Electric Power: Revision Notes (Physics C: E&M 11.4)

Short on time? This page is the recap. For the derivation, Figures 1–3 and the worked examples, use the full study guide.

Recap

Power is the rate of energy transfer. For charge dq crossing a potential difference ΔV, dU = ΔV dq, so P = IΔV.
1 W = 1 J/s = 1 V·A. 1 kWh = 3.6 × 10⁶ J is a unit of energy.
Sources (batteries, generators) deliver power; resistors, bulbs and motors receive it. Total delivered = total received.
A resistor dissipates electrical energy as thermal energy at the rate I²R.
A bulb’s brightness increases with its power, so compare powers to compare bulbs.
A motor converts electrical energy to mechanical energy, with some thermal loss: IΔV = P_mech + P_thermal.

Situation	Relationship	Note
Any element	P = IΔV	Includes motors and batteries
Resistor	P = I²R = ΔV²/R	Uses ΔV = IR
Same current (series)	P ∝ R	Larger R, more power
Same potential difference (parallel)	P ∝ 1/R	Smaller R, more power
Fixed resistor	P ∝ ΔV², P ∝ I²	Double ΔV → 4P
Ideal battery, emf ℰ	P = ℰI	Power delivered
Lifting at constant speed	P_mech = mgv	g = 9.8 m/s²
Power varies with time	E = ∫P dt	Area under P–t graph

Resistors and bulbs are ohmic unless stated, so R does not change as they warm up.
Batteries are ideal (no internal resistance) unless stated.
Connecting wires dissipate negligible power.

“More resistance, more power” without checking what is fixed.
Saying bulbs use up current. They transfer energy; the current in and out is the same.
I²R for a motor’s input. That is only the thermal part; the input is IΔV.
Averaging I or ΔV before squaring. Integrate P dt instead.
Doubling ΔV doubles P. For a fixed R it quadruples P.
kWh as power. It is energy.

A current of 3.0 A passes through a 5.0 Ω resistor. What is the power? (45 W)
The potential difference across a fixed resistor is doubled. By what factor does the power change? (4)
How much energy does a 60 W device transfer in 2.0 minutes? (7.2 × 10³ J)