Question

Calculate the temperature at which the resistance of a conductor becomes 20% more than its resistance at 270C.
The value of the temperature coefficient of resistance of the conductor is .

Answer 1

SOLUTION:-

GIVEN BY:-
the resistance at 270oC is  R.
the temperature be T at which resistance become 20% more.
then ,
$(R + \frac{20R}{100} ) = ( \frac{100R +20R)}{100} = \frac{6R}{5}$
$R _{t} = R _{0}( 1 + kt) \\ here \: k \: is \: the \:temprature \: cofficiant \\ R _{270} = R _{0}(1 + k \times 270) = R ...........(1) \\ R _{t} = R _{0}(1 + kt) = \frac{6R }{5} ...................(2) \\ dividing \: eq(1) \: and \: eq(2). \\ \frac{R _{0}(1 + k270)}{R _{0} + kt} = \frac{R }{ \frac{6R }{5} } = \frac{5}{6} \\ \frac{1 + k270}{1 + kt} = \frac{5}{6} \\ t = (\frac{6(1 + kt)}{5} - 1) \\ = ( \frac{6(1 + 2 \times {10}^{ - 4} \times 270 )}{5} - 1) \\ = ( \frac{6 (1.0540)}{5} - 1) \times \frac{1}{2 \times {10}^{ - 4} } \\ = \frac{0.2648}{2 \times {10}^{ - 4} } = 1324c$

hence, t = 1324C

■temperature coefficient of resistance of the conductor is 2 x 10^-4 / K


■I HOPE ITS HELP■

Answer 2

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.