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Question

The reisistance of thermistors A & B both at 298K is 1kΩ and the ratio of their resistance at 398K is 0.75. The difference in the material constants of thermistor A & B (in K) is _________?


  1. 341

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Solution

The correct option is A 341
The resistance of thermistor at any temperature T is expressed as:-

RT=R0 exp β(1T1T0)

Where RT & R0 are resistances at temperature T & T0 and β is materials constant.

Hence,
RA(T)=R0 exp βA(1T1T0)
and
RB(T)=R0 exp βB(1T1To)

Their Ratio = RARB(T)=0.75

0.75=exp βA(1T1To)expβB(1T1To)

0.75=exp(βAβB)[1T1To]

ln(0.75)=ln [exp(βAβB)(1T1To)]

0.2876=(βAβB)(1T1To)

(βAβB)=0.2876(1T1To)

=0.2876(1T01T)

In the given problem T = 398 K & T0= 298 K

(βAβB)=0.2876(1298K1398K)=341

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