TopMyGrade

GCSE/Combined Science/OCR

P3.3Series and parallel circuits: rules for current, p.d. and resistance

Notes

Series and parallel circuits (P3.3)

Gateway A loves a circuit-rules question with mid-circuit diagrams. Get the four rules in your head and you can answer almost anything.

Symbols you must know

SymbolComponent
────Wire / conductor
Two parallel lines (one short, one long)Cell
──[──]Resistor
──(L)──Lamp
──(A)──Ammeter (placed in series, low resistance)
──(V)──Voltmeter (placed in parallel, high resistance)
Saw-toothVariable resistor
⏞⏟Switch
(T)Thermistor
(D)LDR (light-dependent resistor)
Diode

Series circuits — the four rules

In a series circuit, all components are on the same loop; current has only one path.

  1. Current is the same everywhere. I_total = I₁ = I₂ = I₃.
  2. Total p.d. = sum of p.d.s. V_total = V₁ + V₂ + V₃.
  3. Total resistance = sum of resistances. R_total = R₁ + R₂ + R₃.
  4. Adding more components in series increases total resistance → reduces current.

Parallel circuits — the four rules

In a parallel circuit, components are on separate branches; current has multiple paths.

  1. Total current = sum of branch currents. I_total = I₁ + I₂ + I₃.
  2. P.d. is the same across each branch. V_total = V₁ = V₂ = V₃.
  3. Total resistance is LESS than the smallest branch resistance. Adding more parallel branches decreases total resistance.
  4. Each branch operates independently — break one and the others still work. (This is why house wiring is parallel.)

Why parallel resistance decreases

Imagine a single resistor of 6 Ω. Now add another 6 Ω in parallel. You now have two paths for current — the total resistance is halved. Two paths = double the cross-sectional area = half the resistance.

For two equal resistors in parallel, R_total = R / 2.

Ohm's law (recap from P3.2)

V = I × R

Where V is potential difference (V), I is current A and R is resistance (Ω).

Worked exampleWorked example — series

Three resistors of 4 Ω, 6 Ω and 10 Ω are in series with a 12 V cell.

  • R_total = 4 + 6 + 10 = 20 Ω.
  • I = V / R = 12 / 20 = 0.6 A (the same through every resistor).
  • p.d. across 4 Ω = 0.6 × 4 = 2.4 V.
  • p.d. across 6 Ω = 0.6 × 6 = 3.6 V.
  • p.d. across 10 Ω = 0.6 × 10 = 6.0 V.
  • Sum = 2.4 + 3.6 + 6.0 = 12 V ✓ (matches the cell).

Worked exampleWorked example — parallel

Two 6 Ω resistors are in parallel with a 12 V cell.

  • p.d. across each branch = 12 V.
  • I in each branch = 12 / 6 = 2 A.
  • Total current from cell = 2 + 2 = 4 A.
  • Total resistance = V / I = 12 / 4 = 3 Ω.
  • Note: 3 Ω is half of 6 Ω, as expected.

Mains wiring is parallel

Houses wire all sockets and lights in parallel because:

  • Each appliance has the full mains p.d. across it (~230 V).
  • One appliance failing or being switched off does not affect the others.
  • Different appliances draw different currents independently.

Common Gateway-paper mistakes

  1. Adding parallel resistances directly (1/R rule for parallel — easier to remember "parallel = less than the smallest").
  2. Using V = IR with the wrong V (use the V across that specific component, not the cell e.m.f.).
  3. Forgetting that ammeters are placed in series and voltmeters in parallel.
  4. Drawing voltmeters in series — that breaks the circuit.
  5. Saying "current is used up" through a resistor — current is conserved, only energy is transferred.

Try thisQuick check

Two 4 Ω resistors are in parallel, connected to a 6 V cell.

  • p.d. across each: 6 V.
  • I in each branch: 6 / 4 = 1.5 A.
  • Total current: 3 A.
  • Total resistance: V / I = 6 / 3 = 2 Ω. ✓ (Half of 4 Ω.)

AI-generated · claude-opus-4-7 · v3-ocr-combined-science

Practice questions

Try each before peeking at the worked solution.

  1. Question 15 marks

    Series totals

    Three resistors of 5 Ω, 8 Ω and 12 Ω are connected in series with a 12.5 V cell.

    (a) Calculate the total resistance. [1]
    (b) Calculate the current through the circuit. [2]
    (c) Calculate the p.d. across the 8 Ω resistor. [2]

    [5 marks]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-ocr-combined-science

  2. Question 24 marks

    Parallel totals

    Two resistors of 4 Ω and 6 Ω are in parallel with a 12 V cell.

    (a) State the p.d. across each resistor. [1]
    (b) Calculate the current through each resistor. [2]
    (c) Calculate the total current drawn from the cell. [1]

    [4 marks]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-ocr-combined-science

  3. Question 33 marks

    Identify circuit type

    A circuit contains a 6 V cell, a switch, and three lamps. When one lamp is unscrewed, the other two stay lit.

    (a) Are the lamps in series or parallel? [1]
    (b) Justify your answer in terms of how the circuit operates. [2]

    [3 marks]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-ocr-combined-science

  4. Question 44 marks

    Why house wiring is parallel

    Explain why the lights and sockets in a house are wired in parallel rather than series.

    [4 marks]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-ocr-combined-science

  5. Question 53 marks

    Predicting total resistance

    A circuit has a 10 Ω resistor in series with a parallel combination of two 12 Ω resistors.

    (a) Calculate the resistance of the parallel pair. [2]
    (b) Calculate the total resistance of the circuit. [1]

    [3 marks]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-ocr-combined-science

  6. Question 64 marks

    Ammeter and voltmeter placement

    (a) Where in a circuit must an ammeter be placed, and why? [2]
    (b) Where in a circuit must a voltmeter be placed, and why? [2]

    [4 marks]

    Ask AI about this

    AI-generated · claude-opus-4-7 · v3-ocr-combined-science

Flashcards

P3.3 — Series and parallel circuits: rules for current, p.d. and resistance

12-card SR deck for OCR Combined Science (J250) topic P3.3

12 cards · spaced repetition (SM-2)