Engineering Asked by DarthCavader on March 3, 2021
I wanted to derive the formulae for active, reactive and apparent power.
While I understand the concept of phasors, I think I would find an analytical proof that doesn’t involve phasors, more intuitive. I recently came across an article that explained how active power is basically the time average of the product of instantaneous voltage and current through the load. I could make sense of it.
I was wondering if there are similar derivations for reactive and apparent powers as well, that do not employ phasors.
What I thought of:
The derivation of active power led me to think that I could perhaps derive the other formulae too, using integrals.
I had the following equations with me:
$$begin{align}
I(t) &= I_m sin(omega t + phi)
V(t) &= V_m sin(omega t)
end{align}$$
When I decomposed $I(t)$ in terms of $sin(omega t)$ and $cos(omega t)$, I immediately saw that while finding the average of $V(t)I(t)$ over a time period, the term with $cos(omega t)$ would not contribute.
So, the term $I_m sin(phi)cos(omega t)$ would have something to do with the reactive power. However, I still am unable to logically arrive at the pre-established formula for reactive power: $dfrac{V_m I_m sin(phi)}{2}$.
I was unable to make a headway with the derivation of apparent power, either.
Any help or pointers would be greatly appreciated.
$$begin{align} p &= V(t) I(t) p &= V_m sin(omega t) I_m sin(omega t + phi) p &= V_m I_m sin(omega t) sin(omega t + phi) end{align}$$
Trigonometric identity:
$$ sin alpha sin beta = frac {cos(alpha - beta) - cos(alpha + beta)}{2} $$
$$begin{align} p &= frac {P_m} {2} [{ cos(omega t - omega t - phi) - cos(omega t + omega t + phi)}] p &= P_{avg} cos(-phi) - P_{avg} cos(2 omega t + phi) end{align}$$
First component is Real Power (constant), Second Reactive Power (sinewave at twice frequency of voltage or current). Switch $alpha$ and $beta$ and negative goes away.
Answered by StainlessSteelRat on March 3, 2021
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