`0^(@)C` पर दो हालको का प्रतिरोध समान है परन्तु इनके प्रतिरोध ताप गुणांक `alpha_(1)" तथा "alpha_(2)` हैं। इनके श्रेणी क्रम ब समातर क्रम के प्रतिरोध ताप गुणांक लगभग है

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

A

Explanation

माना `R_(0)` दोनों प्रतिरोधों का प्रारंभिक प्रतिरोध है `theta` ताप पर इनके प्रतिरोध होंगे
`R_(1)=R_(0) (1+alpha_(1) theta)" तथा " R_(2)=R_(0) (1+alpha_(2) theta)`
श्रेणी क्रम में `R_(S)=R_(1)+R_(2)`
`R_(50) (1+a_(s) theta)=R_(0) (1+alpha_(1) theta) +R_(2) (1+alpha_(2) theta)`
`2R_(0) (1+a_(2) theta)=R_(0) (1+alpha_(1) theta)+R_(2) (1+alpha_(2))`
जहा `R_(50)=R_(0)+R_(0)=2R_(0)`
समांतर क्रम में, `R_(P)=(R_(1)R_(2))/(R_(1)+R_(2))`
`R_(p0) (1+alpha_(p) theta)=(R_(2) (1+alpha_(1) theta) R_(0) (1+alpha_(2) theta))/(R_(0) (1+alpha_(1) theta) R_(0) (1+alpha_(2) theta))`
जहाँ `R_(p0)=(R_(0)R_(0))/(R_(0)+R_(0)=R_(0)/2`
जहाँ `R_(p0)=(R_(0)R_(0))/(R_(0)+R_(0))=R_(0)/2`
या `R_(0)/2 (1+alpha_(p) theta) =(R_(0)^(2) (1+alpha_(1) theta) +alpha_(2) theta+ alpha_(1) alpha_(2) theta^(2))/(R_(0) (2+alpha_(0) theta+alpha_(2) theta))`
`alpha_(1)" तथा "alpha_(2)` छोटे रशिया है
`alpha_(1),alpha_(2)` नगण्य है
`alpha_(p)=(alpha_(1)+alpha_(2))/(2+(alpha_(1)+alpha_(2))theta)=(alpha_(1)+alpha_(2))/(2) [1-((alpha_(1)+alpha_(2))/(2)) theta]`
`(alpha_(1)+alpha_(2))^(2)` नगण्य है