Question

A bulb rated $240V,100W$ is connected to a $120V$ supply. Which of the following is found to be correct?
A. the bulbs fuses
B. the bulb is lighted but is dim
C. the bulb lights up to normal brightness
D. the bulb lighted initially then fuses away

Answer 1

Hint: Voltage is the potential difference between the points in an electrical circuit in order to push or pull the charges into the lesser potential side. This causes the charges to flow in the electrical circuit. And the resistance is the opposition provided by the material of wire through which current is flowing. As the resistance is higher, the current flow will be lesser and as the resistance is lower, then the current flow will be maximum. If the resistance is the same then the current flow will be the same. These all may help us to solve this question.

Complete step-by-step solution:
The rating on the bulb is given as,
$V=240V,P=100W$
Here we can see that the maximum potential difference possible for this bulb is given as
$240V$. In the question, it is clearly mentioned that the voltage of the supply to which the bulb is connected is only half of the maximum value. That is,
$V=120V$
From this, we can say that the rated voltage is found to be greater than the supplied voltage. Therefore the bulb will be dim. And also as there is some amount of potential existing, the bulb will be definitely lighted up.
As the resistance of the bulb is not changing and also the voltage has been decreased, the power dissipation will be lesser by the formula,
$H=\dfrac{{{V}^{2}}}{R}t$
Where $t$ be the time taken and \[R\] be the resistance. As the power dissipation is low, the bulb will glow with lesser intensity.

Note: A bulb is an electrical device which is using filaments in order to produce a lighting effect with the use of electric current. There are four types of bulbs. The incandescent bulb, halogen bulb, compact fluorescent bulb, and a light-emitting diode.