In a DC circuit, power is one number: $P = VI$. In AC steady state it splits into three, and the whole chapter is about why. Some of the energy the source pushes into the load never comes back — that is real power $P$, measured in watts, and it is what the utility bills you for. Some of it sloshes out into inductors and capacitors and returns every half cycle without ever being consumed — that is reactive power $Q$, measured in var. The wires, breakers and transformers have to carry both, and what they see is apparent power $S$ in VA. Everything in this chapter — the power triangle, the power factor, complex power, and why factories bolt giant capacitors onto their walls — falls out of that one split.

1. Instantaneous power, and why we average it

Nothing new is being assumed here. Power is always $p(t) = v(t)\,i(t)$, in AC exactly as in DC. What changes is that in AC both factors are sinusoids, so their product wobbles at high speed. Take a load carrying

We put the phase lag on the current, so $\phi = \theta_v - \theta_i$ is the angle by which the current lags the voltage. That choice is not cosmetic — it is the sign convention the rest of the chapter runs on, and it is what makes $\phi$ come out positive for an inductive load. Then

Now the one step that matters. Use the product-to-sum identity $\cos A\cos B = \tfrac{1}{2}\left[\cos(A-B) + \cos(A+B)\right]$ with $A = \omega t$ and $B = \omega t - \phi$, so $A - B = \phi$ and $A + B = 2\omega t - \phi$:

Read that carefully, because the entire chapter is sitting in it. Instantaneous power is a constant offset plus a sinusoid at twice the source frequency. The offset does not depend on $t$ at all. The ripple, being a pure sinusoid, integrates to zero over any whole number of periods. So averaging over one period $T$:

Finally substitute the RMS values $V_{rms} = V_m/\sqrt{2}$ and $I_{rms} = I_m/\sqrt{2}$, so $V_m I_m / 2 = V_{rms}I_{rms}$:

This is exactly why RMS is defined the way it is: it is the value that makes the AC average-power formula look like the DC one, with a single correction factor $\cos\phi$ bolted on the end.

2. Three powers, one triangle

The ripple amplitude in $p(t)$ is $V_{rms}I_{rms}$ regardless of $\phi$ — that is the size of the energy exchange the wires must support. Splitting it by how much rides in phase with $v$ and how much rides in quadrature gives the standard trio:

Because $P$ and $Q$ are the in-phase and quadrature components of the same $S$, they form the legs of a right triangle whose hypotenuse is $S$ and whose angle at the origin is $\phi$. That is not an analogy; it is just $\cos^2 + \sin^2 = 1$ drawn out.

3. Complex power: packaging $P$ and $Q$ into one phasor arithmetic

Having to compute $\cos\phi$ and $\sin\phi$ separately is clumsy, and worse, it loses the ability to add loads. Complex power fixes both. Write the RMS phasors $\mathbf{V}_{rms} = V_{rms}\angle\theta_v$ and $\mathbf{I}_{rms} = I_{rms}\angle\theta_i$. The obvious guess, $\mathbf{V}\mathbf{I}$, is wrong:

Shift your time origin and both $\theta_v$ and $\theta_i$ change by the same amount, so their sum moves — but the power delivered to a load obviously cannot depend on when you started your stopwatch. Only the difference $\theta_v - \theta_i = \phi$ is physical. Conjugating the current is exactly the operation that turns a sum of angles into a difference:

Now expand that polar form into rectangular:

So one multiplication recovers all four quantities at once: $P = \operatorname{Re}\{\mathbf{S}\}$, $Q = \operatorname{Im}\{\mathbf{S}\}$, $S = |\mathbf{S}|$, and $\phi = \angle\mathbf{S}$ so $\text{PF} = \cos(\angle\mathbf{S})$. Three equivalent forms are worth having at your fingertips (using $\mathbf{I}_{rms} = \mathbf{V}_{rms}/\mathbf{Z}$ and $\mathbf{Z} = R + jX$ with signed reactance, $X_L = \omega L$ and $X_C = -1/\omega C$):

That last pair is the sign convention made automatic: feed in a signed $X$ and $Q$ comes out with the right sign every time, with no lagging/leading decision to get wrong. And because $\mathbf{S}$ is a complex number, complex powers add — total $P$ is the sum of the $P$'s, total $Q$ is the sum of the $Q$'s. Apparent powers do not add: $S_{total} \ne S_1 + S_2$ unless every load happens to share the same $\phi$. This is the single most common conceptual slip in the topic.

4. Play with it: the power triangle

Hold the supply at $120$ V rms and vary the current the load draws and the angle by which that current lags (or leads) the voltage. Watch which leg of the triangle grows, and watch the equivalent load $\mathbf{Z} = R + jX$ that would produce it. Push $\phi$ to $\pm 90^\circ$ and note that $P$ collapses to zero while the current — and therefore the $I^2R$ losses in the supply wires — does not.

5. Power factor correction: the reason any of this is a business problem

A utility must size its conductors and transformers for the current the customer actually draws, $I_{rms} = S/V_{rms}$, but it can only bill for $P$. A factory full of motors runs at a PF around $0.6$ lagging, which means the utility hauls $1/0.6 \approx 1.67$ times the current that the billed watts would suggest, and eats the extra $I^2R$ line losses. So industrial customers are penalized for low PF, and the fix is cheap: bolt a capacitor bank in parallel with the load.

Parallel is essential. In parallel the capacitor sees the same voltage as the load, so the load's operating point is untouched — $P$ is unchanged and the load still gets its rated $120$ V (or whatever it is). The capacitor simply injects negative $Q$ that cancels part of the inductive $+Q$, shrinking the vertical leg of the triangle and with it the hypotenuse $S$. The recipe:

where $\phi_1 = \cos^{-1}(\text{PF}_1)$ is the original angle and $\phi_2 = \cos^{-1}(\text{PF}_2)$ the target. $P$ appears in both because correction never changes the real power — an ideal capacitor dissipates nothing.

Practice problems