Wall outlet current draw

I have built a transformer, that will take 120v, and step it down to 2.5v. I used all the right parts, and the correct rated wire for this. My only issue, is that when i connect the primary coil, to a call outlet, at 120v, it draws more current then my 5A fuse can handle, and the power goes out in my home. I need to get 1.5 amps, @120v going through my primary coil, in order to get the desired output on the secondary. I was wondering, how things "draw" current from the wall, as well as if i where to use a resistor, wouldn't it still draw too much amperage, and just dissipate it in heat, to give the desired output (i cannot use, because it draws more than 5 amps). i could not find a resistor out their, that could handle over 100 watts, so i was also wondering how things like vacuum cleaners only draw a small amount of current. Would i need some sort of current regulator (if that's even a thing)? if you had a light-bulb, and a battery, would the circuit the light-bulb is attached to drain less current, if their was a resistor, and thus having the battery last longer, or would it draw the same amount, and just dissipate energy through the resistor in the form of heat, to give the light-bulb its self less current?

sort by: active | newest | oldest
1-10 of 34Next »
gmoon1 year ago

It's complicated, but this stuff can be understood...

Any device has characteristics, specs, etc. This includes a rated operational voltage, and a specific amount of current that's drawn at that voltage (it may not be fixed, as some devices such as motors draw more current when under physical load).

We could say that the "nature" of the device dictates the amount of current required. A 6V motor generally draws more current than a 6V lamp. If the device does more work, it will draw more current. Current draw is a characteristic of the device itself.

We can learn a lot with Ohms Law. Let's look at the specifics of your case: a 5A fuse is pretty small. When I had a fusebox, we'd normally use 15, 20 and 30W fuses. Power (wattage) is V * A, so the maximum wattage your house circuit can handle is 120 * 5, or 600 watts. An unusually small amount of power capacity for a household circuit.

Now, look at your transformer--a good capable step-down transformer will convert voltage with more than 90% efficiency (somewhat higher, generally).

So, what is your 2.5V device, and how much current should it draw? Does it sound right to you that the 2.5V device and the transformer draw more than 5A @ 120V (600W)? Or is there a problem somewhere, such as with your transformer...maybe a short?

(It's somewhat possible to limit current, but Ohms Law WILL be satisfied, no matter what. The electrical energy is all converted to heat (eventually), and this too cannot be cheated.)

merlinj (author)  gmoon1 year ago

The 5A fuse is not the one in my fuse box, my apologies, it is the one in the power strip I am using. I am using the power strip because it has a 5A fuse. It is in place both because I do not want my transformer drawing more current then my house circuit breakers can handle, as well as that my transformer can not handle anything above 5A. so for the circuit is basically the wall outlet (power source), and and inductor (primary coil). How I would have set up my circuit, is to have a 80 ohm resistor (resistance to give me the 1.5 amps) and then the inductor (primary coil) the only issue with this is that I would need a resistor that could handle at least 180 watts, at 80 ohms, which does not exist. Even if it did, wouldn't my circuit still draw over 5 amps of power before it reached the resistor, and the resistor dissipates the energy as hear bring the amperage at my inductor (primary coil) to 1.5 amps? how can things such as phone charges, that are very small have a resistance so great, at a tolerance so great, for their size (resistors that can handle high wattage are huge). How then if I don't use a resistor could I limit the current?

gmoon merlinj1 year ago

Unless I misunderstand, what does adding a resistor to the primary (in series?) have to do with the load on the secondary? And the total current draw?

I.E., why do that?

You may drop the voltage across the primary, but if you're trying to adjust the output voltage like that, you're using the wrong transformer for your application. Trying to change the secondary (voltage or current) with a series resistor/inductor pair primary probably isn't going to work how you think. If you measured the resistance of the primary, the inductance (AC resistance) of the transformer doesn't equal it's DC resistance. In all likelihood, your transformer can't work with any efficiency like that -- it's being prevented from doing it's function.

To muddy things more, devices in series (like an inductor/resistor pair) always draw the same amount of current.

Again -- can't tell from your description if the resistor is in series with the primary...

Remember, whatever current your rig is drawing depends on the load, not the supply. You don't (can't) control the current draw with the power supply (well, under some circumstances, but Ohms Law can't be violated and that generally means the voltage is dropped to compensate) -- a power supply amperage rating is it's capacity only. Voltage you can set "supply side" but not current.

Seems some assumptions are being made here about how to reach your goal (which itself is still foggy). Information still incomplete here...

merlinj (author)  gmoon1 year ago

I just want to say thanks for all the help so far, it has been very helpful to me. My issue is not getting the correct current in the secondary coil of the transformer, my issue is that the primary coil (the one directly attached to the wall), draws more current then my 5A fuse can handle. This I believe is because the resistance between the two leads on the primary coil have a resistance of less than 1 ohm, and thus should draw more than 120 amps. I don't understand how the load on the secondary coil, can effect the amount of current going through the primary coil, because they are two separate coils of wire, that are not attached to each other in any way. It is because the resistance is so low, that I said I would put a resistor in series with the primary coil, to increase the resistance, and thus decrease the current. The main concept I don't understand is if my load has a resistance of less than 1 ohm, and my voltage is 120v, how is it possible for it to draw any current less than 120v? doesn't that break ohms law, because isn't the current determined by the voltage divided by the resistance no matter what?

gmoon merlinj1 year ago

There's your first flawed assumption: that the DC resistance of the primary is the working impedance. It's not.

