A resistance R of thermal coefficient of resistivity `alpha` is connected in parallel with a resistance 3R having thermal coefficient of resistivity `2alpha`. Find the value of `alpha_(eff)`

Answer

`alpha_(eff)=5/4 alpha`

Explanation

`R_(0)=(R_(1)R_(2))/(R_(1)+R_(2))impliesR_(0)=(3R^(2))/(4R)=3/4R`
`R_(t)=R_(0)(1+alpha_(eq)Deltat)....(1)`,
For parallel combination `R_((t))=(R_(1(t)).R_(2(t)))/(R_(1(t))+R_(2(t)))`
`R_((t))=((R+1+alphaDeltat)xx3R(1+2alphaDeltat))/(R(2+alphaDeltat)+3R(1+2alphaDeltat))`
`=(3R^(2)(1+alphaDeltat)(1+alphaDeltat)^(2))/(4R+7RalphaDeltat)`
`=(3R^(2)(1+alphaDeltat)^(2))/(4R(1+7/4RalphaDeltat))`
`R_((t))=3/4R((1+alphaDeltat)^(3))/((1+alphaDeltat)^(7//4))=3/2R(1+alphaDeltat)^(3-7//4)`
`=3/4 R(1+alphaDt)^(5//4)`
`R_(t)=3/2R(1+5/4 alphaDeltat)...(2)`
From `(1)` and `(2)`
`alpha_(eq)=5/4 alpha`