Answer to Question #347242 in Electric Circuits for Nuel Fudalan

Question #347242

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 20o 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.


1
Expert's answer
2022-06-02T10:26:23-0400

Given:

dd=0.1 mm=0.0001 m;

t1t1=200 C;

R1R1=25 Ω, resistance at 200 C temperature;

R0R0=1.0 x 10-7 Ω.m, resistivity at 200 C temperature;

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

R/R1=1.35;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=π(d/2)2;R1=R0(L/A); L=(R1/R0)A; A= \pi(d/2)^2;

L=(R1/R0)π(d/2)2=(25Ω/10(7)Ω.m)3.14((0.0001m)/2)2=1.96m.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+α(tt1));R=R1(1+\alpha(t-t1)); t=(1/α)(R/R11+αt1)=(1/3.9210(3))(1.351+3.9210(3)20)=109C;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.

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment