Introduction When current flows through a conductor, heat energy is generated in the conductor. The heating effect of an electric current depends on three factors:
Hence the heating effect produced by an electric current, I through a conductor of resistance, R for a time, t is given by H = I2Rt. This equation is called the Joule’s equation of electrical heating. Electrical energy and power The work done in pushing a charge round an electrical circuit is given by w.d = VIt So that power, P = w.d /t = VI The electrical power consumed by an electrical appliance is given by P = VI = I2R = V2/R Example
Solution
Solution E = Pt = V2/R *t = (2402 *5*60)/500 = 34,560J
{ans. 244.9488V, 1.8*107J} Solution I = (2500/24)1/2 =10.2062A V=IR= 10.2062 * 24 = 244.9488V
OR E= VIt = 244.9488 * 10.2062 * 2 * 60 * 60 = 1.8 * 107J An electric bulb is labeled 100W, 240V. Calculate: The current through the filament The resistance of the filament used in the bulb.Solution P = VI I = P/V = 100/240 =0.4167A From Ohm’s law, V =IR R=V/I =240/0.4167 = 575.95ΩApplications of heating effect of electric current Most household electrical appliances convert electrical energy into heat by this means. These include filament lamps, electric heater, electric iron, electric kettle, etc. In lighting appliances
In electrical heating
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