question_answer

A silver wire has a resistance of \[2.1\Omega \] at \[27.5\,{}^{o}C\]and a resistance of \[2.7\Omega \] at \[100\,{}^{o}C.\] Determine the temperature coefficient of resistivity of silver.

Or

One billion electrons pass from a point A towards a point B in \[{{10}^{-4}}\] s.  What is the current in milliamperes? What is its direction?

Answer:

Given, resistance of silver wire at \[27.5{}^{o}C={{R}_{27.5}}=21\Omega \] Resistance of silver wire at \[100{}^{o}C={{R}_{100}}=2.7\Omega \] Let the temperature coefficient of silver be \[\alpha .\] \[\therefore \]      \[\alpha ={{R}_{{{t}_{2}}}}-{{R}_{{{t}_{1}}}}/{{R}_{1}}({{t}_{2}}-{{t}_{1}})\] or \[\alpha =\frac{{{R}_{100}}-{{R}_{27.5}}}{{{R}_{27.5}}(100-27.5)}=\frac{2.7-2.1}{2.1\times 72.5}\Rightarrow \alpha =0.0039/{}^{o}C\]Thus, the temperature coefficient of resistivity of silver is \[0.0039/{}^{o}C.\] Or Current associated with the bunch of electrons, \[I=\frac{q}{t}=\frac{ne}{t}=\frac{{{10}^{9}}\times 1.6\times {{10}^{-19}}}{{{10}^{-4}}}\]             \[=1.6\times {{10}^{-6}}A=1.6\times {{10}^{-3}}mA\] The direction of current is from B to A.