Question

Calculate the temperature at which the resistance of a conductor becomes 20% more than its resistance at 27°C.​

Answer 1

Answer:

Given data states that there is a conductor whose resistance is R at a particular temperature of 270 degree Celsius and we need to calculate the temperature at which it becomes 20 percent more resistant.

For that we need new resistance,

R + 20 \frac{R}{100} = 6 \frac{R}{5}.

And, R_t = R_o(1 +k t) where k is the temperature cefficient.

Therefore, initial Resistance at 270 degree celsius

R = R_o ( 1 + k \times 270)

and, New resistance = R_t = 6\frac{R}{5}=R_o ( 1 + k t ).

Divide these both equations by former to latter, we get,

-( R_o[ 1 + k \times 270] / R_o + k t) = \frac{5}{6}

=> 1 + k \times 270 / 1 + k t = 5 / 6

=> t = ( 6 [ 1 + k \times 270 ] / 5-1)

=> put value of Temperature coefficient as 2 \times 10^{-4} / K.

Thereby we get, t = 1324 degree Celsius.

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Answer 2

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