Question

A \[{5^\circ}C\] rise is temperature is observed in a conductor by passing a current, when the current is doubled, the rise in temperature will be nearly
(A) \[{10^0}C\]
(B) \[{16^0}C\]
(C) \[{20^0}C\]
(D) \[{12^0}C\]

Answer 1

Hint: The heat produced during the flow of current through the conductor due to collision of electrons inside the conductor moving under the applied potential difference is absorbed by the conductor which leads to increases in its temperature.
Formula used:
$H = {I^2}Rt$ Where $H$ is heat produced $I$ is the current $R$ is the resistance and $t$ is time.
$H = ms\Delta T$ Where $m$ is the mass of the body $s$ is the body’s specific heat capacity and $\Delta T$ is the change in temperature

Complete step by step solution:
In the question the temperature of the conductor is increasing as current flows through it as heat is produced during the flow of current through the conductor due to collision of electrons inside the conductor moving under the applied potential difference, this heat is absorbed by the conductor which leads to a rise in its temperature.
We know that,
$ \Rightarrow H = {I^2}Rt$
$H$ is heat produced $I$ is the current $R$ is the resistance and $t$ is time.
Let the resistance of wire be $R$ initially current flowing through conductor be $I$and heat produced be${H_1}$and in the second case the current flowing through conductor be $2I$and heat produced be${H_2}$
Hence,
\[ \Rightarrow {H_1} = {I^2}Rt\]
$ \Rightarrow {H_2} = {(2I)^2}Rt$
As the heat is produced during the flow of current is the heat is absorbed by the conductor,
(We know that $H = ms\Delta T$ where$H$ is heat produced, $m$is the mass of the body, $s$ is body’s specific heat capacity, and $\Delta T$ is the change in temperature)
\[ \Rightarrow {H_1} = ms\Delta {T_1}\] ($\Delta {T_1}$is the initial change in temperature)
\[ \Rightarrow {H_2} = ms\Delta {T_2}\]($\Delta {T_2}$is the initial change in temperature)
From the above equations, we can assert that,
\[ \Rightarrow {I^2}Rt = ms\Delta {T_1}\]
\[ \Rightarrow {(2I)^2}Rt = ms\Delta {T_2}\]
Taking the ratio of the above two we get,
\[ \Rightarrow \dfrac{{{I^2}Rt}}{{{{(2I)}^2}Rt}} = \dfrac{{ms\Delta {T_1}}}{{ms\Delta {T_2}}}\]
\[ \Rightarrow \dfrac{1}{4} = \dfrac{{\Delta {T_1}}}{{\Delta {T_2}}}\]
$ \Rightarrow \Delta {T_2} = 4\Delta {T_1}$
It’s given in the question that $\Delta {T_1}$ is\[{5^0}C\].
$ \Rightarrow \Delta {T_2} = {20^0}C$

Hence the correct answer is (C)\[{20^0}C\].

Note: As current flows through the conductor its temperature increases and this also increases the resistance of the wire but this increase is very small thus we have neglected this change in the resistance of the wire in the calculation done above.