A) \[{{\alpha }_{1}}=\,-{{\alpha }_{2}}\]
B) \[{{\rho }_{1}}{{L}_{1}}{{\alpha }_{1}}+{{\rho }_{2}}{{L}_{2}}{{\alpha }_{2}}=0\]
C) \[{{L}_{1}}{{\alpha }_{1}}+{{L}_{2}}{{\alpha }_{2}}=0\]
D) None of these
Correct Answer: B
Solution :
[b] Let initial resistance of the wires are \[{{R}_{1}}\]and \[{{R}_{2}}\]respectively. Then \[{{R}_{1}}'+R{{'}_{2}}={{R}_{1}}+{{R}_{2}}\] \[\Rightarrow {{R}_{1}}(1+{{\alpha }_{1}}\Delta T)+{{R}_{2}}(1+{{\alpha }_{2}}\Delta T)={{R}_{1}}+{{R}_{2}}\] \[\Rightarrow {{R}_{1}}{{a}_{1}}+{{R}_{2}}{{a}_{2}}=0\] \[\Rightarrow \frac{{{\rho }_{1}}{{L}_{1}}}{A}{{\alpha }_{1}}+\frac{{{\rho }_{2}}{{L}_{2}}}{A}{{\alpha }_{2}}=0.\] \[\Rightarrow {{\rho }_{1}}{{L}_{1}}{{\alpha }_{1}}+{{\rho }_{2}}{{L}_{2}}{{\alpha }_{2}}=0\].
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