## 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:
- The resistance, R of the conductor. A higher resistance produces more heat.
- The time, t for which current flows. The longer the time the larger the amount of heat produced
- The amount of current, I. the higher the current the larger the amount of heat generated.

- Hence the heating effect produced by an electric current, I through a conductor of resistance, R for a time, t is given by H = I
^{2}Rt. 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 = I
^{2}R = V^{2}/_{R}

**Example 6.1**

- An electrical bulb is labeled 20W, 240V. Calculate:
- The current through the filament when the bulb works normally
- The resistance of the filament used in the bulb.{
**ans. 0.437A, 575.04Ω**}

Solution- I =
^{P}/_{V}=^{20}/_{240}= 0.437A - R =
^{P}/_{I2}=^{20}/ 0.437^{2}= 575.04Ω or R = V^{2}/_{P}=240^{2}/_{20}= 576Ω

- I =

- Find the energy dissipated in 5 minutes by an electric bulb with a filament of resistance of 500Ω connected to a 240V supply. {
**ans. 34,560J**}

Solution

E = Pt = V^{2}/_{R}× t = (240^{2}× 5 × 60)/_{500}= 34,560J - A 5.5 kW immersion heater is used to heat water. Calculate:
- The operating voltage of the heater if its resistance is 24Ω
- The electrical energy converted to heat energy in 2 hours.

{**ans. 241.9488V, 1.8 × 10**}^{7}J

Solution- P=VI=I
^{2}R

I = (^{2500}/_{24})^{½}=5.2062A

V=IR= 5.2062 × 24 = 241.9488V - E = VIt = Pt = 2500 × 2 × 60 × 60 = 1.8 × 10
^{7}J

OR E= VIt = 241.9488 × 5.2062 × 2 ×60 × 60 = 1.8 × 10^{7}J

- P=VI=I

- An electric bulb is labeled 20W, 240V. Calculate:
- The current through the filament
- The resistance of the filament used in the bulb. {
**ans. 0.437A, 575.95Ω**}

Solution- P = VI

I =^{P}/_{V}=^{20}/_{240}=0.437A - From Ohm’s law, V =IR

R=^{V}/_{I}=^{240}/_{0.437}= 575.95Ω

- P = VI

## Applications 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**