Question

An electric bulb marked $40W$ and $200V$,is used in a circuit of supply voltage $100V$. Now its power is
(A) $10W$
(B) $20W$
(C) $40W$
(D) $100W$

Answer 1

as we all know that the tungsten filament is used in the bulb. It has the property to evaporate in a vacuum and since it evaporates it catches fire as the temperature of the tungsten filament is increased, that is when it becomes hot.

Complete step by step answer:
in this question given is an electric bulb marked with $400w$and $200v$ we have to find that when we supply voltage $100v$, then the power of the bulb is$\dfrac{{{v^2}}}{R}$
But $R$resistance of the bulb is switched,
Now we take,
$\dfrac{{{p_1}}}{{{p_2}}} = \dfrac{{v_1^2}}{{v_2^2}}$
$= \dfrac{{40}}{{{P_2}}} = \dfrac{{{{200}^2}}}{{{{100}^2}}} \\\ \Rightarrow {P_2} = \dfrac{{40 \times 100 \times 100}}{{200 \times 200}} \\\$
${P_2} = 10w$

So, the correct answer is “Option A”.

Additional Information:
We know that incandescent lights like bulbs are all filled with gases that are inert. The gases can be such that it cools the filament and carries away all the heat from it. These types of inert gases reduce evaporation rate of the tungsten filament compared to the condition at which it is operating in a vacuum. If we don’t use these gases, then the temperature of the tungsten filament will increase and the filaments atoms would boil away which ultimately reduces the life of the bulb.

Note:
We also know that nowadays certain halogen bulbs are used in which oxygen is used with a small amount of halogen and the halogen is mainly bromine and iodine. When halogen is evaporated, it contains, with tungsten forming a halogen, tungsten compound which dissociates when comes in contact with the filament of tungsten and it relocates the evaporated tungsten to the filament.