Following the above question the temperature coefficient of resistance of a circuit segment containing A and B in parallel 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

D

Explanation

In parallel combination of resistances
`R_(p)=(R_(1) xxn R_(1))/(R_(1) + nR_(1))=(nR_(1))/(1+n)`
`R'_(p)=(R'_(1)xxR'_(2))/(R'_(1)+R'_(2))=(R_(1)(1+alpha_(1)t)xxnR_(1)(1+alpha_(2)t))/(R_(1)(1 +alpha_(1)t)+nR_(1)(1+alpha_(2)t))`
`=(nR_(1) ( 1+ alpha_(2) t)(1+ alpha_(2) t))/((1+ alpha_(2) t)+ n(1+alpha_(2) t))`.....(i)
If `alpha_(p)` the temp coeff of resistance for parallel combination of resistance then
`R'_(P)= R_(P) ( 1+ alpha_(p)t) = (nR_(1))/((1 + n)) (1 + alpha_(p)t]`.....(ii)
From (i) and (iii)
`(nR_(1))/((1 + n)) (1 + alpha_(p)t) = (nR_(1) (1+alpha_(1)t)( 1+ alpha_(2)t))/( (1 +alpha_(1)t)+n(1+ alpha_(2)t))`
On solving we get `alpha_(P)= (alpha_(2)+n alpha_(2))/(1 + n)`