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Electricity is used in both households and industries. The amount of electricity consumed is determined by these two factors. They are

- The amount of electricity consumed
- The duration of usage

The product of electric power and its usage time is used to calculate the electrical energy consumed.

$\mathit{Electrical}\mathit{energy}=\phantom{\rule{0.147em}{0ex}}\mathit{Electric}\phantom{\rule{0.147em}{0ex}}\mathit{power}\times \mathit{Time}$

If an electric appliance of power \(200\ watt\) is utilised for \(3\ hours\), then the total electric power consumption is

$200\times 3=\phantom{\rule{0.147em}{0ex}}600\phantom{\rule{0.147em}{0ex}}\mathit{watt}\phantom{\rule{0.147em}{0ex}}\mathit{hour}$

Electrical energy consumption is quantified and stated in watt-hours, even though the SI unit is watt-second. But in reality, a larger electrical energy unit is required. The higher units of electrical energy are watt-hour (\(W\ h\)) or kilowatt-hour (\(kW\ h\)).

$\begin{array}{l}1\mathit{watt}\mathit{hour}=\phantom{\rule{0.147em}{0ex}}1\mathit{watt}\phantom{\rule{0.147em}{0ex}}\times \phantom{\rule{0.147em}{0ex}}1\phantom{\rule{0.147em}{0ex}}\mathit{hour}\\ \\ =1\phantom{\rule{0.147em}{0ex}}W\phantom{\rule{0.147em}{0ex}}\times \phantom{\rule{0.147em}{0ex}}\left(60\times 60\right)\phantom{\rule{0.147em}{0ex}}s\\ \\ =3600Ws\end{array}$

In terms of joules,

$1\phantom{\rule{0.147em}{0ex}}\mathit{watt}\phantom{\rule{0.147em}{0ex}}\mathit{hour}\phantom{\rule{0.147em}{0ex}}=3600\phantom{\rule{0.147em}{0ex}}J$

Kilowatt is the larger unit of power that is most commonly used in the devices.

$\begin{array}{l}1\phantom{\rule{0.147em}{0ex}}\mathit{kilowatt}\phantom{\rule{0.147em}{0ex}}\mathit{hour}\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}1\phantom{\rule{0.147em}{0ex}}\mathit{kilowatt}\phantom{\rule{0.147em}{0ex}}\times \phantom{\rule{0.147em}{0ex}}1\phantom{\rule{0.147em}{0ex}}\mathit{hour}\\ \\ =1\phantom{\rule{0.147em}{0ex}}\mathit{kW}\phantom{\rule{0.147em}{0ex}}\times \phantom{\rule{0.147em}{0ex}}\left(60\times 60\right)\phantom{\rule{0.147em}{0ex}}s\\ \\ =1000\phantom{\rule{0.147em}{0ex}}W\phantom{\rule{0.147em}{0ex}}\times 3600\phantom{\rule{0.147em}{0ex}}s\\ \\ =\phantom{\rule{0.147em}{0ex}}3.6\times {10}^{6}\phantom{\rule{0.147em}{0ex}}J\end{array}$

In terms of megajoules,

$1\phantom{\rule{0.147em}{0ex}}\mathit{kW}\phantom{\rule{0.147em}{0ex}}h\phantom{\rule{0.147em}{0ex}}=\phantom{\rule{0.147em}{0ex}}3.6\phantom{\rule{0.147em}{0ex}}M\phantom{\rule{0.147em}{0ex}}J$

The energy consumed for domestic purposes is measured using the electric meter in the units of a kilowatt-hour.

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**Horsepower:**

Another unit called horsepower (\(hp\)) is also used to express electric power. It is one of the units in the foot-pound second (FPS) system or English system.

$1\mathit{horse}\mathit{power}\left(\mathit{hp}\right)=746\mathit{watt}$

**Example:**

Two bulbs have the ratings as \(60\ W\), \(220\ V\) and \(40\ W\), \(220\ V\) respectively. Which one has a greater resistance?

Electric power, $P\phantom{\rule{0.147em}{0ex}}=\frac{{V}^{2}}{R}$

For the same value of \(V\), \(R\) is inversely proportional to \(P\). Therefore, the lesser the power, the greater the resistance. Hence, the bulb with \(40\ W\), \(220\ V\) rating has greater resistance.