Each condition gives one equation of the form ε = V_terminal + I·r — two conditions, two unknowns.

Worked answer

a) The emf is the maximum work done (total energy supplied) by a battery per unit charge passing through it. ✓✓

b) Only the motor is connected, so the voltmeter reads the p.d. across the motor:
$V = I R$ ✓ $= 60 \times 0, 15$ ✓ $= 9, 00$ V ✓

c) $P = I^{2} R$ ✓ $\Rightarrow 24 = I^{2} (6)$ ✓ $\Rightarrow I = 2, 00$ A ✓

d) The two parallel spotlights carry 2 A each: A₁ reads $4, 00$ A ✓

e) Condition 1: $ε = V + I r = 9 + 60 r$ ✓
Condition 2: p.d. across the spotlights $V = I R = 2 \times 6 = 12$ V ✓, so $ε = 12 + 4 r$ ✓
Equating: $9 + 60 r = 12 + 4 r$ ✓ $\Rightarrow 56 r = 3 \Rightarrow r = 0, 05$ Ω ✓
$ε = 12 + 4 (0, 054) = 12, 21$ V ✓

f) (i) DECREASES ✓ — adding the motor in parallel lowers the external resistance, so the total current delivered by the battery increases ✓; the internal volt drop $I r$ increases, and $V = ε - I r$ therefore decreases ✓.
(ii) DECREASES (dimmer) ✓ — the p.d. across the parallel spotlights equals the (now smaller) terminal p.d., so each spotlight dissipates less power ( $P = V^{2} / R$ ). ✓

Modelled on Nov 2025 Paper 1, Question 8 — same skills, new scenario.

Question 4

Series–parallel circuits & switchingApplication2 marksPaper 1

Three identical light bulbs, L₁, L₂ and L₃, are connected to a battery with negligible internal resistance, as shown. L₁ is in series with the battery. L₂ and L₃ are connected in parallel with each other, and switch S is in series with L₃ ONLY. S is initially OPEN.

Switch S is now CLOSED. How will the brightness of L₁ and of L₂ be affected?

This question has a diagram, shown in the app.

Hint

Closing S lowers the total resistance — track the main current and the parallel-section voltage.

Worked answer

B. Closing S halves the resistance of the parallel section (R/2 instead of R), so the total resistance drops from 2R to 1,5R and the main current — the current through L₁ — increases: L₁ is brighter. The parallel section now takes a smaller share of the emf ( $\frac{1}{3} ε$ instead of $\frac{1}{2} ε$ ), so L₂ is dimmer. A ignores the reduced voltage across L₂. C and D wrongly have the main current decreasing when the total resistance has decreased. Modelled on Nov 2025 Paper 1, Question 1.8 — same skills, new scenario.

Open Electric circuits in the app →

Grade 12 Physical Sciences: Electric circuits

Question 1

Question 2

Question 3

Question 4

About this material