Draw the three powers as a right triangle and the whole subject becomes geometry. $P$ is the real power in kilowatts, along the base — the part that turns shafts and makes heat. $Q$ is the reactive power in kilovars (kvar), vertical — the part that only magnetises motor iron and swings back and forth without ever being consumed. $S$ is the apparent power in kilovolt-amperes, the hypotenuse — what the transformer, the cable and the utility actually have to carry.

So $S = \sqrt{P^{2} + Q^{2}}$, read aloud S equals root P squared plus Q squared, and rearranged for the reactive leg, $Q = \sqrt{S^{2} - P^{2}}$. The angle between $P$ and $S$ is the same $\varphi$ from the last lesson, which is why $\text{PF} = \cos\varphi$ and $\varphi = \arccos(\text{PF})$ — phi equals arc-cosine of P F — with $\varphi$ quoted in degrees.

Because it is a right triangle, kW and kvar never add head to tail. 8 kW beside 6 kvar is 10 kVA, not 14 — and the 3-4-5 triangle, scaled, is behind most exam numbers you will meet here. Learn to spot it and half these questions become mental.

Nobody bills you for kvar, yet. What the utility bills is the kVA it had to build for, and every kvar pushes that hypotenuse out. That is the entire commercial argument of this chapter, and the reason a capacitor bank pays for itself.