Eduqas A-Level Physics A420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Conduction of Electricity
Resistance
D.C. Circuits
Capacitance
Solids Under Stress
Electrostatic and Gravitational Fields of Force
Using Radiation to Investigate Stars
Orbits and the Wider Universe
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Nuclear Decay
Particles and Nuclear Structure
Nuclear Energy
Magnetic Fields
Electromagnetic Induction
Option A: Alternating CurrentsOption A
Option B: Medical PhysicsOption B
Option C: The Physics of SportsOption C
Option D: Energy and the EnvironmentOption D
Mathematical Requirements (Appendix C)
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.
Derive the parallel resistance rule from the current and potential difference rules.
Give the shortened form of the parallel rule for exactly two resistors.
Questions from this subtopic
Each one with the model answer underneath.
- 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
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
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
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
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
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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