Using the water analogy, a 10 mm pipe will only allow a certain amount of water to pass, and it will depend on the applied pressure. If you double the pressure, the flow will double.
A certain material will require a certain voltage to flow a certain current. Quoting a nice answer in Physics.SE about gravitation:
Physics [Ohm's law] does not answer existential problems. It gathers data and observations and models them with mathematical equations and functions, and then can explain the data with the model and predict new observations.
The concept of back EMF works well to explain the relationship among speed, load torque and current in a DC motor, but it doesn't work as well for an induction motor. The preferred equivalent circuit is shown below.
R1 and X1 are the resistance of the stator and the part of the inductance that does not provide a useful magnetic field. Bm is the inductance that provides the stator magnetic field. Gc is a resistance that represents the losses in the stator's iron core. X2 represents the inductance of the rotor. R2/s is a variable resistance that represents the resistance of the rotor and also, most importantly, the the mechanical load.
At no load, the rotor of an induction motor turns at a synchronous speed, the speed of the motor's rotating magnetic field. The speed in revolutions per minute (RPM) is RPM = 120f/P where f is the power frequency (Hz) and P is the number of magnetic poles formed by the stator winding configuration. R can be any even number (poles are always north/south pairs). Usually P is 2 or 4, but 6 pole motors are not uncommon, higher numbers of poles are less common and usually found only in large motors.
When the motor is loaded, the operating speed decreases. The difference between the loaded speed and the synchronous speed is called slip (s). Slip is expressed as a fraction of synchronous speed. At rated load, the slip is generally 2 or 3 percent of the synchronous speed and s = 0.02 or 0.03.
When the slip is zero, R2/s in the diagram is R2/0 or infinite. Therefore, at no load, the current in the motor is determined by R1 and X1 in series with the magnetizing circuit, Gc and Xm. As the motor is loaded, the slip increases and R2/s decreases causing the motor to draw more current.
The diagram represents one phase of a three-phase induction motor. Two similar diagrams are used to represent a single-phase motor.
Best Answer
Well, "load" can be anything from a power to a mass if we're talking about what you can load in the back of your truck, so it's a fuzzy term. Its meaning is context-dependent.
OK. The maximum amount of current that can flow through a wire of a certain cross section (like 2.5 mm2) is determined by temperature rise.
If your 2.5 mm2 wire was insulated with enamel (withstands heat well) and in the airflow of a fan, it could carry 50 amps without trouble, probably a lot more.
The wires you'll put in your walls are insulated with PVC, which has a rather low melting point, and they are inside conduits with no airflow. Also the conduits are often inside in-wall insulation like fiberglass. Under such circumstances, they must be very conservatively rated for current, hence the limit for 2.5mm2 is 16A when using fuses, or 20A with circuit breakers, which are more accurate.
So, I was just saying that you define "overload" by selecting the circuit breaker current rating. It's up to you. It should at least protect the wires, and sometimes also the load. For example, if the load is an air extraction fan which draws 20W, it is wise to put in a 2A circuit breaker, because it will trip if the fan is clogged by dust and stalls, whereas a 16A breaker will let the fan catch fire.
Not really, consider a motor for example, it isn't a resistor so it will draw reactive current. If current and voltage are not in phase, you'll have more current that what you get by calculating I=P/V. There is also the power factor. In this case the circuit breaker and wires must be sized according to the current, thus the apparent power in VA, not in Watts.
Also, motors tend to have huge starting currents (which are not overload), huge stall currents (which will result in a fire if the breaker doesn't trip), and mechanical load-dependent current draw (the power factor also changes according to load).
So to define "overload" for a motor is a bit complex. It isn't just "overcurrent" because that happens every time the motor starts. So you'd pick a circuit breaker with an accurate current limit (to trip on stall) but a slow reaction time (to allow the motor to start) like this one:
Notice it has an adjustment knob, so you can set the trip current according to the motor's documentation.