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
Phasor addition in series a.c. circuits
- 3 things to know
- A2 Unit 4 Section B
Overview
Recall questions on Phasor addition in series a.c. circuits.
What you need to know
Answer each one before you reveal it.
A series RL circuit has R = 40 Ω and L = 0.15 H and is driven at 50 Hz. Calculate its impedance.
In a series circuit containing a resistor and an inductor, state the direction of each potential difference phasor relative to the current.
Questions from this subtopic
Each one with the model answer underneath.
- 1
Why is the current chosen as the reference phasor when a series a.c. circuit is analysed?
Model answer
The same current passes through every component of a series circuit, so it is the one quantity common to them all. Drawing it along the reference direction lets each component's potential difference be drawn at its own angle to it.
- 2
In a series circuit containing a resistor and an inductor, state the direction of each potential difference phasor relative to the current.
Model answer
The potential difference across the resistor is drawn along the current direction, in phase with it. The potential difference across the inductor is drawn 90° ahead of the current.
- 3
A series RL circuit has 12 V across the resistor and 5.0 V across the inductor. Calculate the supply potential difference.
Model answer
The two phasors are at right angles, so V = √(12² + 5.0²) = √169 = 13.0 V. Substituting back, 13.0² − 12² = 25, and the square root of 25 is the 5.0 V across the inductor.
- 4
For a series RL circuit with 12 V across the resistor and 5.0 V across the inductor, calculate the phase angle and state which quantity leads.
Model answer
tan φ = 5.0 ÷ 12 = 0.4167, so φ = 22.6°. The supply potential difference leads the current by that angle.
- 5
A series RC circuit has 8.0 V across the resistor and 6.0 V across the capacitor. Calculate the supply potential difference and the phase angle.
Model answer
V = √(8.0² + 6.0²) = √100 = 10.0 V, and tan φ = 6.0 ÷ 8.0 = 0.75 so φ = 36.9°. The current leads the supply potential difference by 36.9°.
- 6
Explain how the potential differences across two components in series can add up to more than the supply potential difference.
Model answer
The two potential differences reach their maxima at different instants, so their peaks never coincide. The supply potential difference is the phasor sum rather than the arithmetic sum, and that sum is smaller.
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