Question

What is the rate of dissipation of joule heat in resistance per unit volume? (E is electric field and $ \rho $ is resistivity)
(A) $ \dfrac{E}{\rho } $
(B) $ \dfrac{{{E}^{2}}}{\rho } $
(C) $ {{E}^{2}}{{\rho }^{2}} $
(D) None of these

Answer 1

Hint: Joule heating, also known as resistive, resistance, or ohmic heating, is the process by which the passage of an electric current through a conductor produces heat. To find the rate of Joule heat dissipation in resistance per unit volume, we need to take the heat generated potential in terms of electric field and resistance in terms of resistivity.

Complete step by step solution
We know that heat radiated/dissipated in resistance is given by:
 $\Rightarrow H=\dfrac{{{V}^{2}}}{R} $
Now rate of heat dissipated per unit volume is $ \dfrac{H}{Vol} $
As Volume = Area $ \times $ Length
Substituting the values of heat dissipated and the volume in the rate term, we get;
 $ \begin{align}
 &\Rightarrow Rate=\dfrac{\dfrac{{{V}^{2}}}{R}}{AL} \\
 &\Rightarrow Rate=\dfrac{{{V}^{2}}}{RAL}\text{ }......................\text{(1)} \\
\end{align} $
Now we know that resistance in terms of resistivity is given by:
$\Rightarrow R=\rho \dfrac{L}{A} $
(Here $ \rho $ is the resistivity)
Also potential in terms of electric field is
 $\Rightarrow V=EL $
Substituting these values of Rand V in equation (1)
 $ \begin{align}
 &\Rightarrow Rate=\dfrac{{{E}^{2}}{{L}^{2}}}{\left( \rho \dfrac{L}{A} \right)AL} \\
 &\Rightarrow Rate=\dfrac{{{E}^{2}}}{\rho } \\
\end{align} $
Therefore, option (B) is the correct answer.

Note
The potential generated by heat dissipation is due to the flow of current through it and so we can also find using the current flow. The joule heating affects the whole electric conductor, unlike the peltier effect which transfers heat from one electrical junction to another. Because this circuit consists of only one resistor, the entire work done goes into energy lost through power dissipation by this resistor, by conservation of energy.