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D.C. Circuits

Resistors in parallel

  • 3 things to know
  • Component 2

Overview

Recall questions on the reciprocal rule and its derivation from the shared potential difference and added currents, worked calculations, why the total is always less than the smallest resistor, and how the current divides between the branches.

What you need to know

Answer each one before you reveal it.

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Key term

State the equation for the combined resistance of resistors connected in parallel.

Key term

Derive the parallel resistance rule from the current and potential difference rules.

Key term

Give the shortened form of the parallel rule for exactly two resistors.

Questions from this subtopic

Each one with the model answer underneath.

  1. 1

    State the equation for the combined resistance of resistors connected in parallel.

    Model answer

    The combined resistance is given by 1/R = 1/R₁ + 1/R₂ + 1/R₃ + … .

  2. 2

    Derive the parallel resistance rule from the current and potential difference rules.

    Model answer

    All the branches have the same potential difference V, and the total current is I = V/R₁ + V/R₂ + … . Since the total current is also V/R, dividing throughout by V gives 1/R = 1/R₁ + 1/R₂ + … .

  3. 3

    Give the shortened form of the parallel rule for exactly two resistors.

    Model answer

    For two resistors the combined resistance is R = R₁R₂/(R₁ + R₂).

  4. 4

    Why is the combined resistance of a parallel network always smaller than the smallest branch resistance?

    Model answer

    Adding a branch provides an extra path for charge, so more current flows for the same potential difference and the effective resistance of the whole network falls.

  5. 5

    Calculate the combined resistance of a 4.0 Ω and a 6.0 Ω resistor in parallel.

    Model answer

    The combined resistance is (4.0 × 6.0) ÷ (4.0 + 6.0) = 2.4 Ω. Substituting back, 1/2.4 = 0.4167 = 1/4.0 + 1/6.0.

  6. 6

    A 4.0 Ω and a 6.0 Ω resistor in parallel are connected across 12 V. Calculate the current in each branch and the total current.

    Model answer

    The branch currents are 12 ÷ 4.0 = 3.0 A and 12 ÷ 6.0 = 2.0 A, giving a total of 5.0 A. Dividing 12 V by the combined 2.4 Ω also gives 5.0 A.

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