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

  • 1
    $\begingroup$ If you can prove it mathematically, you can prove it logically. Math is logical. You might wanna clarify what you mean. Start with your mathematical derviation and tell us what you don;t understand in it. On a sidenote, I think you are referring to the power generated by a resistance under a current of intensity I ($P=RI^2$), but your notations are not very clear or conventional so I'm not 100% sure that's what you're talking about $\endgroup$ Dec 2, 2021 at 12:30
  • $\begingroup$ Do you understand the logic of Ohm's law, i.e., that $V=IR$? $\endgroup$
    – Bob D
    Dec 2, 2021 at 13:12
  • $\begingroup$ Yes mr Bob D, I do $\endgroup$ Dec 2, 2021 at 13:20
  • $\begingroup$ Good, do you understand why the power dissipated in the resistor is the product of the voltage times current, $VI$? $\endgroup$
    – Bob D
    Dec 2, 2021 at 13:41

2 Answers 2


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).

  • $\begingroup$ FYI the OP is in tenth grade. $\endgroup$
    – Bob D
    Dec 2, 2021 at 14:48
  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.

  • $\begingroup$ I may be wrong, but as $P=VI$ and also since $P=I^2R$ and $P=V^2/R$, I think his doubt is why are we telling power is directly proportional to the square of the current when $P=VI$ also exists $\endgroup$ Dec 2, 2021 at 14:36
  • $\begingroup$ @AdilMohammed I'm really no sure what part of the logic behind $I^{2}R$ he doesn't understand, as per the comment by Barbaud, which is why I laid it out step by step. Maybe by doing so he will explain what his issue is. Otherwise we are just guessing. $\endgroup$
    – Bob D
    Dec 2, 2021 at 14:42
  • $\begingroup$ You're right about that $\endgroup$ Dec 2, 2021 at 15:42
  • $\begingroup$ Mr. Bob , p=v²/r and p=I²r , in first power is inversly proportional to r and in 2nd power is directly proportional to r $\endgroup$ Dec 3, 2021 at 12:53
  • $\begingroup$ @VaibhavTiwari The two equations for power are not independent of one another because of Ohms law V=IR. Replace V in the first equation with IR you get the second equation. Replace I with V/R in the second equation you get the first equation. $\endgroup$
    – Bob D
    Dec 3, 2021 at 13:04

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