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

`131.1^(\circ)C`

Step-by-step explanation:

The power delivered in the wire is given by:

`P=(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 = (2)/(3)P_0`, and the new resistance is
`R_1`. Using (1), we can write

`P_0 R_0 = (2)/(3)P_0 R_1\\R_1 = (3)/(2)(P_0 R_0)/(P_0)=(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

`(3)/(2)R_0 = R_0(1+ \alpha \Delta T)`

and we find:

`(3)/(2)=1+\alpha \Delta T\\\Delta T=((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)`