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 series
- 3 things to know
- Component 2
Overview
Recall questions on the rule that resistances add, its derivation from the shared current and added potential differences, worked calculations, why the total always exceeds the largest resistor, and how the potential difference divides between them.
What you need to know
Answer each one before you reveal it.
Why is the combined resistance of a series chain always greater than the largest resistance in it?
For a 4.0 Ω and a 6.0 Ω resistor in series across 12 V, calculate the potential difference across each.
Questions from this subtopic
Each one with the model answer underneath.
- 1
State the equation for the combined resistance of resistors connected in series.
Model answer
The combined resistance is R = R₁ + R₂ + R₃ + …, the simple sum of the individual resistances.
- 2
Derive the series resistance rule from the potential difference and current rules.
Model answer
The same current I passes through each resistor, and the supply potential difference is V = IR₁ + IR₂ + … . Dividing throughout by I gives R = R₁ + R₂ + … .
- 3
Why is the combined resistance of a series chain always greater than the largest resistance in it?
Model answer
Each resistor adds its own opposition to the same single path, so the total can only exceed any one contribution.
- 4
Calculate the combined resistance of a 4.0 Ω and a 6.0 Ω resistor in series.
Model answer
The combined resistance is 4.0 + 6.0 = 10 Ω. Subtracting 6.0 Ω from 10 Ω returns the 4.0 Ω resistor.
- 5
A 4.0 Ω and a 6.0 Ω resistor in series are connected across 12 V. Calculate the current.
Model answer
The current is I = 12 ÷ 10 = 1.2 A. Multiplying 1.2 A by 10 Ω returns the 12 V supply.
- 6
For a 4.0 Ω and a 6.0 Ω resistor in series across 12 V, calculate the potential difference across each.
Model answer
The current is 1.2 A, so the potential differences are 1.2 × 4.0 = 4.8 V and 1.2 × 6.0 = 7.2 V. Those add to the 12 V supply.
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