Two bulbs rated 25 ~W -220 ~V and 100 ~W -220 ~V are connected in series to a 440 V supply. The,

Step-by-step solution

Resistance of a bulb R= V² P R₁= (220)² 25 =1936 ₂= (220)² 100 =484 Since, R₁ and R₂ are in series aligned R_ n e t &=R₁+R₂ &=1936+484=2420 rent I &= V R aligned Hence, current I= V R = 440 2420 = 2 11 ~A Potential difference across 25 ~W bulb aligned V₁ &=I R₁ &= 2 11 1936=352 ~V aligned Potential difference across 100 ~W bulb V₂=I R₂= 2 11 484=88 ~V Potential difference across 25 ~W bulb in this combination is 352 ~V , but is can tolerate only 220 ~V . Hence, it will fuse.