**HEATING EFFECT OF AN ELECTRIC CURRENT **

When current flows, electrical energy is transformed into other forms of energy i.e. light, mechanical and chemical changes.

**Factors affecting electrical heating**

Energy dissipated by current or work done as current flows depends on,

*Current**Resistance**Time*

This formula summarizes these factors as, *E = I ^{2} R t, E = I V t or E = V^{2} t / R*

*Examples *

*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 *

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

*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 *

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

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

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

HEATING EFFECT OF AN ELECTRIC CURRENT

**
**

**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} / R **

The SI unit for power is the watt **(W)**

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

*Examples *

*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.

*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.

HEATING EFFECT OF AN ELECTRIC CURRENT

**Applications of heating of electrical current**

- Filament lamp – the filament is made up of tungsten, a metal with high melting point (3.400
^{0}C). 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. - 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.
- 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.
- 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.

ALL PHYSICS NOTES FORM 1-4 WITH TOPICAL QUESTIONS & ANSWERS

PRIMARY NOTES, SCHEMES OF WORK AND EXAMINATIONS