Heating effect of electric current

Question

I know that

$$H=I² rt$$

I also know it's mathematical derivation , but I can't understand that how the heat is proportional to current squared what must be the logic behind it. I can prove it mathatically but not logically

Answer 1

Drude model view
Let us consider the simplest Drude-like model: electrons are accelerated by electric field $\mathbf{E}$ during time $\tau$ and then lose all their kinetic energy from collisions with the lattice. Between the collisions the electron velocity is governed by Newton's equation: $$m\dot{v}=-eE,$$ so that the average electron velocity is $$v_d = \frac{eE\tau}{m}$$ (I ignore inessential factors and signs.) The electric current density then can be written as: $$j=env_d=\frac{e^2n\tau}{m}E=\sigma E,$$ where $n$ is the electron concentrations and $\sigma$ is the conductivity. This is trivially scaled to a cylindrical wire to produce the Ohm's law, $I= V/R$ (see, e.g., here).

Now, in every collision event the electron loses energy $\frac{mv_d^2}{2m}$, so that the energy lost in the unit volume of the material per unit time is $$w=\frac{mv_d^2}{2m}n/\tau = \frac{e^2n\tau}{m}E^2=\sigma E^2=jE=\frac{j^2}{\sigma},$$ which is the desired result.

As I said before, I am sloppy with the factors and the signs, given the simplicity of the model: it is easily generalized to include distribution of velocities and collision times. Even better the same ideas are implemented via the kinetic equation description. But in a nutshell there is no more to it.

Maxwell equations view
Another way is simply to follow the derivation of the Poynting theorem (see, e.g., here), where the term $\mathbf{j}\cdot\mathbf{E}$ emerges in the energy balance, under rather general conditions.

Work over the circuit
Another approach is to consider the work done by the current: moving one charge over the circuit gives work $qV$, whereas the total number of charges that complete the circuit in a unit of time is $I/q$ (by the definition of current). Hence the work done by all these charges per unit time is $$W=qV\times \frac{I}{q}=IV =I^2R$$ (where the latter equality works, if the Ohm's law apply).

Answer 2

1. Power $P$ is the rate of doing work (watts = J/s)

2. The potential difference $V$ across the resistor is the work required per unit charge to move the charge through the resistor (J/C).

3. Current $I$ is the rate at which charge is moved through the resistor (C/s).

4. Then, $P=VI$ (J/s)

5. From Ohm's law $V=IR$.

6. Substituting for $V$ in (3), $P=I^{2}R$

7. Finally, for the total heat generated by the resistor for time $t$, $P=I^{2}Rt$.

If you understand 1-5, then 6 simply follows from the math.

Hope this helps.