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JEE Main & Advanced AIEEE Solved Paper-2010

  • question_answer
      Two conductors have the same resistance at \[0{}^\circ C\]but their temperature coefficients of resistance are\[{{\alpha }_{1}}\]and\[{{\alpha }_{2}}\]. the respective temperature coefficients of their series and parallel combinations are nearly ?       AIEEE  Solved  Paper-2010

    A) \[\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2},\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}\]        

    B) \[\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2},{{\alpha }_{1}}+{{\alpha }_{2}}\]

    C) \[{{\alpha }_{1}}+{{\alpha }_{2}},\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}\]          

    D) \[{{\alpha }_{1}}+{{\alpha }_{2}},\frac{{{\alpha }_{1}}{{\alpha }_{2}}}{{{\alpha }_{1}}+{{\alpha }_{2}}}\]

    Correct Answer: A

    Solution :

    In series, \[{{R}_{1}}+{{R}_{2}}={{R}_{s}}\] \[R(1+{{\alpha }_{1}}T)+R(1+{{\alpha }_{2}}T)=2R(1+{{\alpha }_{s}}T)\] \[2R+RT({{\alpha }_{1}}+{{\alpha }_{2}})=2R+2R\text{ }{{\alpha }_{s}}T\] \[{{\alpha }_{s}}=\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}\] In parallel\[\frac{1}{Rp}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\] \[\frac{1}{\frac{R}{2}(1+{{\alpha }_{p}}T)}=\frac{1}{R(1+{{\alpha }_{1}}T)}+\frac{1}{R(1+{{\alpha }_{2}}T)}\] \[2(1-{{\alpha }_{p}}T)=1-{{\alpha }_{1}}T+1-{{\alpha }_{2}}T\] \[{{\alpha }_{p}}=\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}\]


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