Heat produced across a resistor can be computed by the following formulae

  1. H=I²Rt
  2. H=(V²/R)t Going by #1, heat is directly proportional to the resistance whereas going by #2 heat is inversely proportional to resistance.

Which is the right formula to be applied to calculate the heat generated across a resistor & why?

  • 1
    \$\begingroup\$ Either formula can be used based on whether you know the current through the resistance or voltage across it. The result for both formulae are supposed to be the same. If a voltage source is connected across the resistance, second formula is easy to use. If a current source is connected across the resistance, first formula is easy to use. \$\endgroup\$
    – AJN
    Commented Aug 13, 2020 at 13:23
  • 1
    \$\begingroup\$ Note that the formulae are only correct for constant I and R or constant V and R. If the current, voltage or resistance is varying, the formula is different. It will look something like \$H = \int i^2(t) \cdot R(t) dt\$. \$\endgroup\$
    – AJN
    Commented Aug 13, 2020 at 13:24
  • 4
    \$\begingroup\$ Nope. I=V/R -> H= (I^2) * R * t = (V^2/R^2) * R * t = V^2 / R * t. They are equivalent. \$\endgroup\$
    – StarCat
    Commented Aug 13, 2020 at 13:25
  • 2
    \$\begingroup\$ In other words (in case it wasn't absolutely clear from StarCat's comment above), the two formula are identical once you learn and apply ohms law. \$\endgroup\$
    – Andy aka
    Commented Aug 13, 2020 at 13:38
  • \$\begingroup\$ Oh my goodness, was it your exam question or what? You reminded me when I learned Ohm's Law in 1st year EE diploma course: (I = V/R, R = V/I, V = I * R, P = V * I = V² / R, E = P * T, ...). I though it easy as middle school algebra and unlike all other class mates, never bothered to do any drill exercises. Then came year exam with loads of questions like yours, and must finish in a short time. I got a bare pass (I always thought the teacher kindly pulled me up from marginally failure). I did graduate with a bare pass (2nd last position in class), started earning a living as a technician, ... \$\endgroup\$
    – tlfong01
    Commented Aug 13, 2020 at 14:00

4 Answers 4


Because of Ohm's law, they are equivalent.

Ohm's law tells us that, for an ideal resistor,

$$V = IR.$$

And, in general for any circuit branch, the power consumed is \$P=IV\$.

Therefore, for an ideal resistor

$$P=IV = I(IR) = I^2R$$


$$P=IV = (\frac{V}{R})V = \frac{V^2}{R}.$$

And the heat generated by the resistor in some amount of time \$t\$ is just \$Q=Pt\$.

You can use whichever expression for power (or heat) is most convenient, which typically depends on the kind of source that is providing power to the resistor.


Both are correct.

  1. is for current source or load ( e.g. fault)

  2. is for voltage source normal operation pulsed R load.


Both are correct.

The power dissipated by a resistor is given by P = VI. So the total energy over a time period t would be H = VIt.

However, Ohms law says that V = IR, or equivalently I = V/R.

Plugging that into H = VIt gives us H = (V²/R)t = I²Rt.



Which is correct? H = I²RT or H = (V²/R)T?


Update 2020aug14hkt1508

Errata and apology

I make a careless mistake of confusing energy E with power P. In other words, all E symbols mentioned below should read P. I am terribly sorry about that.

Many thanks to @AJN for quickly catching my mistake. before I lose too much face.

  1. Firstly, The use of the symbol H is misleading, because if H stands for heat, but heat is not the same "thing" as energy. So we should follow the old guy Einstein and use the symbol E as in E = MC²

  2. Secondly, "Which equation is correct?" is a trick question, because a good answer should argue that both equations are correct as they are equal, proved by using both algebra AND Ohm's Law.

  3. Use of Ohm's Law is crucial in the argument, not using this law should get zero marks. (yes, you cannot say that it is implied, or everybody know that, ...)

  4. Thirdly, Einstein won't say ET = MC²T, because it is stupid to include T which cancels out in the equation. So to make explanation concise, the answer should exclude T.

  5. Now the question becomes: which is correct? E = I²R or E = V²/R

  6. Proof: Substituting Ohm's Law V = IR in second equation, both equations are equal.

  7. QED


(1) Dimensional Analysis - Wikipedia

Appendix A - Dimensional Analysis of the Thermal Energy E, Power P, and Ohm's Law - Wikipedia

Introduction - when I was writing up my answer, I was thinking at the same time of doing a "dimension analysis" to make sure the "units" of the E, I, R, V equation, match, but I was too lazy and also hungry for lunch, to do that.

Now let me see if I still remember how to do dimensional analyisis.

/ to continue, ...

(Just a random note irrelevant to my answer. Please feel free to skip it.)

One very weird thing that I am puzzling about is the following: An hour ago I wrote my answer and then went for my locking down lunch, followed by a cup of tea, and listening to the radio.

But then I somehow (subconsciously) sensed that some thing was wrong with my answer, and I almost immediately "discovered" my mistake of using the term E.

It is very strange that I used to have confidence in my physics and should not be inclined to double check my so simple argument.

Now my guess is that I have some "six sense", my brain multi-tasked and spiritually went to my computer, and spiritually reading AJN's comment which triggered me to have a second thought (sort of twin sisters living far apart can communicate through brain waves.") . It was slightly disturbing, and making me nervous, like I suddenly remember something I forgot to do, ...)

End of answer

  • 1
    \$\begingroup\$ with regards to #4 and #5, letter E implies we are talking about energy, but \$I^2R\$ and \$V^2/R\$ both give result in power which is rate of change of energy. The symbol E is confusing. \$\endgroup\$
    – AJN
    Commented Aug 14, 2020 at 6:57
  • \$\begingroup\$ @AJN, Ha, I just came to apologize my careless conceptual mistake. I should have used P for power, and E = PT. \$\endgroup\$
    – tlfong01
    Commented Aug 14, 2020 at 7:08

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