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Electricity & Circuits

Charge & CurrentVoltageResistanceOhm's LawSeries & ParallelElectrical PowerDomestic Wiring
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⚠️ Common Mistakes — Electricity (O-Level Physics)

📝 Model Answers — Electricity

Q1 (3 marks): Two resistors of 4 Ω and 6 Ω are connected in series to a 12 V battery. Calculate the current through the circuit and the voltage across each resistor.

MODEL ANSWER — 3 marks ✓
"Total resistance = 4 + 6 = 10 Ω (series: R_total = R₁ + R₂)
Current: I = V/R = 12/10 = 1.2 A (1 mark)
Voltage across 4 Ω: V = IR = 1.2 × 4 = 4.8 V (1 mark)
Voltage across 6 Ω: V = IR = 1.2 × 6 = 7.2 V (1 mark)
Check: 4.8 + 7.2 = 12 V ✓"

Examiner note: Always show working clearly. A wrong final answer with correct working can still earn method marks.

Q2 (2 marks): A 60 W bulb operates at 240 V. Calculate the current through it and its resistance.

MODEL ANSWER — 2 marks ✓
"Current: I = P/V = 60/240 = 0.25 A (1 mark)
Resistance: R = V/I = 240/0.25 = 960 Ω (1 mark)"
Two circuit diagrams with rules for current and voltage in series and parallel Series Circuit V R1 R2 Current: same everywhere Voltage: splits V=V1+V2 R_total = R1 + R2 One fails -> all go off Parallel Circuit V R1 R2 Voltage: same in each branch Current: splits I=I1+I2 1/R_total = 1/R1 + 1/R2 One fails -> others stay on
Series vs Parallel Circuits — current and voltage rules

Contents

  1. Charge and current
  2. Voltage (potential difference)
  3. Resistance and Ohm's law
  4. Series and parallel circuits
  5. Electrical power and energy
  6. Domestic wiring and safety
  7. Common exam traps
Topic 9 of 12
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1. Charge and Current

Electric current (I)

The rate of flow of electric charge. Conventional current flows from positive to negative; electrons flow from negative to positive.

I = Q ÷ tI = current (A) · Q = charge (C) · t = time (s) · 1 A = 1 C/s
Worked example

A charge of 90 C flows through a wire in 3 minutes. Find the current.

t = 3 × 60 = 180 s  →  I = 90 ÷ 180 = 0.5 A

Current is the same everywhere in a series circuit. In a parallel circuit, current splits at each junction — the sum of branch currents equals the total current entering the junction.

2. Voltage (Potential Difference)

Potential difference (p.d. / voltage)

The energy transferred per unit charge between two points in a circuit. Measured in volts (V). 1 V = 1 J/C.

In a series circuit, voltages across components add up to the supply voltage. In a parallel circuit, all branches have the same voltage as the supply.

Voltmeter connection

A voltmeter is always connected in parallel across the component being measured. An ammeter is always connected in series.

3. Resistance and Ohm's Law

Resistance (R)

The opposition to current flow. Measured in ohms (Ω).

V = IR  (Ohm's Law)V = voltage (V) · I = current (A) · R = resistance (Ω)
Worked example

A 12 V battery drives a current of 4 A through a resistor. Find the resistance.

R = V ÷ I = 12 ÷ 4 = 3 Ω

Factors affecting resistance of a wire

FactorEffect on resistance
Length ↑Resistance ↑ (proportional)
Cross-sectional area ↑Resistance ↓
Temperature ↑ (for metals)Resistance ↑
Material (resistivity)Different materials have different resistances

Ohmic vs non-ohmic conductors

An ohmic conductor obeys Ohm's law — V/I is constant at constant temperature (e.g. metal wire). A non-ohmic conductor does not have a constant V/I ratio (e.g. filament lamp: resistance increases as it heats up; diode: only conducts in one direction).

4. Series and Parallel Circuits

SeriesParallel
CurrentSame throughout: I_total = I₁ = I₂Splits: I_total = I₁ + I₂
VoltageSplits: V_total = V₁ + V₂Same across each branch: V_total = V₁ = V₂
ResistanceR_total = R₁ + R₂ + …1/R_total = 1/R₁ + 1/R₂ + …
Fault effectOne break stops all componentsOther branches continue to work
Worked example — parallel resistance

Two resistors of 6 Ω and 3 Ω are connected in parallel. Find the combined resistance.

1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2

R = 2 Ω

Note: the combined parallel resistance is always less than the smallest individual resistor.

