Heating Effect of Electric Current: Overview, Questions, Preparation

The heating effect of an electric current is widely seen in our daily lives. Electric iron, kettle, toaster, heaters, etc., are used as alternatives to conventional cooking and laundry methods. The same heating principle applies to electric bulbs, the alternative to conventional lights. The heating effect of an electric current has revolutionized the world for years.

Most household electrical equipment changes electrical energy to heat this way. These include filament lights, electric heaters, electric iron, electric kettle, etc.

In lighting equipment

Filament lights - These are made of tungsten wire covered in glass bulbs from which air has been removed. This is because the air will oxidize the filament. The filament is heated to high temperatures and turns white. Tungsten is used because of the high melting point; 34000 bulbs filled with non-active gases. This problem is minimized by the circular wire and occupies a smaller area that reduces heat loss through convection.

Neon lights - This light is more efficient compared to filament lights and lasts longer. They have mercury vapour in a glass tube that emits ultraviolet radiation when switched on. This radiation causes powder in the tube to shine (fluoresce). Different powders produce different colors. Note that the fluorescent lights are expensive to install, but their costs run far less.

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In electricity heating

Electric stoves - Electric stoves turn hot and produce heat energy, absorbed by cooking pots through conduction.

Electric heaters - Radiation heaters become red at around 900°C, and radiation emitted is directed indoor by the polished reflector.

Electric kettle: The heating element is at the bottom of the kettle, so the liquid is heated and covered. The heat is then absorbed by water and distributed to all fluids with convection.

Illustrated Examples

Electric ball labeled 100W, 240V. Calculate:

1. a) The current through the filament when the bulb functions normally.
2. b) The resistance of the filament used in the bulb.

I = p / v = 100/240 = 0.4167a

R = p / i2 = 100 / 0.41672 = 576.04Ω or r = v2 / p = 2402/100 = 576Ω

Find energy scattered in 5 minutes with an electrical bulb with a 500Ω resistance filament connected to a 240V supply. {ANS. 34.560j}

Electric iron consumes energy at the level of 840W when heating at the maximum level and 360 w when heating is at minimum. The voltage is 220 v. What is the current and Resistance in every case?

We know that power input is

P = v i

So i = p / v at this time

(A) when heating at the maximum level,

I = 840 W / 220V = 3.82 A;

and electric iron resistance is

R = v / i = 220 v / 3.82 a = 57.60 Ω.

(B) When heating at the minimum level,

I = 360 W / 220V = 1.64 A;

and electric iron resistance is

R = v / i = 220 v / 1.6 a = 134.15 Ω.

What is the S.I unit of current?

Ampere (A)