Answer :

Remember one thing whenever there is this type of question we will check the power dissipation in each bulb, the bulb dissipating more power will glow brighter than the bulb with dissipating less power. The dissipation of power depends upon the resistance of the bulb that is more the resistance of the bulb more is the dissipation of the power.

As the bulbs are connected in series that means there is same amount of current flowing through them, so,

Power(P) = potential difference(V)× current(I)

By Ohm’s law, V = I× R

So,

Where V is the rating voltage of the bulbs which a very important concept in is these types of questions.

We will have to take input voltage less than or equal to the V volt or the rating voltage of the bulbs otherwise our bulb will fuse out.

It is the power of 40 W bulb,

Now we have to calculate R_{1} of bulb 1, so

It is the power of 60 W bulb,

Now we have to calculate R_{2} of bulb 2, so

Now as we have got the resistance of bulb 1 and 2, so we will find the current flowing in the circuit , which is

Now let’s calculate power dissipation which is ,

P_{1} =

P_{2} =

As we can see that power dissipated by P_{1} is higher than that of P_{2} ,

So, bulb 1 with 40 W power will glow brighter than bulb 2 with 60 W power.

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