## Factors affecting electrical heating

Energy dissipated by current or work done as current flows depends on,- Current
- Resistance
- Time

**E = I**

^{2}R t, E = I V t or E = V^{2}t / R**Example 1**

An iron box has a resistance coil of 30 Ω and takes a current of 10 A. Calculate the heat in kJ developed in 1 minute.

*Solution*

E = I^{2} R t = 10^{2} × 30 × 60 = 18 × 10^{4} = 180 kJ

**Example 2**

A heating coil providing 3,600 J/min is required when the p.d across it is 24 V. Calculate the length of the wire making the coil given that its cross-sectional area is 1 × 10

^{-7}m

^{2}and resistivity 1 × 10

^{-6}Ω m.

*Solution*

E = P t hence P = E / t = 3,600 / 60 = 60 W

P = V^{2}/ R therefore R = (24 × 24) / 60 = 9.6 &ohms;

R = ρI/ A, I = (RA) / ρ = (9.6 × 1 × 10^{-7}) / 1 × 10^{-6} = 0.96 m

### Electrical energy and power

In summary, electrical power consumed by an electrical appliance is given by;**P = V I**

**P = I**

^{2}R**P = V**

^{2}/ RThe SI unit for power is the watt

**(W)**

**1 W = 1 J/s and 1kW = 1,000 W.**

**Example 3**

What is the maximum number of 100 W bulbs which can be safely run from a 240 V source supplying a current of 5 A?

*Solution*

Let the maximum number of bulbs be 'n'. Then 240 × 5 = 100 n

So 'n' = (240 × 5) / 100 = 12 bulbs.

**Example 4**

An electric light bulb has a filament of resistance 470 Ω. The leads connecting the bulb to the 240 V mains have a total resistance of 10 Ω. Find the power dissipated in the bulb and in the leads.

*Solution*

R_{eq} = 470 + 10 = 480 Ω, therefore I = 240 / 480 = 0.5 A.

Hence power dissipated = I^{2} R = (0.5)^{2} × 470 = 117.5 W (bulb alone)

For the leads alone, R = 10 Ω and I = 0.5 A

Therefore power dissipated = (0.5)^{2} × 10 = 2.5 W.

### Applications of heating of electrical current

**1. Filament lamp**– the filament is made up of tungsten, a metal with high melting point (3.400 0C). It is enclosed in a glass bulb with air removed and argon or nitrogen injected to avoid oxidation. This extends the life of the filament.

**2. Fluorescent lamps**– when the lamp is switched on, the mercury vapour emits ultra violet radiation making the powder in the tube fluoresce i.e. emit light. Different powders emit different colours.

**3. Electrical heating**– electrical fires, cookers e.tc. their elements are made up nichrome ( alloy of nickel and chromium) which is not oxidized easily when it turns red hot.

**4. Fuse**– this is a short length of wire of a material with low melting point (often thinned copper) which melts when current through it exceeds a certain value. They are used to avoid overloading.