Explain electrical energy and power.

(N/A) Let $A$ and $B$ be the end points of a conductor.
Let current $I$ flow through them. The potentials of $A$ and $B$ are $V(A)$ and $V(B)$ respectively. Current flows from $A$ to $B$,hence $V(A) > V(B)$. The potential difference between $A$ and $B$ is:
$V = V(A) - V(B)$ where $V > 0$.
If charge $\Delta Q = I \Delta t$ flows from $A$ to $B$ in time $\Delta t$,then the potential energy of the charge at $A$ is $U_1 = V(A) \Delta Q$ and at $B$ is $U_2 = V(B) \Delta Q$.
The change in potential energy of the charge is:
$\Delta U = U_2 - U_1 = \Delta Q [V(B) - V(A)] = \Delta Q (-V) = -I V \Delta t < 0$.
From the law of conservation of energy,the change in kinetic energy $\Delta K = -\Delta U$. Therefore:
$\Delta K = -(-I V \Delta t) = I V \Delta t > 0$.
If electric charges moved freely under the effect of an electric field,there would be a continuous increase in their kinetic energy. However,charges move with a constant drift velocity and do not undergo accelerated motion.
The reason for this is that during motion in the conductor,electrons collide with ions and atoms. During these collisions,atoms or ions gain energy from the electrons. Consequently,the oscillations of the ions become faster,and the conductor heats up. Thus,the kinetic energy is converted into heat energy. The rate of energy dissipation is power $P = \frac{IV \Delta t}{\Delta t} = IV$.