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.
It's not that multimeters have a high impedance - just voltmeters do.
When measuring voltage, you connect the voltmeter (a multimeter in voltmeter mode) to the nodes between which you want to measure: it is a parallel connection. In order to measure correctly, you need the impedance of the voltmeter to be as high as possible so that very little current goes through it, so that it doesn't disturb the circuit (the current that was going through the, for example, resistance that you are measuring voltage on will continue to be almost exactly the same, thus the voltage drop will as well).
When measuring current, on the other hand, you connect the ammeter in series. This time, to prevent the device from disturbing the measuring, its impedance should be as low as possible, so that the circuit keeps behaving the same way (no voltage drops in the ammeter and the current flowing in that branch of the circuit stays the same).
Best Answer
The conversion from high impedance to low impedance is known as signal conditioning. It's necessary for most sensors in some way to match their output to the desired load.
For theses purposes, high and low impedance are a way of expressing how much load a source can handle while maintaining voltage (and therefore signal integrity.)
A low impedance load draws lots of current, so it needs to be paired with a low impedance source that can sustain it.
Some sensors generate a high signal voltage with very small current. This makes them a high impedance source. If connected to a low impedance load, there would be a significant drop in the source voltage. This is a bad thing because it's the voltage we usually want to measure.
A buffer (and/or other conditioning method) is placed at the sensor output to measure the voltage and reproduce it from a high current (lower impedance) power source. This allows the sensor signal to drive longer runs and larger loads without distortion.
National Instruments has a good tutorial and guide with theory and example circuits for the various methods of signal conditioning.