# transformer as current limiting inductor

Given a transformer of known specification normally associated with a transformer (e.g. voltage of winding, volt-amps capacity, impedance
percentage), is it possible to go from that to figure the inductance to determine the absolute impedance and thus the maximum current it would pass when wired as an inductor? It would seem to me that this should be possible, because given these specifications, the maximum current when used as a transformer is known. But I don't know the relationship between that and open circuit inductance. Also, there are a couple possible ways to do the wiring: use one of the windings or use all of them in series with the appropriate connection order.
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| Phil Howard KA9WGN (ka9wgn.ham.org) / Do not send to the address below |
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On 5/6/07 6:56 PM, in article snipped-for-privacy@news5.newsguy.com,

You are talking about the first things you learn in a course on ac electric machines. Essentially, you can represent a transformer by its equivalent circuit. It a three terminal network. A tee of three inductors does a decent job. From the hot input, there is a primary leakage inductor connected to a secondary leakage inductor to a hot output terminal. From the junction of these inductors, there is another magnetizing inductor connection going to the common neutral of the network.
From this network, you should be able to figure out how to determine the inductor values from measurements at the terminals using shorts and opens on primary and secondary. You do have to throw in compensating factors for any step up and step down.
Bill -- Fermez le Bush--about two years to go.
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| On 5/6/07 6:56 PM, in article snipped-for-privacy@news5.newsguy.com,
| |> Given a transformer of known specification normally associated with a |> transformer (e.g. voltage of winding, volt-amps capacity, impedance |> percentage), is it possible to go from that to figure the inductance |> to determine the absolute impedance and thus the maximum current it |> would pass when wired as an inductor? It would seem to me that this |> should be possible, because given these specifications, the maximum |> current when used as a transformer is known. But I don't know the |> relationship between that and open circuit inductance. Also, there |> are a couple possible ways to do the wiring: use one of the windings |> or use all of them in series with the appropriate connection order. | | You are talking about the first things you learn in a course on ac electric | machines. Essentially, you can represent a transformer by its equivalent | circuit. It a three terminal network. A tee of three inductors does a decent | job. From the hot input, there is a primary leakage inductor connected to a | secondary leakage inductor to a hot output terminal. From the junction of | these inductors, there is another magnetizing inductor connection going to | the common neutral of the network. | | From this network, you should be able to figure out how to determine the | inductor values from measurements at the terminals using shorts and opens on | primary and secondary. You do have to throw in compensating factors for any | step up and step down.
How do you "measure" it without an actual transformer? If I described a specific transformer, could you tell me the equivalent Henries its primary and secondary windings have?
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| Phil Howard KA9WGN (ka9wgn.ham.org) / Do not send to the address below |
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You generally have a handle on the % Impedance which is the short circuit impedance as a % of the rated V/rated I A 100/1000V 10KVA transformer will typically be about 5% so that with the 1000V winding shorted the impedance seen at the 100V winding will be 0.05 (100)^2/(10000) =0.05 ohms If the 100V winding is shorted the impedance seen from the 1000Vwinding will be 100 times this or 50 ohms. In either case the current at rated voltage will be such as to let magic smoke out so you will be limited to rated current on whichever winding you energise- unless you like replacing stinky burnt out transformers. If one winding is open then the current at rated voltage on the other winding will be in the order of 2-3% of rated current for that winding (say 2 to 3A for the 100V winding at 100V and 0.2 to 0.3A at 1000V on the high Voltage winding. If you try to get more current at a higher voltage- you will but the current waveform will be distorted and excessive core heating will occur because of saturation.
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Don Kelly snipped-for-privacy@shawcross.ca
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On 5/7/07 7:31 PM, in article snipped-for-privacy@news2.newsguy.com,

You measure it the same way you measure resistance of a resistor without having a resistor.
Bill -- Fermez le Bush--about two years to go.
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| On 5/7/07 7:31 PM, in article snipped-for-privacy@news2.newsguy.com,
| |> How do you "measure" it without an actual transformer? If I described a |> specific transformer, could you tell me the equivalent Henries its primary |> and secondary windings have? | | You measure it the same way you measure resistance of a resistor without | having a resistor.
I've long since forgotten the color band codes. Transformers have them, too?
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| Phil Howard KA9WGN (ka9wgn.ham.org) / Do not send to the address below |
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Short circuit current equals full load current divided by p.u. impedance (4.7%IZ = 0.047). Same as for a utility transformer when you are determining available fault current at the secondary based on transformer rating and percent impedance. Note that 'current' is about the same as 'power' in p.u. because voltage is 1 p.u. (|S|=V*I=1*I=I).
If you have multiple impedances in series, add the let through kVA like you would add admittances. If the utility txf is 1 MVA with 4%IZ, secondary available kVA with infinite primary is 25000kVA. If your 600D:600Y txf is 1 MVA at 6%IZ, it's max kVA let through is 16667 kVA. 1/(1/25000+1/16667)000kVA is the total available at the 600:600 secondary. Here the 4% and 6% add up nice to 10% (and agree with 1MVA/10%000kVA) but that only works because I happened to use the same txf rating (1 MVA) thereby giving the p.u. a common base.
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Sure.
But remember how a transformer "works."
Without a load but with normal voltage on the primary, the current is a small fraction of any expected load current. That "excitation" current would be the current used with the input voltage to calculate the impedance.
It you expect a transformer to "drop" significantly voltage than it's rating, it will "saturate" and the current will only be limited by the winding resistance.
"inductors tend to have "softer" iron (in the magnetic sense) that requires much more ampere turns to saturate.
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