# Electrical Power

## What is Electrical Power?

** Electrical power ** is defined as the rate of using electrical energy

The unit of electrical power is watt [W].

A bulb of 60W converts 60J of electrical energy into light and heat energy per second

Formula for Electrical Power

$P=\mathrm{VI}$

P: Electrical power [W]

V: Potential difference [V]

I: Current [A]

### Example 1

A 12V battery is giving off a current of 2A to a resistor. Find the power dissipated in the resistor

Solution

$P=\mathrm{VI}$

Given that

V = e.m.f. = 12V

I = 2A

P = ?

$P=12\times 2$

Answer: $P=\mathrm{24W}$

### Example 2

A p.d. of 12V is applied across the 4Ωresistor. Find the power dissipated in the resistor.

$P=\mathrm{VI}$

Given that

V = 12V

R = 4Ω

I = ?

P = ?

Note that to find power(P) we have to find current(I) first

$R=\frac{V}{\mathrm{I}}$

$4=\frac{\mathrm{12}}{\mathrm{I}}$

$\mathrm{4I}=\mathrm{12}$

$\frac{\mathrm{4I}}{4}=\frac{\mathrm{12}}{4}$

$I=\mathrm{3A}$

Therefore we can now apply the formula for power

$P=12\times 3$

Answer: $P=\mathrm{36W}$

## Cost of electrical energy

When you use the electricity supplied by an electrical utility company, you have the electricity meter. In the meter, you can find the unit of kilowatt-hours (kWh). By using this unit, the cost of electrical energy is calculated. 1kWh is called 1 unit

### How to calculate the cost of electrical energy

1. Calculate the energy consumed in kwh.

$E=\mathrm{Pt}$

E: Energy consumed [kWh]

P: Power of electrical components [kW]

t: time taken [hr]

2. Calculate the cost of electrical energy by crossmultiplication.

#### Example 1

If electrical energy costs K50 per unit, what is the total cost by using 1000kwh?

Solution

1unit ----> K50

1000unit ----> x

$x=5\times 1000$

Answer: $x=\mathrm{k5000}$

#### Example 2

A light bulb of 100W is used for 7hours. What is the energy cost if the energy costs K50 per unit?

Solution

First you have the energy of electricity

$E=\mathrm{Pt}$

Given that

P = 100W = 0.1kW

t = 7hrs

E = ?

$E=\mathrm{0.1}7$

$E=\mathrm{0.7kWh}$

Therefore equate the energy of electricity (units) to the amount spent in order to find the cost

1unit ----> K50

0.7unit ----> x

$x=50\times 0.7$

Answer: $x=\mathrm{K35}$

#### Example 3

4-security lights of 120W are turned on for 30days. What is the energy cost if it costs K60 per unit?

Solution

Convert 30 days to hours and find the total power for the 4 lights

Given that

P = 120W×4 = 480W = 0.48kW

t = 30days = 30days×24hrs = 720hrs

E = ?

$E=\mathrm{0.48}\times 720$

$E=\mathrm{345.60kWh}$ or E = 345.60 units

Hence we can then equate the units found to the cost

1unit ----> K60

345.60unit ----> x

$x=60\times 345.60$

Answer: $x=\mathrm{K20736}$

## Use of electricity in the house

When we use an electrical appliance in our house, a plug of the appliance is connected to a soket. Zambia uses three pin plugs. (Japan uses two pin plugs.)

- Live wire is a brown wire. It supplies the electrical energy to the appliance. The line has a high voltage. If you touch this line, it is dangerous because the current flows through your body. You may die of electric shock.

- Neutral wire is a blue wire. It makes the circuit complete in the electrical appliance.

- Earth wire is a green or yellow wire. This wire is connected to the metal casing of an electrical appliance. If live wire is in contact with the metal casing due to any accident, the user gets the electric shock from the metal casing. If earth wire is connected to the metal casing, the current eacapes from the earth wire. It protects the user from the electric shock.

## Dangers of electricity

Contacting electricity (especially the live wire) is dengerous and causes some accidents.

- It causes the electric shock to human beings. A large current can be fatal (die).

- It may cause fires or burns in an electrical appliance, the plug and the socket.

### Dangers of electricity can be caused by three cases shown below

#### Damaged insulation

The electrical wires (cables) are insulated. If those insulators are removed by the deterioration, the live wire can be contacted to somewhere and it can cause electric shock and fire.

#### Overheating of cables

If a large current flows in the wires or components, it can cause overheating. Then it can melt the insulation and start a fire. A short circuit or overloading is easy to cause this accident.

#### Damp condition

In damp condition such as a wet bathroom, the current flows through the human body easily. Because the body’s resistance depends on whether the skin is wet or dry.

### Safe use of electricity in the house

To use the electricity safely, there are some electrical components. They are shown below.

- Earth wire

- Some electrical appliances are ** double insulated **. It makes the leakage of current difficult.

- The function of ** switch ** is to turn on or off the electrical appliance. In the case of leakage, the switch can be used as the
safety device to cut off the current. The switch should be installed on the live wire so that the electrical appliance is
disconnected from high voltage when the switch is open.

If too much current flows through an electrical component, the component can overheat or start a fire. The ** fuse **
prevents too much current from flowing through it. If too much current flows through a fuse, a wire in the fuse melts and it intercepts too much current from live wire. Therefore it is installed on the live wire.

### Fuse rating

The ** fuse rating ** is the maximum current that the fuse can carry without melting. We should choose a proper fuse rating.

- If we choose a large fuse rating, it allows too much current to flow.

- If we choose a small fuse rating, the electrical appliance doesn’t work.

- The fuse rating should be slightly larger than the working current of an appliance under normal operation.

- Available fuse ratings are 3A, 5A, 13A, 15A or 30A.

#### Example

A refrigerator is rated at 240V 480W. Which fuse should be used, 3A or 13A?

Solution

$P=\mathrm{VI}$

V = 240V

P = 480W

I =?

$\mathrm{480}=\mathrm{240I}$

$\frac{\mathrm{240I}}{240}=\frac{\mathrm{480}}{240}$

$I=\mathrm{2A}$

Answer: 3A is a proper fuse rating.