Answer to Question #347242 in Electric Circuits for Nuel Fudalan

Question

A platinum resistance thermometer consists of a coil of 0.10-mm-diameter platinum wire wrapped in a coil. (a) What is the length of the wire needed so that the coil’s resistance at 20^o C is 25 Ω. The resistivity of platinum at this temperature is 1.0 x 10^-7 Ω.m. (b) What is the temperature of this platinum wire when its resistance increases 35% of its initial temperature? The temperature coefficient of resistivity for platinum is 3.92 x 10^-3 /^o C.

Answer 1

Given:

$d$=0.1 mm=0.0001 m;

$t1$=20⁰ C;

$R1$=25 Ω, resistance at 20⁰ C temperature;

$R0$=1.0 x 10^-7 Ω.m, resistivity at 20⁰ C temperature;

$\alpha$=3.92 x 10^-3 /^o C, temperature coefficient of resistivity for platinum;

$R/R1=1.35;$ here R- resistance of platinum wire at the final temperature;

Find:

a) length of the wire;

b) temperature of the wire;

Solution:

a) $R1=R0(L/A); L=(R1/R0)A; A= \pi(d/2)^2;$

$L=(R1/R0)\pi(d/2)^2=(25\Omega/10^(-7) Ω.m)*3.14*((0.0001m)/2)^2=1.96m.$

b) $R=R1(1+\alpha(t-t1));$ $t=(1/\alpha)*(R/R1-1+\alpha*t1)=(1/3.92*10^(-3))*(1.35-1+3.92*10^(-3)*20)= 109C;$

Answer:

a) L=1.96m;

b) t=109⁰C.