HEATING EFFECT OF AN ELECTRIC CURRENT

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,

1. Current
2. Resistance
3. Time

This formula summarizes these factors as, E = I² R t, E = I V t or E = V² t / R

Examples

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

• = I² R t = 10² × 30 × 60 = 18 × 10⁴ = 180 kJ

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 m² and resistivity 1 × 10^-6 Ω m.

Solution

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

P = V² / 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 = I² R

            P = V² / R

The SI unit for power is the watt (W)

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

Examples

1. 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.

2. 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 = I² 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

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.

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

PRIMARY NOTES, SCHEMES OF WORK AND EXAMINATIONS