Question

5. If a silver wire has a resistance of 10.0 Q at 20.0 °C, what resistance does it have at 40.0 °C? ( 3.8x10)

Answer 1

$\text{Solution : }\\ \text{Relation between resistance with temperature is as following;}\\ R_T= R_o \ \left[1+ \alpha (T-T_o) \right]\\ \text{Given :}\\ *T_o \text{ is a reference temperature. It is taken as } T_o = 20^oC\\ *R_o \text{ is a reference resistance at temperature } 20^oC. \ \ \ \ \ \ R_o = 10.0 \ \Omega \\ *\text{Final temperature } \ \ T = 40^o C\\ *\text{Temperature coefficient } \ \ \alpha = 3.8\times 10^{-3}\ ^oC^{-1}\\ \ \\ \text{Plugging in these values in the equation};\\ R_T = 10.0\Omega \ [1+ 3.8\times 10^{-3}\ ^oC^{-1}\times (40^oC-20^oC)]\\ R_T = 10.76\ \Omega$

Answer 2

SOLUTION :

Temperature coefficient of resistance of silver “alpha” is 0.0038 per ºC .

Resistance increases with increase in temperature and it is given by the following formula :

R = Ro ( 1 + alpha * temperature difference)

=> R40 = 10 ( 1 + 0.0038 * (40 - 20))

=> R55 = 10.76 Ω    (ANSWER).