5. Electrical Power and Energy

P = IV = I²R = V²/RP = power (W) · I = current (A) · V = voltage (V) · R = resistance (Ω)
E = Pt = IVtE = electrical energy (J) · P = power (W) · t = time (s)
Worked example

A 230 V kettle draws 8 A. Find (a) the power, (b) the energy used in 3 minutes.

(a) P = IV = 8 × 230 = 1840 W

(b) E = Pt = 1840 × 180 = 331 200 J ≈ 331 kJ

6. Domestic Wiring and Safety

WireColourFunction
LiveBrownCarries the alternating high voltage (230 V in Singapore)
NeutralBlueCompletes the circuit; at ~0 V
EarthGreen/yellow stripeSafety wire — connected to metal casing of appliance

Fuse and circuit breaker

A fuse contains a thin wire that melts if current exceeds the rated value, breaking the circuit. A circuit breaker does the same using an electromagnet or bimetallic strip, but can be reset. Both protect against overheating and fire.

Why appliances are earthed

If a fault causes the live wire to touch the metal casing, the earth wire provides a low-resistance path. A large current flows → the fuse blows → circuit breaks → user is protected from electric shock.

Key Electricity Equations
I = Q/t  |  V = IR  |  P = IV = I²R = V²/R  |  E = Pt
I in amps, V in volts, R in ohms, P in watts, t in seconds, Q in coulombs
Must-Know for Exam

7. Common Exam Traps

Trap 1 — Parallel resistance is always smaller

Adding resistors in parallel always reduces total resistance below the smallest individual value. If your answer is larger than the smallest resistor, you have made an error.

Trap 2 — Voltmeter vs ammeter connection

Voltmeter → parallel (across component). Ammeter → series (in line). Swapping them will give wrong readings or damage the meter.

Trap 3 — P = IV vs P = I²R

Use P = IV when both I and V are given. Use P = I²R when I and R are given. Use P = V²/R when V and R are given. Memorise all three forms.

Trap 4 — Conventional vs electron current direction

Conventional current flows from + to −. Electrons flow from − to +. In most circuit questions, use conventional current direction unless specifically asked about electron flow.

Key Terms — Flashcard Review

Tap each card to reveal the definition.

Current
Rate of flow of charge. I = Q/t. Unit: ampere (A). In circuits: conventional current flows from + to -.
Voltage (p.d.)
Energy transferred per unit charge. V = W/Q. Unit: volt (V). Measured with voltmeter in PARALLEL.
Resistance
Opposition to current flow. R = V/I (Ohm's Law). Unit: ohm. Increases with temperature for metals.
Series circuit
Components in single loop. Same current throughout. Voltages add up. Total R = R1+R2+...
Parallel circuit
Components in separate branches. Same voltage across each. Currents add up. 1/R = 1/R1+1/R2+...
Electrical power
P = IV = I2R = V2/R. Unit: watt (W). Energy per unit time transferred by electric current.

🎯 Practice Quiz — Test Yourself

8 O Level-style questions on this topic. Select an answer to see instant feedback.

Question 1 of 8
In a series circuit, current through each component is:
Explanation: Series circuit: one path → same current flows through every component.
Question 2 of 8
If voltage doubles and resistance stays the same (V = IR):
Explanation: I = V/R. Double V, same R → double I.
Question 3 of 8
Two 6 Ω resistors in parallel. Combined resistance:
Explanation: 1/R = 1/6 + 1/6 = 2/6 → R = 3 Ω. Parallel always gives lower resistance than either resistor alone.
Question 4 of 8
A 60 W bulb at 240 V. Current through it:
Explanation: P = VI → I = P/V = 60/240 = 0.25 A.
Question 5 of 8
A fuse protects against:
Explanation: Fuse melts when current exceeds its rating → breaks circuit → prevents wires overheating → prevents fire.
Question 6 of 8
Two resistors of 6 ohm and 3 ohm are connected in parallel. The total resistance is:
Explanation: 1/R = 1/R1 + 1/R2 = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2. So R = 2 ohm. Total resistance in parallel is ALWAYS less than the smallest individual resistance. Here the smallest is 3 ohm, so total must be less than 3 ohm.
Question 7 of 8
A 230 V appliance draws a current of 4 A. Its power is:
Explanation: P = IV = 4 x 230 = 920 W. Alternatively P = V2/R = 2302/57.5 = 920 W. This is a high-power appliance (nearly 1 kW), typical of a kettle or hairdryer.
Question 8 of 8
In a series circuit with two resistors, adding a third resistor in series will:
Explanation: In series, R_total = R1 + R2 + R3. Adding a resistor always increases total resistance. By V = IR (V constant from battery), higher R means lower total current. This current is the same through all components in series.
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Original study notes for Singapore students. Not affiliated with MOE, SEAB or Cambridge.