Topic Heating Effect of Electric Current Estimated reading: 3 minutes 79 views Introduction to 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 heatingEnergy dissipated by current or work done as current flows depends on,CurrentResistanceTimeThis formula summarizes these factors as, E = I2 R t, E = I V t or E = V2t / RExample 1An 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.SolutionE = I2 R t = 102 × 30 × 60 = 18 × 104 = 180 kJExample 2A 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.SolutionE = P t hence P = E / t = 3,600 / 60 = 60 WP = V2/ 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 mElectrical energy and powerIn summary, electrical power consumed by an electrical appliance is given by;P = V IP = I2RP = V2/ RThe SI unit for power is the watt (W)1 W = 1 J/s and 1kW = 1,000 W.Example 3What is the maximum number of 100 W bulbs which can be safely run from a 240 V source supplying a current of 5 A?SolutionLet the maximum number of bulbs be ‘n’. Then 240 × 5 = 100 nSo ‘n’ = (240 × 5) / 100 = 12 bulbs.Example 4An 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.SolutionReq = 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 ATherefore power dissipated = (0.5)2 × 10 = 2.5 W.Applications of heating of electrical current1. 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.Tagged:Form 3Heating Effect of Electric CurrentNotesPhysics KE Topic - Previous Electrostatics II Next - Topic Quantity of Heat