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Q.

A parallel connection to an electrical supply is made up of three incandescent bulbs with respective ratings of 40 W,60 W,and C 100 W. Which of the choices below is true?


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a

Brightness of all the bulbs will be equal

b

Brightness of bulb B will be more than that of C

c

Brightness of bulb A will be the highest

d

Brightness of bulb A will be less than that of B 

answer is D.

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Detailed Solution

If in an electric circuit, three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source then brightness of bulb B will be more than that of A.
It is given that the bulbs of power PA=40 W,  PB=60 W, PC=100 W are connected in parallel.
It is known that the power drawn (P) by a component can be given as
P=V2R            Where V is the voltage and R is the resistance of the component.
It can also be written as
R=V2P
Substituting the values of the powers of the bulb A, B and C, we get their respective resistances as
RA=V240
RB=V260
RC=V2100
where Ra, Rb, Rc are the resistances of A, B and C respectively.
Therefore, from the above relations, it can be said that RA>RB>RC.
Since power is inversely proportional to resistance, the bulb with lower resistance will glow brighter. Hence, the brightness of bulb B will be more than that of A.
 
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