A resistance measures 4 Ω at 40°C and 6 Ω at 80°C at T = 0°C the resistance will measure-

Question

Question

This question was previously asked in

RSMSSB JE Electrical (Diploma) 19 May 2022 Official Paper

Answer 1

Temperature coefficient of resistance

It calculates a relative change of resistance per degree of temperature change.

The value of the resistance at temperature 't2' is given by:

$R_t=R_o[1+α (t_2-t_1)]$

where, Rt = Final resistance at temperature 't2'

Ro = Initial resistance at temperature 't1'

α = Temperature coefficient of resistance

Calculation

Given, $R 40 = 4Ω$ at 40^oC

$R 80 = 6 Ω$ at 80oC

Using the resistance-temperature formula:

$4=R_{0}[1+α (40-0)]$ .....(i)

$6=R_{0}[1+α (80-0)]$ .....(ii)

Multiplying equation (i) with 4 and equation (ii) with 2, we get:

$16=R_{0}[4+160\alpha]$ .......(iii)

$12=R_{0}[2+160\alpha]$ ........(iv)

Subtracting (iv) with (iii), we get:

$2R_o=4$

$R_o=2\Omega $