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P4 · 4.4Paper 1✅ Yr10

Electric Circuits

Current, voltage, resistance, series and parallel circuits, component characteristics. IV characteristics appear every year.

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Key facts

Key definitions and equations — fill in the table

TermMeaningUnit
Current (I)
Potential difference / voltage (V)
Resistance (R)
Ohm's law
Charge (Q)

Series and parallel circuits — fill in the table

Circuit typeCurrentVoltageTotal resistance
Series
Parallel

Component IV characteristics — fill in the table

ComponentBehaviour
Ohmic resistor
Filament lamp
Diode
Thermistor (NTC)
LDR (light-dependent resistor)

Power equations

  • P = IV (power = current × voltage).
  • P = I²R.
  • P = V²/R.
  • Energy = Power × time (E = Pt).
  • Units: power in watts (W), energy in joules (J), time in seconds (s).
Exam questions — 3 questions · 10 marks
⚡ Extended questions included. Show all working for calculation questions. The AI marks against the real AQA mark scheme.
1 markOhm's LawAQA 8463 P1 style

A resistor has a potential difference of 12 V across it and a current of 3 A through it. What is its resistance?

A 36 Ω
B 4 Ω
C 0.25 Ω
D 9 Ω
1 markIV characteristicAQA 8463 P1 style

The IV graph for a component shows a curve that increases steeply in one direction and stays close to zero in the other direction. What is the component?

A Ohmic resistor
B Filament lamp
C Diode
D Thermistor
4 marksSeries vs parallel analysisAQA 8463 P1 style

Two identical 6 Ω resistors are connected (a) in series and (b) in parallel to a 12 V battery. For each, calculate the total resistance and the total current from the battery. (4 marks)

Hint: Series: R_total = R1 + R2. Parallel: 1/R_total = 1/R1 + 1/R2 (or use product/sum for two resistors). Then I = V/R.
+40 XP
3 marksPower calculationAQA 8463 P1 style

A kettle operates at 230 V and draws a current of 9 A. Calculate (a) the power of the kettle and (b) the energy transferred in 3 minutes. (3 marks)

Hint: P = IV. Then E = Pt (convert minutes to seconds first).
+30 XP
Quick recall flashcards

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