Electronic – Current input and output of transformer

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.

Lenz's law is all about maintaining the magnetic field constant.

Let's assume Lenz's law is wrong in order to justify it for a transformer: -

So, if a current is taken from a secondary winding it will produce a magnetic flux and that flux would be additive with the flux produced by the primary and more secondary current would flow which produces more flux and the transformer would eventually destroy itself!

What happens is that the flux in a transformer (ignoring leakage inductance and DC resistance) remains exactly the same irrespective of how much load current is taken. I know this sounds non-intuitive but it is correct.

Maybe consider these following scenarios: -

This is a single turn coil around a core. A small magnetization current flows due to the inductance of the winding (for a power transformer the inductance is quite large because many turns are used). Now, imagine splitting that coil into two parallel coils: -

The magnetization current is shared between the two coils. Next, what if one coil were open circuited: -

You'd get primary and secondary voltages that are in-phase. If you don't believe that consider the repercussions in the second picture above when the two coils are connected. Finally connect a load to the secondary: -

Magnetization current still flows but the load current in the secondary produces a load current in the primary of equal and opposite value i.e. if turns are the same then ampere-turns are opposing leaving only flux produced by the magnetization current.