Question

Rate of dissipation of Joule’s heat in resistance per unit volume is (symbols have useful meaning).
(A) $\sigma E$
(B) $\sigma J$
(C) $JE$
(D) None

Answer 1

Hint : Here, the rate of dissipation is asked for a resistor. So, we have to understand the concept of rate of dissipation that the dissipation is the result of the irreversible process that takes place in homogeneous thermodynamic systems. Heat flows through a thermal resistance in a conductor because of electrical current flow through an electrical resistance (Joule heating).This dissipation rate we have to calculate for a resistor.

Complete Step By Step Answer:
Consider the resistor to obtain the rate of dissipation of joule’s heat in resistance per unit volume with length $L$ and resistance $R$ also the area of that resistor as $A$ as shown in figure:

The potential applied to the resistance with length $L$ and the direction of the current is given by $E$ .
Thus, the voltage is given by $V = EL$
Heat in the resistor is given by
$P = \dfrac{{{V^2}}}{R}$
$\Rightarrow P = \dfrac{{{{\left( {EL} \right)}^2}}}{{\left( {\dfrac{{\rho L}}{A}} \right)}}$ …. (use $V = EL$ and $R = \dfrac{{\rho L}}{A}$ )
$\Rightarrow P = \dfrac{{{E^2}AL}}{\rho }$ …. $(1)$
Now, let $v$ be the volume of the resistor such that $v = AL$
In equation $(1)$ we have to put the volume $v$
$\Rightarrow P = \sigma {E^2}v$ …. (Since, $\sigma = \dfrac{1}{\rho }$ )
Thus, the rate of dissipation of joule’s heat per unit volume is given by:
$\Rightarrow \dfrac{P}{v} = \sigma {E^2}$
$\Rightarrow \dfrac{P}{v} = \sigma E.E = JE$ …. (Since, $J = \sigma E$ )
Here, $J$ is the current density.
Therefore, the answer we have obtained is that the rate of dissipation per unit volume is given by:
$\dfrac{P}{v} = JE$
Therefore, the answer is option C.

Note :
Here, we have discussed the rate of dissipation of heat in the resistor as above. So, we have calculated the heat per unit volume rate as $JE$ and here, $J$ is the current density of the conductor and $E$ is the electrical energy produced in the resistor. We must know the concept of heat and energy in thermodynamics.