
In an electrical circuit three incandescent bulbs A, B, and C rating \[40W\],\[60W\], and \[100W\] respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?
A) Brightness of all the bulbs will be the same.
B) Brightness of bulb A will be the maximum.
C) Brightness of bulb B will be more than that of A.
D) Brightness of bulb C will be less than that of B.
Answer
586.2k+ views
Hint: Every electrical appliance will have rated or design values like wattage, voltage printed on it. These values give information about resistance and allowable current etc.., let $P$ and $V$ be the power and voltage rating on a bulb.
We have $P= V.I= \dfrac{V^2}{R}$
Complete step by step solution:
As the three bulbs are connected in parallel, the voltage applied across them would be the same, let us consider as $V$ and current through each of them is different.
Now, due to the fact that the power dissipated by the three bulbs is different, also the current flowing through them would be different and thus they will have different brightness.
We know that the brightness of the bulb is directly proportional to the power it uses or the amount of current it draws.
Given, power of each bulbs be ${P_A} = 40W$, ${P_B} = 60W$ and ${P_C} = 100W$ respectively are connected in parallel to an electrical source.
From the given things we can observe that, ${P_C} > {P_B} > {P_A}$
Bulb C is having more power; its brightness is highest compared to the remaining two. And least brightness is bulb A. thus, bulb B is in between these two.
$\therefore $ Bulb B have more brightness than the bulb A. Option (C) is correct.
Additional information:
A device that produces light energy from electricity is known as a light bulb. The light bulb turns the light by sending current through a thin wire known as filament. The filament emits light when the electricity is passed through it and the filament is made up of Tungsten.
Note:
When appliances are connected in parallel then the current flowing through them is different but the potential difference or voltage across them is the same. And in the case of series connection, the current flowing through them is the same and the potential difference is different.
In a parallel connection, if one bulb gets fused the other will work.
We have $P= V.I= \dfrac{V^2}{R}$
Complete step by step solution:
As the three bulbs are connected in parallel, the voltage applied across them would be the same, let us consider as $V$ and current through each of them is different.
Now, due to the fact that the power dissipated by the three bulbs is different, also the current flowing through them would be different and thus they will have different brightness.
We know that the brightness of the bulb is directly proportional to the power it uses or the amount of current it draws.
Given, power of each bulbs be ${P_A} = 40W$, ${P_B} = 60W$ and ${P_C} = 100W$ respectively are connected in parallel to an electrical source.
From the given things we can observe that, ${P_C} > {P_B} > {P_A}$
Bulb C is having more power; its brightness is highest compared to the remaining two. And least brightness is bulb A. thus, bulb B is in between these two.
$\therefore $ Bulb B have more brightness than the bulb A. Option (C) is correct.
Additional information:
A device that produces light energy from electricity is known as a light bulb. The light bulb turns the light by sending current through a thin wire known as filament. The filament emits light when the electricity is passed through it and the filament is made up of Tungsten.
Note:
When appliances are connected in parallel then the current flowing through them is different but the potential difference or voltage across them is the same. And in the case of series connection, the current flowing through them is the same and the potential difference is different.
In a parallel connection, if one bulb gets fused the other will work.
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