3.4 - Electrical Energy and Power

Electrical Energy

Electrical appliances convert electrical energy into other forms; a lamp produces light and heat.
From $V = E / Q$ , electrical energy is $E = V Q$ . Substitute $Q = I t$ :

E = V I t

$E$ is energy in joules, $J$ ; $V$ is potential difference in volts; $I$ is current in amperes; $t$ is time in seconds.
1 J is used when 1 A flows through a device with 1 V across it for 1 s.

P = \frac{E}{t} = \frac{V I t}{t} = V I

P = V (\frac{V}{R}) = \frac{V ^{2}}{R}, P = (I R) I = I^{2} R

An appliance’s labelled voltage is its working voltage, required for normal operation. At this voltage, it uses the power stated on its label.
A rice cooker rated 240 V, 700 W uses 700 J each second when supplied with 240 V.

t = 60 s, E = V I t = 6 (0.7) (60) = 252 J

R = \frac{V ^{2}}{P} = \frac{24 0 ^{2}}{32} = 1800 Ω

I = \frac{P}{V} = \frac{32}{240} \approx 0.13 A

t = 3 (60) (60) = 10800 s

E = Pt = 32 (10800) = 345600 J = 345.6 kJ

Consumption is measured in kilowatt-hours, kWh; 1 kWh is 1 unit of electricity.
Use power in kW and time in hours when calculating $E = Pt$ in kWh.
For several identical appliances, total daily energy equals quantity × power in kW × hours used per day. Add the energy used by all appliances for household consumption.
For a given cost per unit:

Cost = energy used in kWh \times cost per unit

Worked comparison: a 40 W CFL and 12 W LED produce equal brightness and operate for 12 hours daily for 30 days. Assume RM0.218 per unit for this calculation.

Lamp	Energy and cost
CFL	Daily: $0.04 (12) = 0.48 kWh$ . Cost: $30 (0.48) (0.218) = RM 3.139$ .
LED	Daily: $0.012 (12) = 0.144 kWh$ . Cost: $30 (0.144) (0.218) = RM 0.942$ .

For equal brightness and operating time in this comparison, the LED uses less energy and has higher efficiency.

Close doors and windows while using air conditioning; keep its filter clean for faster, more efficient cooling.
Switch off appliances when unused.
Run the washing machine with a full load.
Use energy-saving lamps.