Two electic bulbs marked `25 W -220 V` and `100 W - 220 V` are connected in series to a 440 V supply. Which of the bulbs will fuse?

Answer

`25 W` bulb will fase, both the bulbs will fase

Explanation

Let `l_(1) ,l_(2)` be the resistance allowed current in the two bulbs and `R_(1),R_(2)` their respectively resistance
`:. R_(1) = ((220)^(2))/(25) = 1936 Omega`
`i_(1) = (P_(1))/(V) = (25)/(220) = 0.114 A`
`R_(2) = ((220)^(2))/(100) = 484 Omega`
`i_(2) =(P_(2))/(V) = (100)/(220) = 0454 A`
Case (a) When bulb are connected is series net resistance
`R_(1) R_(1) + R_(2) = 1936 + 484 = 2420 Omega`
Current to the circuit
`i = (V')/(R_(1)) = (440)/(2420) = 0.18 A`
it means a current `0.18 A` passes through bulb the bulb connected in series .By comparing this value of current with the value of `i_(1), i_(2)` we note that `i gt i` but less than `i_(2)` it shown `25 W` bulb will fase
Case (b) When bulb are connected is parallel let the currets through `l` and if bulb `i_(1)` and `i_(2)` In parallel combination the voltage across each bulb in the same
`:. i_(1) = (440)/(R_(1)) = (440)/(1936) = 0.23 A`
`and i_(2) = (440)/(R_(2)) = (440)/(1484) = 0.91 A`
As` t_(1) gt t_(2)` and `i_(1) gt i_(2)` hence in this case both the bulbs will fase