# How does a domestic appliance “know” how much current to draw?

I have a small wall plug which takes 240V 50Hz AC from the mains, and can give me 500mA 5V DC. I suspect there is a transformer stepping the voltage down and a bridge rectifier converting the AC to DC.

BUT... the amount of current which flows... how is it controlled?

I know that if I plug in a 3KW heater, that the electrical cable gets quite warm from all the current flowing through it, but the cable on a phone charger does not.

Is it to do with every mains socket being a parallel load? i.e. if everyone on the national grid unplugged every appliance at the same time, apart from my charger, would my charger explode?

• If the entire load on the national grid were to be removed all at once, it's scary to think what might happen. – user28910 Nov 20 '13 at 14:36
• – pjc50 Nov 20 '13 at 15:13
• For "load removed all at once", compare what happens when a section of the grid goes down: en.wikipedia.org/wiki/Northeast_blackout_of_2003 – pjc50 Nov 20 '13 at 15:14
• This is a very common question/concept on the site here. I'm sure there are many duplicates. – JYelton Nov 21 '13 at 0:28

Electrical loads such as appliances have resistance. That, together with Voltage determines the current they draw (see Ohms Law).

Above the rated current, the wires in your wall plug transformer overheat and burn (if a fuse is not present).

The current provided by a switch-mode power supply on the other hand is generally limited by its design. Usually if you draw too much current, the voltage starts to sag.

Finally, if everyone unplugged their appliances at once, the voltage in the grid would want to go up, but the generation side will back off to keep it stable at the rated voltage. I'm told stories about the 3pm Tea-Kettle hour in England which used to be a stressful time for the power authority as everyone in the country plugged in their kettles, the power authority folks needed to balance the demand by increasing power generation and keeping the voltage at the nominal rate.

• So to hypothetically limit the current from the wall socket further, if I had a lossless voltage transformer which dropped it to 1V, then a resistor, then a lossless transformer which raised it up to 5V, then this would reduce the current from the outlet further (whilst keeping the same voltage of 5V)? – xxjjnn Nov 20 '13 at 15:44
• Well, no. With V=IR, if you keep the resistance the same, and current changes, the voltage will have to change (the second transformer draws more current if current on the output increases, the voltage across the resistor will drop. Limiting current is tricky, otherwises fuses would go out of business :) – MandoMando Nov 20 '13 at 20:16

Electricity is made of an invisible fluid, called charge. There's a force that acts on these charges, called voltage.

If you want to understand how this invisible fluid works, simply think of a system where the fluid isn't invisible.

Pipes are filled with a fluid, called water. There's a force that acts on this fluid, called pressure.

How does the sprinkler know how much water to squirt? If all the sprinklers in the world were suddenly shut off, except yours, would your sprinkler explode?

$V=IR$ is a good place to start answering your question, along with $P=IV$. Every device has an impedance which defines the current it will draw when you put a voltage across it. The voltage and current define the power it will use.

The appliance "knows" nothing; it's just a physical characteristic. Keep in mind that a rated current or power does not define the current , but is often times the maximum current that the device will use. It might mean the maximum you'd see when turning the device on (e.g. a motor, which is inductive in nature and will have a transient when turned on), or it might be continuous (e.g. an electric blanket is just a resistor in a blanket, and resistors have no transient response).

Put simply the voltage(V) provided is fixed at 240V. Your device has a particular resistance(R) due to its design. So the current it draws(I) can be found from Ohm's law: V=IR