A 100 W and a 500 W bulbs are joined in series and connected to the mains. Which bulb will glow brigher ?

Explanation

Let `R_(1) and R_(2)` be the resistance of the two bulbs. If each bulb is connected separately to the mains of voltage V, then `P_(1)=(V^(2))/(R_(1)) and R_(2)=(V^(2))/(R_(2))`
`therefore" "(P_(1))/(P_(2))=(R_(2))/(R_(1)) (or) (R_(1))/(R_(2))=(P_(2))/(P_(1))=(500)/(100)=5`
If the two bulbs are in series with the mains, the same current .i. flows through each of them.
Let `P_(1) and P_(2)` be the powers dissipated by two bulbs, turn `(P_(1)^(1))/(P_(2)^(1))=(R_(1))/(R_(2))=5 or P_(1)^(1)=5P_(2)^(1)`
Since 100 watt bulb dissipates more power, it glows brigher