Answer:

Step-by-step explanation:
Lets take
Resistance of bulb 1 =R₁
Resistance of bulb 2 =R₂
As we know that power P
P= ΔV²/R
Given that voltage difference is same for both bulbs
So
P₁R₁= ΔV² --------1
P₂R₂= ΔV² -----------2
When these resistance are connected in series then equivalent resistance R
R=R₁+ R₂
The new power P'
P'=ΔV²/R
P'R=ΔV² ------3
From equation 1 ,2 and 3
P'(R₁+ R₂) = ΔV²

