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)
Resistance
Electrical power and energy transfer
- 3 things to know
- Component 2
Overview
Recall questions on P = IV and the derived forms I²R and V²/R, choosing the right form for the data given, worked power and energy calculations, the kilowatt hour, and why the power dissipated quadruples when the current doubles.
What you need to know
Answer each one before you reveal it.
Show how P = IV follows from the definitions of current and potential difference.
State the two forms of the power equation that use resistance.
Questions from this subtopic
Each one with the model answer underneath.
- 1
State the equation for the power dissipated in a component in terms of current and potential difference.
Model answer
The power is P = IV, where P is in watts (W), I is the current in amperes (A) and V is the potential difference in volts (V).
- 2
Show how P = IV follows from the definitions of current and potential difference.
Model answer
Power is energy per unit time, W/t, and W = QV, so P = QV/t. Since Q/t is the current, P = IV.
- 3
State the two forms of the power equation that use resistance.
Model answer
The two forms are P = I²R and P = V²/R.
- 4
Derive P = I²R from P = IV.
Model answer
Substituting V = IR into P = IV gives P = I × IR, so P = I²R.
- 5
Derive P = V²/R from P = IV.
Model answer
Substituting I = V/R into P = IV gives P = (V/R) × V, so P = V²/R.
- 6
A 6.0 Ω resistor is connected across a 12 V supply. Calculate the power dissipated in it.
Model answer
The power is P = V²/R = 144 ÷ 6.0 = 24 W. The current is 2.0 A, and I²R = 4.0 × 6.0 gives the same 24 W.
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