We know that electric power can be written as $P=VI$, or $P=\frac{V^2}{R}$, or $P=I^2R$.

But when to use which one? Sometimes two different formulas give different results! Please explain with some examples.I 'm feeling very confused!

  • $\begingroup$ Please, specify the case when two different formulas give different results. It will help to clear the situation. $\endgroup$
    – headcrabby
    Oct 12, 2015 at 9:22
  • $\begingroup$ @headcrabby I meant the cases when OHM's law is not applicable.Which is the most general form of the equation for power is such cases? $\endgroup$
    – user74370
    Oct 12, 2015 at 9:25
  • $\begingroup$ Basic formula there is P = V * I, since the work is the energy you spend to move the charge q from potential u1 to potential u2 (and voltage V = u2-u1). If the voltage is constant over time, then power (as first derivative of work over time) will be voltage * (first derivative of charge over time) = voltage * current. $\endgroup$
    – headcrabby
    Oct 12, 2015 at 9:36
  • $\begingroup$ @headcrabby What if voltage is not constant over time?Is P=VI still valid? $\endgroup$
    – user74370
    Oct 12, 2015 at 9:38
  • 1
    $\begingroup$ For the Alternate Current (AC) there are several different power types (see for instance here: ee.ic.ac.uk/hp/staff/dmb/courses/ccts1/01400_AcPower.pdf). P=V * I will also work for AC, but it will be instantaneous power (which apparently depends on time). So the most general definition of the power is as first derivative of the work (changings of energy?) over time. From this definition you can specify the case you need. $\endgroup$
    – headcrabby
    Oct 12, 2015 at 9:51

2 Answers 2


The product of the instantaneous voltage across and current through a circuit element gives the instantaneous power delivered to the circuit element (assuming passive sign convention).

$$p(t) = v(t) \cdot i(t)$$

This holds regardless. For particular circuit elements, one can eliminate one of the variables, e.g.

Resistor: $v = Ri$

$$p(t) = Ri^2(t) = \frac{v^2(t)}{R}$$

Inductor: $v = L\frac{di}{dt}$

$$p(t) = Li\frac{di}{dt}$$

Capacitor: $i = C \frac{dv}{dt}$

$$p(t) = Cv\frac{dv}{dt} $$

  • $\begingroup$ For L and C this is extremely misleading for the OP. Averaged over time perfect inductors (L) and capacitors(C) do not dissipate power. For example, consider $I=I_0 sin(\omega t)$, then ${{dI}\over {dt}}=I_0 \omega cos(\omega t)$. Now integrate over one cycle and get 0 because the $sin$ and $cos$ are orthogonal. Your instantaneous $p(t)$ for L and C are correct, but just show energy is going into and out of the magnetic field of the L and the electric field of the C. When someone builds a circuit they are interested in the power actually dissipated. $\endgroup$ Oct 12, 2015 at 19:17
  • $\begingroup$ @GaryGodfrey, I've clearly denoted that it is instantaneous power, I do not consider you a spokesman for the OP, and your judgement of what one is interested in when one builds a circuit is of no interest or value to me. $\endgroup$ Oct 13, 2015 at 1:20
  • $\begingroup$ OK cancel the "extremely misleading" part of my comment ... poor choice of words on my part. Again, the physics of your answer is correct, and I had no intent to upset you with my additions. $\endgroup$ Oct 13, 2015 at 2:55

All 3 formulas are true for a $V$ and $I$ that are constant in time. Imagine there is a constant voltage $V$ across a resistor $R$. By Ohms Law this causes a current $I={V\over R}$ to flow through the resistor. The power being dissipated by the resistor is $$ P=IV $$ Ohms Law says $I={V\over R}$ so substitute this into the first equation for $I$ to get $$ P={{V^2}\over R} $$ Ohms Law also says $V=IR$ so substitute this into the first equation for $V$ to get $$ P=I^2 R $$ Now, for a time varying voltage, look at a single frequency $V=V_0 cos(\omega t)=RealPart(V_0e^{j\omega t})$ and $I=I_0 cos(\omega t-\phi)=RealPart(I_0e^{j(\omega t-\phi)})$ . The angle $\phi$ is the phase shift between $V$ and $I$ caused by the complex $Z=|Z|e^{j\phi}$ in Ohms Law. The formulas for Ohms Law and time-averaged (dissipated power) are still true just in slightly modified form (the hats indicate complex numbers) and the brackets <> mean time-averaged. $$ \hat {V}= \hat{I} \hat{Z} $$ $$ <P>={1\over2}I_0 V_0 cos(\phi) $$ $$ <P>={1\over 2}{{V_0^2}\over {|Z|}} cos(\phi) $$ $$ <P>={1 \over 2}I_0^2 |Z| cos(\phi) $$ where Z is the complex impedance of the circuit you want the power dissipation for. You make Z by adding up complex impedances $R$, $j\omega L$, and ${1 \over {j\omega C}}$ of components as if they were resistors using your series and parallel formulas. The ${1\over2}cos(\phi)$ come from integrating the product of two cosines when doing the time- average.

Complex impedances are described in many text books and web links ... here is one link that also derives the $<P>$ formulas.

  • $\begingroup$ All three are false whenever the impedance is dynamic. $\endgroup$ Oct 12, 2015 at 12:57
  • $\begingroup$ The question as originally asked by the OP appeared to be for V and I being constant in time. If they are not, this is usually handled by going into the frequency domain and replacing R with the complex impedance Z. I have added to my answer above to cover this case. $\endgroup$ Oct 12, 2015 at 19:26

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