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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?

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  • \$\begingroup\$ did you at least read the wiki page. There must be enough inductance to be high impedance \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 5 '16 at 22:58
  • \$\begingroup\$ I'm sorry I don't know what you mean. Wiki page? \$\endgroup\$ – Trevor Dec 5 '16 at 22:59
  • \$\begingroup\$ What @TonyStewart.EEsince'75 said, plus there must be a high enough saturation current (related to core volume) so that it can withstand the volt-seconds impressed across it without saturating. \$\endgroup\$ – John D Dec 5 '16 at 23:00
  • \$\begingroup\$ Ok so how does a transformer become saturated? \$\endgroup\$ – Trevor Dec 5 '16 at 23:02
  • \$\begingroup\$ I think John meant enough flux to couple the windings not saturate it for normal use. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Dec 5 '16 at 23:29
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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.

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  • \$\begingroup\$ Does the secondary affect the inductance? \$\endgroup\$ – pjc50 Dec 6 '16 at 10:45
  • \$\begingroup\$ @pjc50 I'm not sure if you are prompting me to add something to my answer (given that you are a long term member and might have spotted some error). Taking the comment on face value, the magnetization inductance of the transformer (primary winding) remains constant and is unaffected by the secondary (loaded or unloaded) but I feel you probably already knew that! If you were to ask if the secondary current "appears" to affect the primary inductance then yes, certainly it appears so. \$\endgroup\$ – Andy aka Dec 6 '16 at 10:50
  • \$\begingroup\$ I was genuinely unclear on that as my EM theory is rusty and it seemed like it might be relevant. \$\endgroup\$ – pjc50 Dec 6 '16 at 10:54
  • \$\begingroup\$ Hopefully I've covered this then? Ignoring losses and leakage inductance, a real transformer transfers power to the secondary and also takes an extra current (through the primary) that is due to the raw inductance of the primary not being infinite. \$\endgroup\$ – Andy aka Dec 6 '16 at 10:57

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