Question

Tungsten has a temperature coefficient of resistivity of 0.0045 (c°)-1. a tungsten wire is connected to a source of constant voltage via a switch. at the instant the switch is closed, the temperature of the wire is 21 °c, and the initial power delivered to the wire is p0. at what wire temperature will the power that is delivered to the wire be decreased to 2/3 p0?

Answer 1

Answer:

$131.1^{\circ}C$

Explanation:

The power delivered in the wire is given by:

$P=\frac{V^2}{R}$

where V is the voltage of the battery and R is the resistance of the wire.

Since the voltage of the battery is constant, we can rewrite this equation as follows:

$V^2 = PR=const.$ (1)

At the beginning, the initial resistance is $R_0$, and the power delivered is $P_0$. Later, when the temperature increases, the power becomes $P_1 = \frac{2}{3}P_0$, and the new resistance is $R_1$. Using (1), we can write

$P_0 R_0 = \frac{2}{3}P_0 R_1\\R_1 = \frac{3}{2}\frac{P_0 R_0}{P_0}=\frac{3}{2}R_0$ (2)

So, the new resistance must be 3/2 of the initial resistance.

We know that the resistance increases linearly with the temperature, as

$R_1 = R_0 (1+\alpha \Delta T)$

where

$\alpha = 0.0045 ^{\circ}C^{-1}$ is the temperature coefficient

$\Delta T$ is the change in temperature

Using (2), we can rewrite this equation as

$\frac{3}{2}R_0 = R_0(1+ \alpha \Delta T)$

and we find:

$\frac{3}{2}=1+\alpha \Delta T\\\Delta T=\frac{\frac{3}{2}-1}{\alpha}=111.1 ^{\circ}$

So, the new temperature of the wire must be

$T_f = 21^{\circ}+111.1^{\circ}=132.1^{\circ}$