Question

When current is passed through an electric bulb, its filament glows, but the wire leading current to the bulb does not glow because:
(A) Less current flows n the leading wire as compared to that in the filament
(B) The leading wire has more resistance than the filament
(C) The leading wire has less resistance than the filament
(D) Filament has coating of fluorescent material over it

Answer 1

Resistance is the opposition to the flow of current. If the material has more resistance then the current flowing through it, will be lost to the surrounding as heat. This concept is used as the principle of the electric bulb, the filament glows to produce light.

Formula used:

The formula of the Joule’s law of heating is given as

$H = {I^2}Rt$

Where $H$ is the heat loss, $I$ is the current flowing through the circuit, $R$ is the resistance of the circuit and $t$ is the time taken for the flow of current.

Complete step by step solution:
The filament of the bulb is made up of tungsten and the wire is made up of copper or aluminum. The tungsten offers great resistance to the flow of current through it, but the wire allows the good conduction of the current with the very least resistance. Let us consider the Joule’s heating law,

$H = {I^2}Rt$

$H\alpha R$

From the above formula, it is clear that the heating loss is directly proportional to the resistance. Hence if the resistance of the conductor increases, then the heating loss will also be higher. So the filament will produce more and more heat due to its higher value of the resistance when compared to that of the leading current wire. Due to the more heat loss, the filament glows but the wires do not glow.

Thus the option (C) is correct.

Note: The Joule’s law is also specified by the term, heating effect of electric current. Due to the reason that high resistance produces more loss of current, metals like copper etc. with less resistance are used for carrying the current.