The power company uses a unit called kilowatt hour (kWh) which is the energy transformed by a kW appliance in one hour. 1 kW = 1,000 W × 60 × 60 seconds = 3,600,000 J. The meter used
for measuring electrical energy uses the kWh as the unit and is known as joule meter.

#### Examples

1. An electric kettle is rated at 2,500 W and uses a voltage of 240 V.

a) If electricity costs Ksh 1.10 per kWh, what is the cost of running it for 6 hrs?

b) What would be its rate of dissipating energy if the mains voltage was dropped to 120 V?

**Solution**
a) Energy transformed in 6 hrs = 2.5 × 6 = 15 kWh. Cost = 15 × 1.10 × 6 = Ksh 99.00

b) Power = V

^{2}/ R = 2500. R = (240 × 240) /2500 = 23.04 Ω.

Current = V / R = (240 × 2500) / (240 × 240) = 10.42 A

Power = V I = (2500 × 120) / 240 = 1,250 W.

2. An electric heater is made of a wire of resistance 100 Ω connected to a 240 V mains
supply. Determine the;

a) Power rating of the heater

b) Current flowing in the circuit

c) Time taken for the heater to raise the temperature of 200 g of water from 23°C to95°C. (specific heat capacity of water = 4,200 J Kg^{-1}K^{-1})

d) Cost of using the heater for two hours a day for 30 days if the power company charges Ksh 5.00 per kWh.

**Solution**
a) Power = V

^{2}/ R = (240 × 240) / 100 = 576 W

b) P = V I =>> I = P / V = 576 / 240 = 2.4 A

c) P × t = heat supplied = (m c θ) = 576 × t = 0.2 × 4200 × 72.

Hence t = (0.2 × 4200 × 72) / 576 = 105 seconds.

d) Cost = kWh × cost per unit = (0.576 × 2 × 30) × 5.0 = Ksh 172.80

3. A house has five rooms each with a 60 W, 240 V bulb. If the bulbs are switched on fro 7.00 pm to 10.30 pm, calculate the;

a) Power consumed per day in kWh

b) Cost per week for lighting those rooms if it costs 90 cents per unit.

**Solution**
a) Power consumed by 5 bulbs = 60 × 5 = 300 W = 0.3 kWh. Time = 10.30 – 7.00 = 3½

hrs.Therefore for the time duration = 0.3 × 3½ = 1.05 kWh

b) Power consumed in 7 days = 1.05 × 7 = 7.35 kWh. Cost = 7.35 × 0.9 = Ksh 6.62