Energy transfers in everyday appliances
Every electrical appliance transfers energy. Two key equations let you calculate the rate of transfer (power) and total transferred (energy).
Power equations
$P = VI$ (power = potential difference × current)
$P = I^2 R$ (power = current² × resistance)
- $P$ in watts (W).
- $V$ in volts.
- $I$ in amperes.
- $R$ in ohms.
The first form is the most general; the second comes from substituting $V = IR$.
Energy equations
Energy transferred = power × time:
$E = Pt$
Or directly from charge and p.d.:
$E = QV$ (which combined with $Q = It$ gives $E = VIt = Pt$, all consistent.)
- $E$ in joules (J).
- $Q$ in coulombs.
- $t$ in seconds.
✦Worked example— Worked example 1
A 2.0 kW kettle is plugged into the 230 V mains. Find the current.
- $P = VI \Rightarrow I = P/V = 2000/230 \approx 8.7$ A.
✦Worked example— Worked example 2
A 60 W lamp runs for 5 minutes. Find the energy transferred.
- $t = 300$ s.
- $E = Pt = 60 \times 300 = 18000$ J = 18 kJ.
✦Worked example— Worked example 3
A 0.50 A current passes through a 100 Ω resistor for 2.0 minutes. Energy?
- $P = I^2 R = 0.25 \times 100 = 25$ W.
- $E = Pt = 25 \times 120 = 3000$ J.
Why a kettle has a thick cable
High-power appliances draw large currents. By $P = I^2 R$, even a small cable resistance dissipates significant heat at high I. A thicker cable has lower R, reducing wasted heating and the risk of fire.
Cost of running an appliance
Energy companies charge in kilowatt-hours (kWh), where 1 kWh = 3.6 × 10⁶ J.
- Energy in kWh = power (kW) × time (h).
- Cost = energy (kWh) × price per kWh.
A 2 kW heater running 3 hours uses 6 kWh. At 30 p/kWh, that's £1.80.
Efficiency reminder
Power input ≠ useful power out. Efficiency = (useful output power)/(total input power). A 60 W lamp may convert only 5 W to light and 55 W to heat.
⚠Common mistakes
- Using P = I²R when only V is given — convert first.
- Forgetting kW → W (multiply by 1000) or hours → seconds (×3600) before substituting.
- Confusing energy (J) and power (W) — they aren't the same.
- Adding currents on a multi-appliance plug instead of summing powers (P = sum of P; I_total = sum of I assuming same V).
➜Try this— Quick check
A 1.2 kW microwave runs for 90 seconds. Energy?
- P = 1200 W.
- E = Pt = 1200 × 90 = 108 000 J ≈ 108 kJ.
AI-generated · claude-opus-4-7 · v3-deep-physics