Transformers are essentially "electromagnetic machines" with nearly ideal characteristics. Their primary impedance changes with the load (on the secondary), drawing more current from the source only when a larger load is connected.

Impedance is "AC resistance" and for inductors is dependent on the rising and falling magnetic flux in the core (and other stuff like frequency). The impedance (working resistance) with AC is much higher than you think, in fact that's not present at all with a DC test. Brings home that fact that transformers only work with AC...

So... The impedance of the primary drops when a load is applied (to the secondary); the greater the load, the more the impedance drops. Devices in series always draw equal current; as your transformer tries to work, your series resistor steals an equal amount of current -- the harder the transformer works, the more the resistor takes. AND it's preventing the transformer from performing it's task efficiently. This may be limiting the current on the secondary, but by multiplying current on the primary.

Assumption two: that a way exists to limit current to a motor that's performing a large amount of work (on a household circuit, no less).

Yes, you can limit the current. But it cannot do the work without the current. Work is work. It's not free.

Volts and amps, amps and volts. Put them together and you get power. It's unusual for a large motor to be low voltage/high amperage, even though it can perform work equal to a high voltage/low amperage motor. Why? Because high-power transformers are as expensive as the motor itself. Easier to work with existing voltage supplies.

Was there a custom PS for this motor, like a dedicated engine/generator? If so, replacing that with a transformer might be more expensive than replacing the motor....which is why you made your own transformer....

I (we'd) like to see your DIY transformer! Sounds cool, even it it's not equal to the task... Good luck, BTW.

merlinj (author)  gmoon1 year ago

Thanks, I did not realize the difference between impedance and resistance. Is their a way to calculate the impedance of a circuit? My apologize for not saying what I was connecting it to I am connecting it to a Bridge rectifier, and the output of the transformer is actually like 5v, so that the bridge rectifier brings it down to around 2.5 volts dc. Even if I connect the transformer with nothing on the secondary coil (aside from my voltmeter), it blows the fuse. I think this is similar to putting a paperclip in the wall outlet, accept the wire can handle more than 5A. Shouldn't it draw very little current on the primary coil if the secondary coil has nothing connected to it? and isn't connecting the primary coil to the wall outlet, the same as just connecting the two wall pins with a long wire (as in it draws the same amount of current)? So the load on the secondary coil determines the current through the primary coil? and if so why if their is no load on the secondary coil does the fuse blow? Thank you allot everyone for your help its really appreciated.

gmoon merlinj1 year ago

Yeah, the impedance can be calculated theoretically or measured directly. But as discussed above, the load on the secondary creates a "reflected impedance" on the primary. So the inherent impedance of a transformer doesn't matter much compared to it's impedance when loaded.

You are correct, a properly designed transformer shouldn't draw much current without a load. Think of the magnetic flux field that builds (and collapses 'cause it's AC) as a force that resists current flow. Convert some of that field (induce current) back into electricity on the secondary, and the impedance on the primary drops (begins to drift toward the DC resistance value).

I think you have a problem somewhere in the transformer. Maybe the secondary has a short. A well-designed power has closely coupled mutual inductance between the primary and secondary. Maybe there's a design issue.

In any case, your 80 ohm resistor shouldn't be necessary. Without a load drawing current through the primary, not much should flow. If it does, you'll have to track down the problem...

merlinj (author)  gmoon1 year ago

Thanks, but should connecting the primary coil with nothing on it be the same as basically putting a paperclip in the wall, and directly connecting the hot wire with the other lead? and doesn't that draw more than the circuit breakers in your house handle?

gmoon merlinj1 year ago

You weren't wrong about an 80 ohm resistor drawing about 1.5A with 120V... But the 5A breaker tripped anyway, which shouldn't have happened, if our assumptions are correct. Something isn't quite right (already). And I don't think that resistor is helping.

I'd say remove the resistor, and anything connected to the secondary and try it. The breaker should prevent any serious issues, and it's backed up with the breaker in the circuit. The circuit breakers in your house are there to prevent the house from burning down (if someone sticks a fork in the outlet).

If it trips, there's a problem somewhere within your transformer. Additionally, maybe test the breaker in the strip with the 80 ohm resistor only. Do that first, if you don't trust it.

And seriously be careful around this stuff. We haven't stressed safety here. 120V mains are dangerous!

merlinj (author)  gmoon1 year ago

Im sorry, did not test it with a 80 ohms resistor, because I do not have one that can handle 180 watts, they are large and expensive. However, without any sort of resistor, or load on the secondary coil, the fuse blows anyway, so does that mean that their is something wrong with the transformer? (if so what), and isnt connecting just a primary coil (basically just a long piece of wire) similar to putting a fork into the wall (in how much current it draws)?

1-10 of 34Next »