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5.1.12Chemical Energetics

Series Circuits — Cambridge IGCSE Chemistry 0620 Revision Notes & Topical Questions

5 min read 5 parts Lesson 5 of 12 Unit quiz

📚 Cambridge IGCSE Chemistry 0620 Hub / Topical Revision
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Part 1 of 5Rules for Series Circuits

5.1.12 Series Circuits
Current
  • The current is the same at every point of a series circuit, because there is only one path for the charge.
Voltage
  • The potential difference of the supply is shared between the components in series.
Resistance
  • The total resistance equals the sum of the individual resistances, so R = R₁ + R₂ + R₃ and so on.
Adding components
  • Adding more components increases the total resistance, which reduces the current for a fixed supply voltage.

Part 2 of 5Changing a Series Circuit

1
Increasing the supply voltage drives more current round the same set of components.
2
Adding a further resistor in series increases the total resistance, so the current falls.
3
If one component is removed or breaks in a series circuit, the circuit is broken and no current flows through any of the others.
4
Identical lamps in series share the supply voltage, so each lamp is dimmer than it would be if it were connected alone.

Part 3 of 5Series Calculations

Series Calculations
Missing resistance
  • Find a missing resistance by subtracting the known resistances from the total resistance.
Ohm's law
  • Use V = I × R on the whole circuit or on one component, since the current is the same everywhere.
Voltage split
  • The voltage across each resistor is proportional to its resistance, and the voltages add up to the supply voltage.
Check
  • The potential differences across all the components in series should add up to the supply potential difference.

Part 4 of 5Worked Example

Question

  • A series circuit has a 6 V supply and two resistors of 4 Ω and 8 Ω.
  • Calculate the current in the circuit and the potential difference across the 8 Ω resistor.

  1. Step 1: Total resistance R = 4 Ω + 8 Ω = 12 Ω.
  2. Step 2: Current I = V ÷ R = 6 V ÷ 12 Ω = 0.5 A.
  3. Step 3: Potential difference across the 8 Ω resistor: V = I × R = 0.5 A × 8 Ω = 4 V.
    What the symbols mean: V = potential difference (V) · I = current (A) · R = resistance (Ω)

Answer The current is 0.5 A, and the potential difference across the 8 Ω resistor is 4 V (2 V + 4 V = 6 V).

Part 5 of 5Exam Checklist & Tips

5.1.12 – Key Points for the Exam

  • Series circuit A circuit in which all components lie on a single path for the current.
  • Total resistance The sum of all the resistances in a series circuit.
  • Supply potential difference The voltage provided by the source, which is shared between components in series.
  • Ohm's law The relationship V = I × R that links potential difference, current and resistance.

Common Mistakes

  • Saying that current is used up as it passes through components; the current is the same at every point.
  • Using the parallel resistance formula for resistors in series.
  • Assuming each component receives the full supply voltage; in series the potential difference is shared.
  • Treating the total resistance of the circuit as the resistance of one component.

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