Constructing a transformer: how to limit the primary side current?

Question

I am confused on how a transformer works. I know it uses a changing magnetic field to induce current into another coil that has a set turn ratio to achieve a specific voltage.

My question involves the current draw of a transformer. I understand that a wires resistance is very small and coiling it does not change its resistance. So how does the transformer not trip the fuse every time it's plugged in.

I have tried creating a primary coil before and have only had two results: I trip the fuse, or my coil starts on fire...

Could someone explain to me how I am able to make a coil not draw too much current?

Answer 1

I know it uses a changing magnetic field to induce current into another coil

No, it doesn't do that; the changing magnetic field induces a voltage into the other coil. Any current in that coil is due to that induced voltage driving current through any load connected to that winding. Current cannot be induced, only voltage.

So how does the transformer not trip the fuse everytime it's plugged in.

1 turn of wire wound on a bobbin might have 1 uH inductance. At 50 Hz, the impedance of that 1 turn is 0.000314 ohms (an inductor's impedance is \$2\pi f L\$). Clearly that is far too low to connect to any 50 Hz AC voltage source. So you wind more turns and what you find is that inductance increases as the turns are squared so, ten turns yields 100 uH inductance and 1000 turns yields 1 henry of inductance.

1 henry has an impedance of 314 ohms at 50 Hz and, with a little bit of generalistic hand waving and over-simplification, a transformer primary aims to be about 10 henry (3142 ohms).

If you applied 230 V 50 Hz to this primary winding you would find that the current is about 73 mA and won't cause a fire.

So, it's got nothing to do with the coil's resistance. It's all to do with getting enough inductance.