Question

Two mateeial Ge and Al are cooled from 300k to 600k. What will be its effect on their resistivity

Answer 1

Question #78081:

Two mateeial Ge and Al are cooled from 300k to 600k. What will be its effect on their resistivity?

Solution:

$$T _ {1} = 3 0 0 K$$ $$T _ {2} = 6 0 0 K$$ $$\alpha = \frac {R _ {2} - R _ {1}}{R _ {1} (T _ {2} - T _ {1})}$$ $$R _ {2} - R _ {1} = \alpha R _ {1} (T _ {2} - T _ {1})$$ $$R _ {1} \left(\frac {R _ {2}}{R _ {1}} - 1\right) = \alpha R _ {1} (T _ {2} - T _ {1})$$ $$\frac {R _ {2}}{R _ {1}} - 1 = \alpha (T _ {2} - T _ {1})$$

The value $\left(\frac{R_2}{R_1} - 1\right)$ indicates how many times the resistance change

For Germanium (Ge) $\alpha_{Ge} = -0.05$ (Tabular value from the directory)

$$\left(\frac {R _ {2}}{R _ {1}} - 1\right) = - 0.05 \cdot (600 - 300) = - 15 \text{ or } - 1500 \%$$

The value has a minus sign, and hence the resistance will increase with cooling

For Aluminum (Al) $\alpha_{Al} = 0.00429$ (Tabular value from the directory)

$$\left(\frac {R _ {2}}{R _ {1}} - 1\right) = 0.00429 \cdot (600 - 300) = 1.287 \text{ or } 128.7 \%$$

The value has a plus sign, and hence the resistance will decrease with cooling

Answer:

For Ge resistance will increase 15 times (1500%), and for Al resistance will decrease 1,287 times (128,7%) after cooling (nonlinearity of the temperature coefficient is not taken into account in the calculations)