Explain electrical energy and power.

Let $\mathrm{A}$ and $\mathrm{B}$ are end point of conductor.

Let current I flow through them. Potential of $\mathrm{A}$ and $\mathrm{B}$ are $\mathrm{V}(\mathrm{A})$ and $\mathrm{V}(\mathrm{B})$ respectively. Current flows from $A$ to $B$ hence V(A) > V(B) and potential difference between $A$ and $B$.

$\mathrm{V}=\mathrm{V}(\mathrm{A})-\mathrm{V}(\mathrm{B})$ $\therefore \mathrm{V}>0$

If charge $\Delta \mathrm{Q}=\mathrm{I} \Delta t$ flows from $A$ to $B$ in $\Delta t$ time then potential energy of charge at$ A$,

$\mathrm{U}_{1}=\mathrm{V}(\mathrm{A}) \Delta \mathrm{Q}$ and at $B$ $\mathrm{U}_{2}=\mathrm{V}(\mathrm{B}) \Delta \mathrm{Q} .$

$\therefore$ Change in potential energy of charge,

$\Delta \mathrm{U}=\mathrm{U}_{2}-\mathrm{U}_{1}$

$\therefore \Delta \mathrm{U}=\Delta \mathrm{Q}[\mathrm{V}(\mathrm{B})-\mathrm{V}(\mathrm{A})]$

$\quad=\Delta \mathrm{Q}(-\mathrm{V})$

$\quad=-\mathrm{IV} \Delta t<0$

From law of conservation of charge,

$\Delta \mathrm{K}=-\Delta \mathrm{U}$ $\therefore \Delta \mathrm{K}=-[-\mathrm{I} \Delta+\mathrm{V}]$

$\therefore \Delta \mathrm{K}=$ IV $\Delta t>0$

If in conductors, electro charge moves freely under effect of electric field then there will be increase in kinetic energy of electric charge. But charge moves with constant drift velocity and do not have accelerated motion.

Reason behind this can be explained as follows.

During motion in conductor electron collide with ions and atoms. During this collision atoms or ions gain energy from electron. Hence, oscillation of ion become faster and conductor heats up.

Thus, kinetic energy is converted into heat energy.