`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.

Explanation

Let the voltage supply be `V` . Let resistance of each bulb br `R`.
Total resistance of circuit is `nR`. Power illuminated by all bulbs,
`P_(1) = ( V^(2))/( nR )`
Power illuminated by one bulb ,
`P_(2) = ( P_(1))/( n) = (V^(2))/( n^(2) R )`
After one bulb is fused, the powers are
`P'_(1) = ( V^(2))/(( n-1) R ) , P'_(2) = ( V^(2))/(( n-1)^(2) R)`
(a) Fractional change in the illumination of all the bulbs is
`f_(1) = (p'_(1) - P_(1))/( p_(1)) = (P'_(1))/( P_(1)) -1 = ( n) /( n-1) -1 = (1) /( n-1)`
(b) Fractional change in the illumination of one bulb is
`f_(2) = ( P'_(2) - P_(2))/( P_(2) = ( n^(2))/((n-1)^(2)) -1 = ( 2n -1) /(( n-1)^(2))`