JEE MAIN - Physics (2010 - No. 16)
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
$${{{\alpha _1} + {\alpha _2}} \over 2},\,{\alpha _1} + {\alpha _2}$$
$${\alpha _1} + {\alpha _2},\,{{{\alpha _1} + {\alpha _2}} \over 2}$$
$${\alpha _1} + {\alpha _2},\,{{{\alpha _1}{\alpha _2}} \over {{\alpha _1} + {\alpha _2}}}$$
$${{{\alpha _1} + {\alpha _2}} \over 2},\,{{{\alpha _1} + {\alpha _2}} \over 2}$$
Explanation
$${R_1} = {R_0}\left[ {1 + {\alpha _1}\Delta t} \right];$$
$${R_2} = {R_0}\left[ {1 + {\alpha _2}\Delta t} \right]$$
$$R = {R_1} + {R_2}$$
$$ = {R_0}\left[ {2 + \left( {{\alpha _1} + {\alpha _2}} \right)\Delta t} \right]$$
$$ = 2{R_0}\left[ {1 + \left( {{{{\alpha _1} + {\alpha _2}} \over 2}} \right)\Delta t} \right]$$
$${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$$
In Parallel, $${1 \over R} = {1 \over {{R_1}}} + {1 \over {{R_2}}}$$
$$ = {1 \over {{R_0}\left[ {1 + {\alpha _1}\Delta t} \right]}} + {1 \over {{R_0}\left[ {1 + {\alpha _2}\Delta t} \right]}}$$
$$ \Rightarrow {1 \over {{{{R_0}} \over 2}\left( {1 + {\alpha _{eq}}\Delta t} \right)}} = {1 \over {{R_0}\left( {1 + {\alpha _1}\Delta t} \right)}} + {1 \over {{R_0}\left( {1 + {\alpha _2}\Delta t} \right)}}$$
$$2\left( {1 - {a_{eq}}\Delta t} \right) = \left( {1 - {\alpha _1}\Delta t} \right)\left( {1 - {\alpha _2}\Delta t} \right)$$
$$\therefore$$ $${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$$
$${R_2} = {R_0}\left[ {1 + {\alpha _2}\Delta t} \right]$$
$$R = {R_1} + {R_2}$$
$$ = {R_0}\left[ {2 + \left( {{\alpha _1} + {\alpha _2}} \right)\Delta t} \right]$$
$$ = 2{R_0}\left[ {1 + \left( {{{{\alpha _1} + {\alpha _2}} \over 2}} \right)\Delta t} \right]$$
$${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$$
In Parallel, $${1 \over R} = {1 \over {{R_1}}} + {1 \over {{R_2}}}$$
$$ = {1 \over {{R_0}\left[ {1 + {\alpha _1}\Delta t} \right]}} + {1 \over {{R_0}\left[ {1 + {\alpha _2}\Delta t} \right]}}$$
$$ \Rightarrow {1 \over {{{{R_0}} \over 2}\left( {1 + {\alpha _{eq}}\Delta t} \right)}} = {1 \over {{R_0}\left( {1 + {\alpha _1}\Delta t} \right)}} + {1 \over {{R_0}\left( {1 + {\alpha _2}\Delta t} \right)}}$$
$$2\left( {1 - {a_{eq}}\Delta t} \right) = \left( {1 - {\alpha _1}\Delta t} \right)\left( {1 - {\alpha _2}\Delta t} \right)$$
$$\therefore$$ $${\alpha _{eq}} = {{{\alpha _1} + {\alpha _2}} \over 2}$$
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