"Question: Since the tap is to earth from the transformer, why isn't there a short or something at the tap?"
Current flow occurs when there is a difference in potential. As long as there is only one grounded point, there is no short. It is actually only a reference point. A transformer could be connected with any one of the three connection points to ground without it being a short AS LONG AS THERE IS ONLY ONE POINT GROUNDED. Of course, the load (house) would have to be wired accordingly.
"So there is no direction for the current going from negative to positive because there is no positive as in DC"
Actually there is. When you refer to AC, you are referring to a field that reverses polarity. There is a negative and positive relationship: it just reverses at 60 times a second. Current is not consumed...current is the flow of electrons. Think of the X1 and X3 terminals of a transformer as reversing polarity at 60 cycles. X1 is postitive in regard to X3, and then X3 is positive in regard to X1. The current goes back and forth at each change. But there is a voltage drop across the load as this happens.
Think of a water driven paddle wheel. The water itself is the current (amps) and the gravitational drop or pressure across the paddles is the voltage. The pressure of the water across the paddles does the work. The water itself is not consumed.
"For 120V applications, it seems that the current only moves in one direction; since the neutral doesn't have the same potiential as a negative "hot" to carry current because it is tied back to earth at the panel. It seems that the electricity in this example continually moves toward the neutral from the hot, and the voltage is only gained from the upper half of the sine wave created by the AC generator. As opposed to the top and bottom halves of the sine wave creating 240V between them. So, the only current that the neutral carries is the unbalanced load"
I really don't understand exactly what you are saying here, other than the last sentence is correct.
"Speaking of unbalanced loads, in the case of 120V applications, where is the unbalanced load comming from? Also, why does it vary in amperage?"
Think again of the transformer with its three connections (X1, X2 or ground, and X3). Let's freeze time at one point in the cycle. If X3 is positive in regard to X1, then X2 is also positive in regard to X1, but negative in regard to X3. It's at a "mid-potential" between the other two.
So in a load connected from X3 to X2, the curent flows from X3 to X2. In a load connected from X2 to X1, (at this same instant in time) the current flows from X2 to X1. If a load is connected across each, the current flows from X3 to X1 and only the imbalance goes inot X2 (as you noted).
By Ohms law, the current through a load is proportional to its voltage and impedance. Lets say X3 to X2 is 120 volts across a 12 ohm resistance.
120 volts/12 ohms = 10 amps
X2 to X1 is 120 volts across 24 ohms.
120 volts / 24 ohms = 5 amps.
In this case 10 amps is leaving X3, with 5 amps returning through X2 and 5 amps returning through X1. So far so good.
But inside the transformer, the winding from X2 to X3 is carrying twice the current that the X1 to X2 half is.
Let's exaggerate a little. If the load between X3 and X1 was 1 ohm, there would be 120 amps flowing out of X3 and 115 going in X2. So now half the transformer has 115 amps on it, and the other half only has 5. Hold this thought for a minute.
A 10 KVA transformer at 240 volts is rated for about 41 amps. Since the internal windings are in series, each can carry 41 amps. Remember that the copper in this winding can only carry 41 amps without exceeding its current rating. If you put all 41 amps onto one leg, that winding is at capacity. But since it is at 120 volts (120 v X 41 amps = 5 Kva), you are only utilizing one half of the total capacity of the transformer. By splitting the load between the two windings, you can utilize the entire transformer.
Probably went off on a tangent or two here, but I hope it helps.
