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CBSE Class 12-science Answered

Two conductors have resistance R1 and R2 at 0°C. Their temperature coefficients of resistances are a1 and a2 respectively.  Show that the temperature coefficients of equivalent resistances when joined in series and in parallel are respectively  (R1a1+R2a2)/(R1+R2) and (R1a2+R2a1)/(R1+R2)
Asked by mridulabarua05 | 08 Feb, 2019, 10:10: AM
answered-by-expert Expert Answer
Resistances at temperature t°C  are R1(1+a1t)  and R2(1+a2t)
 
when connected in series,  R1(1+a1t) + R2(1+a2t) = (R1 + R2) ( 1+ aeqt) ......................(1)
 
where aeq is tempertaure coefficient of equivalent resistance
 
By simplifying eqn (1) , we get  aeq = a1 R1/(R1 + R2 )  + a2 R2/(R1 + R2 )  
 
if R is the resistance at 0ºC and a is the temperature coefficient , then we have resistance Rt at temperature t  is given by
 
Rt = R0(1+a t) .......................(2)
 
hence temperature coefficient is written as,  a = begin mathsize 12px style 1 over R fraction numerator partial differential R over denominator partial differential t end fraction end style
when two resistances R1 and R2 are parallel, then equivalent resistance  R is given by,  begin mathsize 12px style 1 over R space equals space 1 over R subscript 1 plus 1 over R subscript 2 end style  .........................(3)
by partial differentiation of (3) we get  begin mathsize 12px style 1 over R squared fraction numerator partial differential R over denominator partial differential t end fraction space equals space fraction numerator 1 over denominator R subscript 1 superscript 2 end fraction fraction numerator partial differential R subscript 1 over denominator partial differential t end fraction space plus space fraction numerator begin display style 1 end style over denominator begin display style R subscript 2 superscript 2 end style end fraction fraction numerator begin display style partial differential R subscript 2 end style over denominator begin display style partial differential t end style end fraction end style .................................(4)
hence eqn.(4) is written as  begin mathsize 12px style a subscript e q end subscript over R space equals space a subscript 1 over R subscript 1 space plus space a subscript 2 over R subscript 2 end style  .............................(5)
by substituting R = R1R2 / (R1 + R2) in (5),  we get,   aeq = a1 R2/(R1 + R2 )  + a2 R1/(R1 + R2 )
Answered by Thiyagarajan K | 08 Feb, 2019, 03:48: PM

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