Electronic – approximate impedance Z% to X% neglecting resistance in > 25MVA transformers

Diagram!

This is for a complex impedance:

\$Z = R + \dfrac{1}{j \omega C} \$

Resistance \$R\$ is in phase with the applied voltage, so the vector points in the same X direction. The impedance of a capacitor is almost completely reactive, i.e. its resistive part is much smaller than the \$\dfrac{1}{j \omega C} \$. The \$j\$ causes a \$\theta\$ = 90° rotation, and since the \$j\$ (= \$\sqrt{-1}\$) is in the denominator the angle is negative \$\left( \dfrac{1}{j} = -j \right) \$.
To calculate the current \$ I = \dfrac{U}{Z} \$, we note that when dividing by an impedance with angle \$\theta\$ we subtract the angle from our reference, so that the angle's sign is inverted.
The result shows how for a capacitive load the current leads the voltage by an angle \$ \theta\$, where \$ 0 \le \theta \le 90°\$.
For inductive loads a similar diagram can be drawn, only \$ j \omega L\$ points in the opposite direction of \$\dfrac{1}{j \omega C} \$, and the current will trail the voltage.

edit (after your edit of the question)
So, resistance will cause the current to be in phase with the voltage. If there's an imaginary term (the \$j\$) then that term represents the reactance, either capacitive or inductive, and

Resistance + Reactance = Impedance

In an ideal world, if you don't have capacitors or coils you wouldn't have reactance either. But a circuit may have parasitic impedance: the length of a PCB trace will cause an inductive reactance (it behaves as a coil), and two adjacent traces will have a capacitive reactance (they behave as a capacitor). Parasitic impedances are unintentional, and most of the time a nuisance, though sometimes a designer can make good use of them.
You can measure components impedances with an RLC-meter, which will give you resistance in series or parallel with a reactance (inductive or capacitive).
Reactance will show as a phase shift in voltage or current. This phase shift can be shown on an oscilloscope in X-Y mode; a zero phase shift will show a straight line, a 90° phase shift will show a circle, anything in between will give you an ellipse.

You're right, the current does depend on the impedance connected to the secondary coil. I think you're getting confused about what rules to follow when. Here are the rules that always apply to ideal transformers:

$$\frac {V_P}{V_S} = \frac {N_P}{N_S}$$

$$\frac {I_P}{I_S} = \frac {N_S}{N_P}$$

$$\frac {Z_P}{Z_S} = \left( \frac {N_P}{N_S} \right)^2$$

$$P_P = P_S$$

Just because your power source can supply 1000 volts at 1 amp doesn't mean it will under all circumstances. What we can say is that if the source is supplying 1000 V and 1 A into the primary side of an ideal transformer, then:

• The current on the secondary side is \$1\cdot\frac {N_P}{N_S}\$ amps.
• The voltage on the secondary side is \$1000 \cdot\frac {N_S}{N_P}\$ volts.
• The apparent input impedance on the primary side has a magnitude of 1000 ohms.
• The actual load impedance on the secondary side has a magnitude of \$1000\left(\frac {N_S}{N_P}\right)^2\$ ohms.
• The input power on the primary side and the output power on the secondary side are both 1000 watts.

If you change the turns ratio \$\frac{N_P}{N_S}\$, the apparent impedance of the primary side will change, just as if you had changed the actual load impedance. The voltages and currents on both sides will change accordingly.

EDIT: Yes, the primary impedance (also known as the reflected impedance) depends on the turns ratio due to mutual inductance. The actual load impedance on the secondary does not -- it's the physical impedance connected across the wires, what you're calling "appliances". But just as with Ohm's Law, if you know the voltage and current on the primary side and the turns ratio, you can calculate the impedance on the secondary side. That's what I'm doing above.

To put it another way: the voltage applied to the primary side determines the voltage seen on the secondary side. The secondary voltage and the load impedance determine the secondary current. The secondary current determines the primary current. The primary voltage and primary current give you the reflected (primary) impedance.

Think of the reflected impedance as a Thevenin equivalent. If you connect a 22k resistor to the secondary side of a transformer where \$\frac {N_P}{N_S} = 10\$, then the primary side of the transformer acts just like a 220 ohm resistor.

See also this question.

EDIT 2: I'm not an expert on generators, but I'll try to answer your question anyway. :-)

As a rule of thumb, the voltage of a generator depends on its speed. Turning the generator at a constant speed produces a constant voltage (an AC voltage, in this case). When applied to a load, this voltage causes current to flow. The current acts as an electromagnet and puts a torque on the generator, opposing its motion. Overcoming that torque consumes mechanical energy. The mechanical energy consumed is equal to the electrical energy produced. (I'm ignoring friction and inertia, and only talking about the steady state.)

If you feed a constant amount of mechanical power into the generator, its voltage will vary based on the load resistance, keeping the electrical power equal to the mechanical power. But in practice, we normally want a generator to produce a constant voltage. So the mechanical power is varied to maintain the voltage.