Question

11. Consider the circuit given below and answer the questions that follow:

a) What is Joule's heating effect?
b) Find the effective resistance of the circuit.
c) Find the value of current 'i' in the above diagram.
d) How much heat would be produced in the 4 ohm resistor in 5 seconds?​

Answer 1

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 = 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

An electrical bulb is labeled 100W, 240V. Calculate:

a)The current through the filament when the bulb works normally

b)The resistance of the filament used in the bulb.

Solution

I = P/V = 100/240 = 0.4167A

R = P/I2 = 100/ 0.41672 = 576.04Ω or R = V2/P =2402/100 = 576Ω

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 = V2/R *t = (2402 *5*60)/500 = 34,560J

A 2.5 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. 244.9488V, 1.8*107J}

Solution

P=VI=I2R

I = (2500/24)1/2 =10.2062A

V=IR= 10.2062 * 24 = 244.9488V

E = VIt = Pt = 2500*2*60*60 = 1.8 * 107J

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 = 575.95Ω

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

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