>Why not do some simulations, or good old pad-and-pen analysis, and figure this out yourself?
Well the answer is in the title of the forum I am a beginner in electronics and a push in the right direction from experts will be invaluable.
Well... what would you need to know, to figure it out?
The definition of power:
P = V*I
The instantaneous product of voltage and current. Power varies over time. For example, a sine wave V = Vpk sin(ωt), and similarly for I, has P varying as a product of sines, which is also a sine at twice the frequency, due to a trig identity.
So, you need algebra and either trig or geometry (for sine waves) or calculus (for general waves) to work it through.
We don't normally measure or work with instantaneous power, directly; normally, average power is what we mean. That is, the average level of P.
Then you'll need the definition of the resistance, inductance and capacitance.
We could use the most general definition (instantaneous values), but this will be much easier for the sinusoidal case. That is, at a given frequency F of a sine wave, the properties are:
Resistor: V = I R
Inductor: X = 2 π F L
Capacitor: X = -1 / (2 π F C)
X is pure reactance, and does not dissipate power. R is pure resistance, and dissipates only power.
Reactance works the same as resistance, but rotated 90 degrees. The angle only matters when you put X and R together (i.e., a lossy inductor or capacitor). In that case, you construct a right triangle, with side lengths R and X, and the length of the hypotenuse gives the impedance Z. You can measure this by applying a current and measuring the voltage (V = I Z).
Doing this by geometry would get rather complicated to handle, and so far, you've learned nothing about how power works with an impedance.
By convention, we write down Z as its two parts: Z = R + i X. Think of 'i' as a label denoting reactance, and then the unlabeled part is resistance. Like you might label points on graph paper with x and y coordinates, saying P = 10x + 20y, or dropping one label because it's redundant and labeling just the other, P = 10 + 20i.
i also has the property that i * i = -1. So, it participates in algebra, and anywhere you have products, it can be simplified. Let's do a multiplication, then a division.
You wouldn't normally multiply impedances, willy nilly. For context, let's say we're solving for two impedances in parallel. The same parallel formula applies as always:
Ztot = Z1 * Z2 / (Z1 + Z2)
This has a product in the numerator, and is divided by the sum.
Start by labeling the components...
Z1 = a + bi
Z2 = c + di
Ztot = (some resistance part) + i (some reactance part)
We always write the result in this separated form, because it's easier to work with.
First, the denominator. This is easy, nothing happens, it's just all written together. We still group the terms, just by convention.
Zdenom = (a + bi) + (c + di)
Zdenom = (a + c) + i (b + d)
Now the product. The numerator is:
Znum = Z1 * Z2
Znum = (a + bi) * (c + di)
Znum = ac + bdi^2 + bci + adi
We're just distributing terms. Simplify i^2 = -1 and group,
Znum = (ac - bd) + (bc + ad)i
And that's multiplication!
To divide, we use a trick. Aside: consider,
(x + yi) * (x - yi)
what does this simplify to?
Use the above result, substituting a = x, b = y, c = x and d = -y.
(x + yi) * (x - yi) = (xx - yy) + (yx - xy)i = x^2 + y^2
the i part cancels out!
Why is this important? It means we don't need to come up with some weird definition to divide by this composite number -- we can reduce the problem to multiplication by a well-spotted term, then division by pure, ordinary variables.
So we can finally write the quotient,
Zquot = Z1 / Z2
again we use the same Z1 and Z2 for show. We must later substitute Znum and Zdenom for a, b, c and d, to get Ztot. It'll be easier doing it later, than dragging all those terms along right now!
Zquot = (a + bi) / (c + di)
Note the denominator is c + di, so multiply top and bottom by (c - di) and simplify.
Zquot = (a + bi) (c - di) / [ (c + di) (c - di) ]
Zquot = [ (ac + bd) + (bc - ad)i ] / (c^2 + d^2)
Zquot = (ac + bd) / (c^2 + d^2) + i(bc - ad) / (c^2 + d^2)
Okay, that last step might be less simplified, but the point is to show it's always separable into two terms, a resistance part (no i) and a reactance part (only one i).
Incidentally, c^2 + d^2 = hypotenuse^2 is the Pythagorean theorem. So what's actually being done is dividing by the length of the hypotenuse, of that triangle we constructed -- the divisor's triangle. Well, dividing by it twice. (Units must always be balanced -- we're multiplying on top, so we're dividing on bottom. This is a good way to spot algebra mistakes!)
