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

Options

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

Answer

C

Explanation

(c ) `R_(t_(1)) = R_(1) (1 + alpha_(1) r)` and `R_(t_(2)) = R_(2) (1 + alpha_(2) t)`
Also `R_(eq.) = R_(t_(1)) + R_(t_(2)) implies R_(eq) = R_(1) + R_(2) + (R_(1) alpha_(1) + R_(2) alpha_(2))t`
`implies R_(eq) = (R_(1) + R_(2)) {1 + ((R_(1) alpha_(1) + R_(2) alpha_(2))/(R_(1) + R_(2))) t}`
So `alpha_(eff) = (R_(1) alpha_(1) + R_(2) alpha_(2))/(R_(1) + R_(2))`