A) \[\frac{{{\alpha }_{1}}+{{\alpha }_{2}}}{2}\]
B) \[\sqrt{{{\alpha }_{1}}{{\alpha }_{2}}}\]
C) \[\frac{{{\alpha }_{1}}{{R}_{1}}+{{\alpha }_{2}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\]
D) \[\frac{\sqrt{{{R}_{1}}{{R}_{2}}{{\alpha }_{1}}{{\alpha }_{2}}}}{\sqrt{R_{1}^{2}+R_{2}^{2}}}\]
Correct Answer: C
Solution :
\[{{R}_{{{t}_{1}}}}={{R}_{1}}(1+{{\alpha }_{1}}t)\] and \[{{R}_{{{t}_{2}}}}={{R}_{2}}(1+{{\alpha }_{2}}t)\] Also \[{{R}_{eq.}}={{R}_{{{t}_{1}}}}+{{R}_{{{t}_{2}}}}\Rightarrow {{R}_{eq}}={{R}_{1}}+{{R}_{2}}\]\[+({{R}_{1}}{{\alpha }_{1}}+{{R}_{2}}{{\alpha }_{2}})t\] Þ \[{{R}_{eq}}=({{R}_{1}}+{{R}_{2}})\left\{ 1+\left( \frac{{{R}_{1}}{{\alpha }_{1}}+{{R}_{2}}{{\alpha }_{2}}}{{{R}_{1}}+{{R}_{2}}} \right).t \right\}\] So \[{{\alpha }_{eff}}=\frac{{{R}_{1}}{{\alpha }_{1}}+{{R}_{2}}{{\alpha }_{2}}}{{{R}_{1}}+{{R}_{2}}}\]
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