Chapter 18, Problem 9Q

To determine

To explain:

The reason why the light bulbs tend to burn out immediately after being switched on

Answer to Problem 9Q

Solution:

This happens because the electrons of connecting wires are pushed in the bulb just as it is turned on.

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,

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

Explanation of Solution

When a light bulb is connected to a voltage source through the connecting wires, it does not mean that it will start burning out when the energetic electrons from the voltage source reaches the resistive environment of the electric bulb.

Actually, because the connecting wires are made-up of conducting material, the electrons are spread in the entire path from the voltage source till the electric bulb. So, when we switch on the circuit, the negative terminal of the voltage source pushes the electrons existing just out-side the source, in the conducting wire and that electron pushes the next neighboring electron in the conducting wire, and so on.

In this way, the electrons existing just outside the electric bulb, receive a push from their neighboring electrons in the conducting wire and they start entering into the bulb just at the moment when the circuit is switched on. Therefore, the light bulbs almost always burn out just as they are turned on and not after they have been on for some time.