Applications of Thermal Effects of Current

1.0Electric Bulb

The electric heating or joules heating is used in producing light in ‘electric bulb’. Here, the filament must retain as much of the heat generated as possible, so that it gets very hot and emits light. It must not melt at such high temperature. Thus, an electric bulb consist of a filament of a strong metal with high melting point such as ‘tungsten’ (melting point 3380°C) sealed in a glass bulb. The bulbs are usually filled with chemically inactive ‘nitrogen’ or ‘argon’ gases which prevents the oxidation of filament at high temperatures, thereby, increasing the life of bulb. Most of the power consumed by the filament appears as heat, but a small part of it is in the form of light radiated. 

If an electric bulb is rated ‘V’ volts. and ‘P’ watts, then its resistance is given by $R=\frac{V^2}{P}$ . For example, for a 40 W, 220 V bulb,$R=\frac{220^2}{40}$ = 1210 Ω.

2.0Electric Fuse and its Action

It is a safety device used to prevent the electric appliances against excessive electric currents. It consists of a piece of wire made of a metal or an alloy of appropriate melting point, for example aluminium, copper, iron, lead etc. Usually, a metallic conducting wire (fuse wire) made of Tin (25%) and Lead (75%) having low melting point is used. The fuse wire is usually encased in a cartridge of porcelain or similar material with metal ends. It is put in series with the electric device in the circuit. If a current larger than the specified value of current capacity flows through the circuit, the temperature of the fuse wire increases. This melts the fuse wire and breaks the circuit and thus, the electric device in the circuit is prevented from getting damaged.

The fuses used for domestic purposes are rated as 1A, 2A, 3A, 5A, 10A, etc.

The current capacity of a fuse is independent of its length, it is proportional to its radius. More the radius, more will be the current capacity and vice-versa.

For an electric iron which consumes 1 kW electric power when operated at 220 V, a current of (1000/220)A, that is, 4.54 A will flow in the circuit. In this case, a 5 A fuse must be used.

Examples

1. 200 J of heat is produced every second in a 8 Ω resistance. Find the potential difference across the resistor.

Solution

Given, heat, H = 200 J  ; resistance, R = 8Ω ; time, t = 1s.

Now, H = I² R t or $I^2=\frac{H}{Rt}=\frac{200}{8\times 1}=25$

or $I=\sqrt{25}=5A$

Now, potential difference, V = I × R = 5 × 8 = 40 V

2. A bulb draws 24 W when connected to a 12 V supply. Find the power drawn if it is connected to a 9V supply.

Solution

Given, Power P = 24 W and voltage, V = 12 volt, R = ?

Now , $P=\frac{V^2}{R}$

$R=\frac{V^2}{P}=\frac{12^2}{24}=\frac{144}{24}=6\Omega$

Now in second case, voltage V' = 9 volt, R = 6Ω, P'= ?

∴ Now Power drawn, $P’=\frac{V^{‘2}}{R}=\frac{9^2}{6}=\frac{9\times 9}{6}W$

$=\frac{27}{2}W=13.5W$

We may think that electrons are consumed in an electric circuit. This is wrong! We pay the electricity board or the electric company to provide energy to move electrons through the electric devices like electric bulb, fan, televisions, refrigerators, etc. We pay for the energy that we consume in the electric devices.

3.0Also Read