electric-currentelectrical-resistanceelectromagnetisminductancevoltage
Assume an ideal transformer, with no load/resistor attached on the Secondary (Open Circuit), it is said that the Primary would act as if it is an open circuit too, thus no current flows through the Primary.
My Question is how does the Primary 'knows' there is nothing connected on the Secondary? (and acts like an open circuit although they're not)
Here is from wikipedia a schematic of a transformer:
If V_S is open no current will flow through the secondary circuit. The primary circuit will have the current appropriate to the resistance and impedance of the simple, now, circuit.
This happens all the time when we have a transformer just sitting there, without anything plugged in at the secondary circuit. The transformer draws energy and gets heated up anyway. That is why we are instructed to disconnect unused electrical appliances, and not just switch them off.
Note: the ratios in the schema should be inverted to avoid spurious infinities due to the zero current in V_s
The step-up transformer boosts voltage on the secondary. Since the power must be preserved on primary and secondary, the current is reduced by the same factor:
$$V_P \cdot I_P = V_S \cdot I_S$$
The loads are rarely characterized as resistances. In reality they have some complex impedance. For example, motors are modelled as inductances, air conditioning devices even have negative impedance etc.
For transmission you should think of it like this: there is some power $P$ required by loads; if the voltage on the transmission line is $V$ then the current will be $I = \frac{P}{V}$. The total power output equals loads plus ohmic losses in the transmission line.
If the load is purely resistive as in your example, then power increases with voltage since $P_R = \frac{U^2}{R}$.
See related discussion about power transmission: https://physics.stackexchange.com/a/682024/149541
Best Answer
It doesn't. A transformer works by magnetic induction between the two coils, it doesn't matter whether the secondary is loaded or not. The primary, if connected to a source of current, conducts just fine.
The alternating magnetic field, created by the primary is always present in the system and the induction happens even in an open secondary. It just doesn't do any work.
Unlike what PhysicsDave said below, adding a resistor just anywhere produces different results. Adding a resistor along the primary reduces the voltage, reaching the induction coil by dissipating some energy as heat. While adding a resistance or load along the secondary makes use of the energy rather than waste it in the case of resistor in primary winding.
Also opening the primary winding's circuit with the secondary unloaded might causes spark to jump across the switch, cause by backflow of current due to self induction of the primary.
Induction in a transformer is a bit tricky I'll say, when a coil is energized and a magnetic field is formed around it, this energy seems to be stored somewhere in space within the field as potential. This energy bank energizes the electron in the secondary winding.
Now, if the secondary is not consuming the energy and the primary circuit is also abruptly discontinued, what happens to the potential energy in the field and the energy in the energized electrons in the secondary winding? It's not in a form of energy that can propagate out into space - heat, sound or electromagnetic -, so if it can't flow outward, then it flows back inward. If self induction never happens in a case like that, then we'll have a cloud of energy floating around in 3D space, and any conductor in its path get toasted (I mean a human).