Answer to Question #174610 in Electric Circuits for BOb

Question

Tungsten has a temperature coefficient of resistivity of 0.0045 oC-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 28 oC, and the initial power delivered to the wire is Po. a) Calculate the temperature at which the power that is delivered to the wire be decreased to ½ Po. (3) b) Calculate by what factor the resistance of the tungsten wire has increased at this new temperature.

Answer 1

GIVEN:

Temperature cofficient of resistivity $=\alpha=0.0045(\degree C)^{-1}$

initial power$=P_o$

final power $={P_0\over2}$

initial temperature $=T_o=25\degree C$

SOLUTION:

As a function of temperature ,the resistance of the wire is given by equation;

$R=R_o[1+\alpha(T-T_o)]$ ,...........(1)

and we no. equation of power $P={V^2\over R}$ ............(2)

combining both equation (1) and (2) we get,

$P={V^2\over R_o[1+\alpha(T-T_o)]}={P_o\over1+\alpha(T-T_o)}$ .

where, $P_o={V^2\over R_o}$.

Now, since the voltage is constant, but power $P ={P_0\over2}$ ,

so we will get,

${P_o\over2}={P_o\over1+\alpha(T-T_o)}\\ \implies2=1+\alpha(T-T_o)$

solving for T;

$T={1\over\alpha}+T_o={1\over 0.0045}+28=\boxed{250\degree C}$

now putting value of T=$250\degree C$ in equation (1),

we can calculate by what factor value of R increased.

i.e,$R\over R_o$

therefore, ${R\over R_o}=[1+0.0045*(250-25)]\\ \boxed{{R\over R_o}= 2.0125}$