# Why real power is average power and not average power times frequency?

## TLDR

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}\$$?

## Explanation

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, $$p(t) = v(t)i(t) =V_m I_m\,\sin \omega t \,\sin(\omega t +\phi)$$

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.

Therefore,

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).

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.

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.

• 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 Aug 26, 2021 at 8:43
• @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 Aug 26, 2021 at 8:44
• Power is energy per second. Frequency is per second. Keep track of how often you have divided by time. Aug 26, 2021 at 8:55

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

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

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.