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Physics

Electrical Power and Energy

PDF
Matthew Williams
|May 20, 2026|4 min read
CSEC PhysicsEfficiencyElectrical EnergyElectrical PowerKilowatt-hourPaper 01Paper 02Section D

Electrical power (P = IV = I²R = V²/R), energy (E = Pt), the kilowatt-hour as a unit of energy, efficiency of electrical appliances, and comparing LED, fluorescent, and incandescent lamps.

Electrical Power

Power is the rate at which electrical energy is transferred. For a component with current III and potential difference VVV:

P=IVP = IVP=IV

Using V=IRV = IRV=IR and I=V/RI = V/RI=V/R, two additional forms follow:

P=I2RP=V2RP = I^2 R \qquad P = \frac{V^2}{R}P=I2RP=RV2​

Unit: watt (W), where 1 W = 1 J s⁻¹.

A 60 W lamp converts 60 J of electrical energy to light and heat every second.

Electrical Energy

Energy is power multiplied by time:

E=PtE = PtE=Pt

where EEE is in joules (J), PPP is in watts (W), and ttt is in seconds (s).

In practice, the kilowatt-hour (kWh) is used for billing electricity:

1kWh=1000W×3600s=3.6×106J=3.6MJ1 \text{kWh} = 1000 \text{W} \times 3600 \text{s} = 3.6 \times 10^6 \text{J} = 3.6 \text{MJ}1kWh=1000W×3600s=3.6×106J=3.6MJ

Example/Electrical energy and cost

A 2000 W electric kettle is used for 5 minutes. The electricity costs $0.30 per kWh.

Energy used in joules:

E=Pt=2000×(5×60)=600 000JE = Pt = 2000 \times (5 \times 60) = 600\,000 \text{J}E=Pt=2000×(5×60)=600000J

Energy used in kWh:

E=2×560=1060≈0.167kWhE = \frac{2 \times 5}{60} = \frac{10}{60} \approx 0.167 \text{kWh}E=602×5​=6010​≈0.167kWh

Cost:

Cost=0.167×$0.30≈$0.05\text{Cost} = 0.167 \times \$0.30 \approx \$0.05Cost=0.167×$0.30≈$0.05

Efficiency of Electrical Appliances

All appliances waste some energy, usually as heat. Efficiency is expressed as a percentage:

efficiency=useful power outputtotal power input×100%\text{efficiency} = \frac{\text{useful power output}}{\text{total power input}} \times 100\%efficiency=total power inputuseful power output​×100%

The LED lamp is more efficient than incandescent or fluorescent alternatives:

Lamp typeApproximate efficiencyNotes
Incandescent5%95% of energy wasted as heat
Compact fluorescent (CFL)15-25%Better than incandescent but contains mercury
LED30-50%Best efficiency; long lifespan; no mercury

In the Caribbean, where electricity is expensive (imported fuel), LEDs offer significant cost savings.

Using the Power Formulas

Example/Power formulas applied

A resistor of 10 Ω carries a current of 2 A.

Power dissipated:

P=I2R=22×10=4×10=40WP = I^2 R = 2^2 \times 10 = 4 \times 10 = 40 \text{W}P=I2R=22×10=4×10=40W

Voltage across the resistor:

V=IR=2×10=20VV = IR = 2 \times 10 = 20 \text{V}V=IR=2×10=20V

Check: P=IV=2×20=40P = IV = 2 \times 20 = 40P=IV=2×20=40 W. ✓

Exam Tip

Three power formulas exist, choose the one that uses the quantities given:

  • P=IVP = IVP=IV: when both current and voltage are known.
  • P=I2RP = I^2RP=I2R: when current and resistance are known (saves computing V first).
  • P=V2/RP = V^2/RP=V2/R: when voltage and resistance are known (saves computing I first).

When converting between J and kWh: 1 kWh = 3.6 × 10⁶ J. Divide joules by 3.6 × 10⁶ to get kWh; multiply kWh by 3.6 × 10⁶ to get joules.

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Electric Circuits
Next in syllabus order
Domestic Electricity