Two wires of resistance `R_(1)` and `R_(2)` have temperature coefficient of resistance `alpha_(1)` and `alpha_(2)` respectively. These are joined in series. The effective temperature coefficient of resistance is

Options

A. `(a_(1)+a_(2))/(2)`
B. `sqrt(a_(1)a_(2))`
C. `(a_(1)R_(1)+a_(2)R_(2))/(R_(1)+R_(2))`
D. `(sqrt(R_(1)R_(2)+a_(1)a_(2)))/(R_(1)^(2)+R_(2)^(2))`

Answer

C

Explanation

`R_(t)` = `R_(1t) +R_(2t)`, also `R_(1t)` = `R_(1)(1+a_(1)t)`
and `R_(2t)` =`R_(2)(1+a_(2)t)`
`R_(t)`=`R_(1)(1+a_(1)t)+R_(2)(1+a_(2)t)`= `(R_(1)+R_(2))+(R_(1)a_(1)+R_(2)a_(1)+R_(2)a_(2))t`=(R_(1)+R_(2))[1+(R_(1)a_(1)+R_(2)a_(2))/(R_(1)_R_(2))t]`
Hence effective temperature co-eficient is -`(R_(1)a_(1)+R_(2)a_(2))/((R)(1)+R_(2))`