Chapter 18, Problem 6Q

What happens when aIightbulb burns out?

To determine

To explain:

The inner mechanism when a light emitting bulb burns out.

Answer to Problem 6Q

Solution:

The rnergy is dissipated in the resistance which appears in form of light energy.

Explanation of Solution

Formula used:

    • The Ohm’s law tells that the ratio of voltage drop (V)and the current (I)passing through a resistance (R)is ideally constant which is equal to the value of that resistance itself, i.e.,

             $R=\frac{V}{I}$

    •The power consumption by a resistor is given as,

             $P=I^{2}R=V\times I=\frac{V^2}{R}$

Actually, when a light bulb is connected in a circuit, the electrons start entering from one side and leave from the other end. The bulb offers a huge resistance to the moving electrons but because the electrons are moving due to the potential difference, maintained constant, of an external voltage source, therefore the electrons enter into the resistive environment of the electric bulb, forcefully.

Now, the resistance of the light bulb works as a dissipative force to the incoming electrons and the energy of the electrons keep on dissipating inside the light-bulb. As, we know that energy can neither be created nor be destroyed, so this dissipated energy is converted into the light energy and we say that the light-bulb is burning out.