At `0^(@)C` the electron of conductor of a conductor `B` is `n` times that of conductor A temperature coefficient of resistance for A and B are `alpha_(1) and alpha_(2)` respectively the temperature coefficient of resistance of a circuit segment constant A and B in series is

Options

A. `(n alpha_(1) + n alpha_(2))/(1 + n)`
B. `(n alpha_(1) - n alpha_(2))/(1 + n)`
C. `(alpha_(1) + n alpha_(2))/(1 + n)`
D. `(alpha_(2) + n alpha_(1))/(1 + n)`

Answer

C

Explanation

Let `R_(1)` and `R_(2)` be the resistance of A and B at `0^(@)C` and `R'_(1)` and `R'_(2)` be the resistances of A and B at `t^(@)C`. Then
`R_(2)=nR_(1),R'_(1)=R_(1)(1+alpha_(1)t), R'_(2)=R_(2)(1+alpha_(2)t)`
When resistances are connected in series, then
`R'_(s)=R'_(1)+R'_(2)=R_(1)(1+ alpha_(1)t),R'_(2)=R_(2)(1+alpha_(2)t)`
or `R'_(s)=R_(1)[(1+n)+(alpha_(1)+n alpha_(2))t]`......(i)
Let `alpha_(s)` be the temp. coeff. of resistance of Standard equation for series combination is
`R'_(s)=(R_(1)+R_(2))(1+alpha_(s)t)=(R_(1)+n R_(1))(1+alpha_(s)t)`
`=R_(1)(1+n)(1+alpha_(s)t)`
`= R_(1) [(1+n)+(1+n) alpha_(s)t)`....(ii)
Comparing (i) and (ii), we have
`(alpha_(1)+n alpha_(2))=(1+n)alpha_(s) or alpha_(s)=(alpha_(1)+nalpha_(2))/(1+n)`