Little transformer question

[JoshG]

Say I wrap a toroidal transformer with 10 windings with as thin a wire as possible for a primary.

Then I wrap 100 windings of the largest diameter wire possible as a secondary.

I energize the primary with 120v AC with a 12 watt light bulb in series giving a 1200 Ohm resistance on the primary side.

Because of the larger diameter wire on the secondary, the resistance on the secondary is very low - say 1.2 ohms. The transformer steps the voltage up by 10x to 1.2kV because of the turn ratio. 1.2kV at 1.2 Ohms give 1.2 megawatts of power at the secondary.

Does this power exist? Does it self destruct? I'm just curious.

Peace,

Josh G.

[Farmhand]

Hi josh,
I'm probably not the best person to have a shot at this but, I think your primaries may not be able to saturate or induce the required forces to get usable current and voltage on the secondary to the degree you speak of.

I hope i'm wrong, at least I get your question back to the top.

Cheers

[citfta]

Hi JoshG,

Yes you can get 1.2 Megawatts from the secondary. All you have to do is figure out how to get 10,000 amps to flow through the thin primary wire without the primary wire melting. Power is volts times amperage. To get 1.2 Megawatts out of the secondary you have to put 1.2 Megawatts into the primary. Sorry, but your very thin wire won't actually allow you to do that. I hope this explanation helped.

Carroll

PS. Was this a test?

[gyula]

Hi Josh,

The biggest current that is able to flow in your 10 turn primary is defined by the 12W lamp you connect in series with it. i.e. 12W/120V= .1A i.e. 100mA
Whatever load you place across your secondary coil, the 100mA current in the primary can vary but a little, depending on instanteneous ratio of the primary coil inductive reactance to the 1200 Ohm bulb resistance. And the primary reactance depends on the secondary's load of course but due to the high transformation ratio even the extrem load changes get transformed down, seen from the primary side, so the result is the 100mA current changes only a few millampers.

rgds, Gyula

PS1: if you need further help in understanding this setup, please google series RL impedance calculation where R is your 1200 Ohm lamp and the L is the 10 turn primary coil inductance. Normally the latter ranges from a few hundred microHenry to a few milliHenry if you choose a regular ferromagnetic core for the transformer. Such an inductance gives but a maximum of few Ohms of inductive reactance at the 120V mains AC frequency, (ZL=2Pi*f*L) completely negligible wrt the lamp's 1200 Ohm.

PS2: In fact, the lamp with the primary coil in series constitutes a voltage divider where the input voltage divides in the ratio of the impedances. Suppose the primary coil's inductive impedance is 2 Ohms, (suppose you placed a certain load across its secondary coil which is transformed back to the primary and the (paralleled) result is 2 Ohms), this 2 Ohms comes in series with the 1200 Ohm lamp impedance, a huge divison ratio happens so that the primary coil can only see approximately .2Volt (2*120/1202= .199V)

[rave154]

gyula to the rescue....definitely worthy of the title "senior member"