Two resistors with temperature coefficients of resistance `alpha_1` and `alpha_2` have resistances `R_(01)` and `R_(02) ` at `0^@C`. Find the temperature coefficient of the compound resistor consisting of the two resistors connected.
a.. In series and
b. in paralllel

Explanation

`R_(01)[1 + alpha_1 theta] + R_(02)[1 + alpha_2 theta] = R_0[1 + alpha_(eq) theta]`
`(R_(01) + R_(02)) + (R_(01) alpha_1 + R_(02) alpha_2) theta`
=`(R_(01) + R_(02)(1 + alpha_(eq) theta)`
`alpha_(eq) = (R_(01) alpha_1 + R_(02) alpha)/(R_(01) + R_(02))`
(b) At `0.^@C`
At `theta^@ C : (1)/(R_(01)(1 + alpha_1 theta)) + (1)/(R_(02) (1 + alpha_2 theta)) = (1)/(R_0(1 + alpha_(eq) theta))`
`((1 + alpha_1 theta)^-1)/(R_(01)) + ((1 + alpha_2 theta)^-1)/(R_(02)) = ((1 + alpha_(eq))^-1)/(R_0)`
`alpha_1, alpha_2, alpha_(eq)` are very small, using binomial theorem
`(1-alpha_1 theta)/(R_(01)) + (1 - alpha_2 theta)/(R_(02)) = (1 - alpha_(eq) theta)/(R_0) = ((1 - alpha_(eq) theta)(R_(01) + R_(02)))/(R_(01).R_(02))`
`R_(02) (1 - alpha_1 theta) + R_(01)(1 -alpha_2 theta) = (R_(01) + R_(02)) - alpha_(eq)(R_(01) + R_(02)) theta`
`alpha_(eq) = (R_(01) alpha_2 + R_(02) alpha_1)/(R_(01) + R_(02))`.

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