WJEC A-Level Physics 1420QS
Basic Physics
Kinematics
Dynamics
Energy Concepts
Solids Under Stress
Using Radiation to Investigate Stars
Particles and Nuclear Structure
Conduction of Electricity
Resistance
D.C. Circuits
The Nature of Waves
Wave Properties
Refraction of Light
Photons
Lasers
Circular Motion
Vibrations
Kinetic Theory
Thermal Physics
Nuclear Decay
Nuclear Energy
Capacitance
Electrostatic and Gravitational Fields of Force
Orbits and the Wider Universe
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)
Option A: Alternating Currents
Rms values, mean power and UK mains
- 3 things to know
- A2 Unit 4 Section B
Overview
Converting between peak and rms current and voltage, mean power in a resistor from rms and peak values, and the frequency, rms and peak voltage of the UK mains.
What you need to know
Answer each one before you reveal it.
Calculate the rms current when peak current = 7.5 A.
State the equation relating rms voltage to peak voltage.
Questions from this subtopic
Each one with the model answer underneath.
- 1
State the equation relating rms current to peak current.
Model answer
Iᵣₘₛ = I₀/√2.
- 2
Calculate the rms current when peak current = 7.5 A.
Model answer
Rms current = 7.5 ÷ 1.41421 = 5.3 A.
- 3
State the equation relating rms voltage to peak voltage.
Model answer
Vᵣₘₛ = V₀/√2.
- 4
Calculate the rms voltage when peak voltage = 187 V.
Model answer
Rms voltage = 187 ÷ 1.41421 = 132 V.
- 5
State the equations for mean power dissipated in a resistor using rms values.
Model answer
P = Iᵣₘₛ²R or P = Vᵣₘₛ²/R.
- 6
Why can the standard DC power equations be used directly with rms values?
Model answer
Because rms values are defined as the equivalent DC values that dissipate the same power in a resistor.
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