Question

The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on, the initial current decreases until the filament reaches its steady-state temperature. The temperature coefficient of resistivity of the filament is 4.5 times 10-3 K-1. The final current through the filament is one- eighth the initial current. What is the change in temperature of the filament?
a. 0100 K
b. 378 K
c. 1627 K
d. 814 K
e. 1162 K

Answer 1

To solve this problem we can apply the concept related to thermal expansion, including the analogy with resistance and final intensity.

The mathematical expression that describes the expansion of a material by a thermal process is given by

$R = R_0\alpha \Delta T$

Where

$R_0$= Initial resistance

$\alpha =$ Thermal expansion coefficient

$\Delta T =$ Change in the temperature

If we want to directly obtain the final value of the resistance of the object, you would simply add the initial resistance to this equation - because at this moment we have the result of how much resistance changed, but not of its final resistance - So,

$R_f = R_0 + L_0\alpha \Delta T$

$R_f = R_0(1 + \alpha \Delta T)$

Re-arrange to find the change at the temperature,

$\Delta T=\frac{1}{\alpha}\frac{R_f}{R_0}-1}$

Since the resistance is inversely proportional to the current and considering that the voltage is constant then

$R \propto \frac{1}{I}$

Then,

$\Delta T=\frac{1}{\alpha}\frac{I_0}{I_f}-1}$

$\Delta T = \frac{1}{4.5*10^{-3}}(\frac{I_0}{I_0/8}-1)$

$\Delta T = \frac{1}{4.5*10^{-3}}(8-1)$

$\Delta T = 1555.5k$

(It is possible that there is a typing error and the value is not 4.5 but 4.3, so the closest approximate result would be 1627K and mark this as the correct answer)