The energy lost in one cycle is the average of \$p(t)\$ from \$0\$ to \$2\pi\$. Since there are \$f\$ cycles in a second and real power is energy lost per second, shouldn't real power be \$ f \times \dfrac{\int_0^{2\pi} p(t) dt}{2\pi}\$?


Real power is the amount of energy lost or consumed by the circuit in a second. If we consider a single-phase \$50 \,hz\$, power supply connected to an RC circuit with power factor \$\cos \phi\$


simulate this circuit – Schematic created using CircuitLab

The instantaneous voltage and current, $$v(t)= V_m\,\sin \omega t \\ i(t) = I_m \,\sin(\omega t +\phi)$$

The instantaneous power is the product of these two, $$p(t) = v(t)i(t) =V_m I_m\,\sin \omega t \,\sin(\omega t +\phi)$$ enter image description here

In one cycle (\$20ms\$, from \$0\$ to \$2\pi\$).

  1. Positive value for \$p(t)\$ implies power is drawn from the source
  2. Negative value for \$p(t)\$ implies power is given to the source
  3. Area under \$p(t)\$ represents the energy lost or gained.


1. From \$\omega t =0\$ to \$\dfrac{3 \pi}{4}\$, the circuit consumes energy since the power is positive. The area under \$p(t)\$ in this interval will give the exact value of energy consumed. Suppose the energy given to the circuit is \$20 J\$ (a random number for illustration). enter image description here

2. From \$\omega t =\dfrac{3 \pi}{4}\$ to \$\pi\$, some energy is given back to the source, again, calculate the area under \$p(t)\$ to obtain the value, let me assume \$5J\$ of energy is given back. enter image description here

The same two things happen again from \$\omega t =\pi\$ to \$2\pi\$

Hence from \$\omega t =0\$ to \$\pi\$, the energy

  • Given to the circuit is \$20J\$
  • Returned to the circuit is \$5J\$
  • Consumed by the circuit is \$ 20J - 5J = 15J\$

The \$15J\$ consumed is lost forever but the \$5J\$ returned is once going to be given the circuit in the next cycle. So from \$\omega t =0\$ to \$ 2\pi \, \$,\$ 30J\$ are consumed and \$10J\$ are returned.

That is, the net energy consumed would be the difference between the positive and negative areas, which is $$\text{Real Energy Consumed per cycle} = \dfrac{\int_0^{2\pi} p(t) dt}{2\pi}$$ So shouldn't the energy consumed in \$50\$ cycles ( since \$f = 50hz\$) be \$ f \times\dfrac{\int_0^{2\pi} p(t) dt}{2\pi}\$?

I've probably made a silly mistake somewhere but I'm not able to spot it.

  • \$\begingroup\$ The calculation of 20J, 5J and 15J doesn't seem correct. If you look visually at the graph, the red area fits more than 10 times in the green area \$\endgroup\$
    – Ferrybig
    Commented Aug 26, 2021 at 8:43
  • \$\begingroup\$ @Ferrybig Yeah they are random numbers, just to understand what's happening. But the integrals evaluated in the respective periods would give the exact value \$\endgroup\$ Commented Aug 26, 2021 at 8:44
  • 2
    \$\begingroup\$ Power is energy per second. Frequency is per second. Keep track of how often you have divided by time. \$\endgroup\$
    – Neil_UK
    Commented Aug 26, 2021 at 8:55

1 Answer 1


Assuming period \$T\$, then the straight-forward expression of average will be: $$ \frac{1}{T}\int_0^T{p(t)dt} $$

Comparing with your original expression:

$$ \frac{1}{2\pi}\int_0^{2\pi}{p(t)dt} $$

you can spot the mistake: expressions would match only if \$T=2\pi\$. This is only true for \$\omega=1\$. For \$\omega\neq1\$, then in fact \$T=2\pi/\omega=1/f\$. Meaning you could rewrite this expression as: $$=\frac{\omega}{2\pi}\int_0^{2\pi/\omega}{p(t)dt}=f\int_0^{1/f}{p(t)dt} $$

Which makes explicit the multiplying frequency factor you had suspected necessary.


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