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Question: The resistance of wire in a heater at room temperature is \[65\Omega \] . When the heater is connect...

The resistance of wire in a heater at room temperature is 65Ω65\Omega . When the heater is connected to a 220V supply the current settles after a few seconds to 2.8A. What is the steady temperature of the wire. (Temperature coefficient of resistance α=1.70×104C1\alpha = 1.70 \times {10^{ - 4}}^\circ {C^{ - 1}} )
(A) 955C955^\circ C
(B) 1055C1055^\circ C
(C) 1155C1155^\circ C
(D) 1258C1258^\circ C

Explanation

Solution

We will first find the resistance when the heater is connected to a 220V supply using the formula R=SupplyvoltageSupplycurrentR = \dfrac{{Supply\,voltage}}{{Supply\,current}} .
Then we use the formula of dependence of resistance with temperature i.e. R2=R1[1+α(T2T1)]{R_2} = {R_1}\left[ {1 + \alpha \left( {{T_2} - {T_1}} \right)} \right] as the resistance vary with the temperature.
After that we will calculate the temperature by putting all the values given in the question.

Complete step by step solution
It is given resistance of wire in a heater at room temperature i.e. R1=65Ω{R_1} = 65\Omega and temperature is given i.e. room temperature T1=27  C{T_1} = 27\;^\circ C .
Now when the heater is connected to a 220V supply resistance R2=SupplyvoltageSupplycurrent{R_2} = \dfrac{{Supply\,voltage}}{{Supply\,current}} , now we put the values as given in question.
So R2=220V2.8A=78.6Ω{R_2} = \dfrac{{220V}}{{2.8A}} = 78.6\Omega
Now, using the relation
R2=R1[1+α(T2T1)]{R_2} = {R_1}\left[ {1 + \alpha \left( {{T_2} - {T_1}} \right)} \right] , where R2{R_2} is the resistance at temperature T2{T_2} , R1{R_1} is the resistance at temperature T1{T_1} and α=1.70×104C1\alpha = 1.70 \times {10^{ - 4}}^\circ {C^{ - 1}} .
we can find T2T1{T_2} - {T_1} from this equation as we know the value of R2,R1{R_2},{R_1} and α\alpha .
So, we solve this equation:
R2=R1+R1α(T2T1){R_2} = {R_1} + {R_1}\alpha ({T_2} - {T_1})
T2T1=R2R1R1×1α{T_2} - {T_1} = \dfrac{{{R_2} - {R_1}}}{{{R_1}}} \times \dfrac{1}{\alpha } , now we put the values and find the value of T2T1{T_2} - {T_1}
T2T1=78.66565×11.7×104{T_2} - {T_1} = \dfrac{{78.6 - 65}}{{65}} \times \dfrac{1}{{1.7 \times {{10}^{ - 4}}}} =1231
So T2=1231+T1{T_2} = 1231 + {T_1} and we know that T1=27  C{T_1} = 27\;^\circ C
So, after putting value of T1{T_1} in T2{T_2} we get:
T2=1258  C{T_2} = 1258\;^\circ C .

So, option D is correct.

Note: Always remember that we will take the temperature of room 27  C27\;^\circ C and also the formula of resistance when a conductor is connected with supply voltage.
Also remember the general rule i.e. resistivity increases with increasing temperature in conductors and decreases with increasing temperature in insulators.