So, division is just like multiplication, except we flip a sign on one reactance term (bc - ad instead of bc + ad), and divide by the squared length of the divisor.
And hey, congrats, you just derived a definition, for a fundamental arithmetic operator, in a new(?) algebra system!
By the way, these impedance numbers, or composite numbers, are actually called "complex numbers", and i is called the "imaginary unit". (A number or term without 'i' is called "real", and a part with, is called "imaginary".) I hate these terms because they make people think there's something intangible or difficult or cryptic about them. They're none of the sort. Well, "complex" just in the observational sense, sure -- you can see it's a lot longer, written out, than a mere "a * b" is! -- but abstracting that away as the multiplication or division that it is, it's no different. (Indeed, this system is more general, because you can take the sqrt of any number, positive or negative, and always get a valid and consistent result. Yes, your calculator lied to you: the sqrt of a negative number is not error; it exists, it's just written with a little i in front!)
(Also, to be perfectly clear, "composite numbers" are actually their own thing, in number theory. I was just using the term descriptively above. Yes, math, like any other domain, has its fair share of badly named things...)
So, after that introduction -- let's handle power.
If we apply some current I, to an impedance Z = a + bi (yes, we're using two i's here, unfortunately -- for this reason, EEs traditionally use 'j' instead of 'i', which I probably should've done in the first place but I'm not going to edit all this now..), we get some voltage V = I Z.
Now we can write that out as V = I (a + bi). So V is also a complex number. We might as well call I a complex number, too; we'll just define it as all-real (zero imaginary) for convenience.
And then we do P = V*I.
For convention, we take the conjugate of I, which I will write I' here. That's the minus-imaginary-part we did before. Well, we're defining I as real, so that's trivial, but keep that in mind when starting with a voltage, and using I = V/Z.
P = (I Z) I' = I I' Z = I^2 (a + bi)
Well, that was rather boring. It's literally just I^2*R over again. As well it should be!
Say we start with real voltage V, and get some current I = V/Z. Then we have,
P = V I' = V (V / Z)' = V V' / Z' = V^2 / Z'
Now we've done something a little different. Again because V is real, the conjugate doesn't matter (for a real number x, x = x'), but be careful when it isn't. We are, however, dividing by conjugate impedance.
So what is the meaning of real and imaginary P?
Real P is real power (oddly enough!). Work. Heat.
Imaginary P is reactive power: energy stored and recycled, back and forth.
We normally write S = P + iQ, where S is called the apparent power (what I've been calling P above), P is the real power (a real number, not complex), and Q is the reactive power (also a real number).
Clearly(?), we only get reactive power from reactive impedance, and real power from real impedance.
Tying it all together, say we apply 10V AC at 20Hz, to a 1uF capacitance.
S = V^2 / Z'
Z = R + X
Z = 0 + -i / (2 π F C)
Z ~= -i7957 Ω
S = (10V)^2 / (-i7957 Ω)'
S = 100 V^2 / i7957 Ω
S = -i 0.01257 VA (note that 1/i = -i by the multiply-both-sides-by-conjugate rule)
S == P + iQ tells us Q = -0.01257 VAR (by convention, we label Q in "volt-amps reactive")
Reactive power is not dissipated power. In the same way that pulling on a load (in the direction of motion) produces work (W = F.x, units of N.m == J), and pulling on a load (perpendicular to the direction of motion) does not (a centripetal force or torque, units of N.m), so too we have watts (same direction) and volt-amps (mixed or perpendicular).
The angle of the triangle constructed from P = S + iQ is called the phase angle, and the cosine of that angle is the power factor (PF). PF = 1 when P is real, PF = 0 when imaginary.
On an instantaneous basis, reactive power looks like current going constantly between positive and negative current draw, in quadrature with voltage going positive and negative. So it's always chasing its tail and no real work gets done. A class D amplifier recycles the reactive power into its supply rail, so that current is drawn at first (and the supply voltage drops momentarily), then returned (so it rises momentarily). This only causes losses to the extent that the draw and return have losses (ESR, dielectric loss, switch resistance).
The downside is a ~constant adder due to switching losses. That is, it's always dissipating some power, necessary to keep the circuit bouncing back and forth. A very slowly varying output (nearly DC) won't be able to recover more reactive power than is lost due to operation alone, and in that case you might as well stick with a linear amplifier (that can be made to have lower idle/bias current).
You can recover some switching loss by simply operating at a lower frequency, in which case you need a proportionally larger inductor, but not as large as operating at 20Hz directly.
Tim