`n` identical bulbs are connected in series and illuminated by a power supply . One of the bulbs gets fused. The fused bulb is removed , and the remaining bulbs are again illuminated by the same power supply. Find the fractional change in the illuminated of (a) all the bulbs and (b) one bulb.

Answer

`10.25 %`,5%`

Explanation

Let `R` be the resistance of each bulb .Total resistance of all the `21` bulb in series `= 21 R`
`:.` Current, `l= (22.//21R)`,
Iliuminating power of each bulb
`p = ((220)/(21R))^(2) xx R = ((220)/(21))^(2) (1)/(R)`
Illuminating power of all the bulb
`p = ((220)/(21r))^(2) XX 21r = ((220)^(2))/(21r)`
When one bulb is fixed, then current
`l = 220//20 r`
Illaminating power of all the bulb
`p' = ((220)/(20R))^(2) xx R = ((220)/(20))^(2) (1)/(R )`
Illaminating power of all the bulb
` ((220)/(20R))^(2) xx 20 R = ((220)^(2))/(20R)`
(i) `%` increase in illiuminating power of one bulb
`= ((p' - p))/(p) xx 100 = ((p')/(p) - 1) xx 100`
`= (((220)^(2)//(20)^(2) R)/((220)^(2)//(21)^(2)R) - 1) xx 100`
`= [((21)/(20))^(2) - 1]xx 100`
`= (441 - 400)/(400) xx 100`
`= 10.25 %`
(ii) `%` increase in illuminating of all the bulbs
`= ((P'' - P)/(P)) xx 100 = ((P'')/(P) - 1) xx 100`
`= (((220)^(2)//20 R)/((220)^(2)//21 R) - 1) xx100`
`= ((21)/(20) - 1) xx 100 = 5%`