A potential difference is applied across the filament of a bulb at t =0, and it is maintained at a constant value while the filament gets heated to its equilibrium temperautre. We find that the final current in the filament is one sixth of the current drawn at t = 0. if the temperature of the filament at t = 0 is `20^@C` and the temperature coefficient of resistivity at `20^@C is 0.0043 ^@C^(-1),` find the final temperature of the filament.

Explanation

At a constant potential difference, since the final current is one-
sixth of the initial current, iit implies that the final resistance is
six times the initial resistance,`i.e., R_f = 6R_(i) = 6R_(20)`
[Initial temperature of filament is `20^(@)C.`]
Let the final temperature be `t^(@)C` so that
`R_f = R_(f)\t = 6R_(20) (i)`
We can express `R_t = R_(20)[ 1 + alpha_(20)(t - 20)],`
where `alpha_(20)` is the temperautre coefficient of resistance at `20^(@)C`
Usinng Eq. (i),
`6R_(20) = R_(20) [1+alpha_(20) (t- 20)]`
or `t-20= (5)/(alpha_(20))=(5)/(0.0043)= 1162.3` or `t=1182.8^(@)C`