`0^(@)C` पर दो चालकों के प्रतिरोध समान है किंतु उनके प्रतिरोध के ताप गुणांक `alpha_(1)` एवं `alpha_(2)` है। उनके श्रेणी एवं समांतर क्रम के ताप गुणांक क्रमशः होंगे-

Options

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

Answer

D

Explanation

`R_("श्रेणी")=R_(1)+R_(2)=R_(0)(1+alpha_(1)t)+R_(0)(1+alpha_(2)t)`
`=R_(0)[2+(alpha_(1)+alpha_(2))t]=2R_(0)[1+(alpha_(1)+alpha_(2))/(2)t]`
`impliesa_("श्रेणी")=(alpha_(1)+alpha_(2))/(2)`
`(1)/(R_("समांतर"))=(1)/(R_(1))+(1)/(R_(2))=(1)/(R_(0)(1+alpha_(1)t))+(1)/(R_(0)(1+alpha_(2)t))`
`=(1)/(R_(0))[(1+alpha_(1)t)^(-1)+(1+alpha_(2)t)^(-1)]`
`=(1)/(R_(0))[1-alpha_(1)t+1-alpha_(2)t]`
`=(2)/(R_(0))[1-((alpha_(1)+alpha_(2))/(2))t]`
`=(2)/(R_(0)[1-(alpha_(1)+alpha_(2))/(2)t]^(-1))`
`=(2)/(R_(0)(1+(alpha_(1)+alpha_(2))/(2)t))`
`therefore R=(R_(0))/(2)(1+(alpha_(1)+alpha_(2))/(2)t)`
`impliesalpha_("समांतर")=(alpha_(1)+alpha_(2))/(2)`