Voltage across short circuit - Page 1
Forum section: electronics / projects

author: pulse
author_rank: Contributor
timestamp: May 31, 2016, 09:16:42 pm
content: Hi!I'm wondering how could I calculate voltage of short circuit, and I'm not talking about perfect short circuit, but about the real life one. Let's say I have transformer with high voltage primary winding and the secondary winding is one turn of 1mm2 wire and that wire is 30cm (1ft) long. Wire is made of copper, so it has about 0.0175 ohms of resistance. One turn represents 1V. If I connect both ends of this wire, to make a short circuit, I get 57A of current through it. (1/0.0175 ~ 57). There is also some voltage in this circuit, although it's "short circuit", but why? Because the wire has some resistance, you might say... But the wire itself is the only resistor in this circuit (it's even the voltage source)... So, shouldn't there be no voltage in this circuit? I just don't get it. I've been thinking about this all day. Also, if I'd want to use ohm's law to calculate voltage, I'd just get open circuit's voltage (1 volt), but there isn't so much voltage in closed circuit, so why is the calculated current still correct? Or is it not? Would I need to substract short circuit voltage from open circuit voltage, before calculating the current?Thanks!

author: Aodhan145
author_rank: Frequent Contributor
timestamp: May 31, 2016, 09:23:37 pm
content: The voltage just drops. The max current is exceeded and the voltage drops down to a small value, its still there, its still measurable. You can use ohms law to calculate it.

author: pulse
author_rank: Contributor
timestamp: May 31, 2016, 10:24:04 pm
content: Hmm, I don't think I can use ohm's law; V=57x0.0175, V=1V, but it can't be 1V, because this is open circuit voltage.BR, pulse

author: IanB
author_rank: Super Contributor
timestamp: May 31, 2016, 10:34:59 pm
content: Your problem is that Ohm's law alone is not sufficient to calculate the voltage profile around the wire. Your shorted secondary turn is a closed loop. If you start at one point and go around that loop, you know what when you get back where you started the total change in voltage must be zero. However, the voltage change implied by Ohm's law is not zero, it is one volt. So you have a contradiction.The solution to this conundrum is that current flowing through the wire is not the only source of potential gradients along the length of the wire. The interacting magnetic field from the transformer is also producing a potential gradient in the wire where it intersects the field, and this potential gradient balances out the voltage drop caused by the current in the loop.If you wanted to make an idealized lumped parameter model, you could consider an ideal 1 V AC source connected in series with your long length of wire. This is not quite the reality since the 1 V source is "spread out" along a section of your loop, but it may help you to think about what is going on.

author: ebclr
author_rank: Super Contributor
timestamp: May 31, 2016, 11:17:37 pm
content: You forgot that transformer have a primary and a magnetic circuit that can saturate, also the primary is connected to another place, that also can make influence,

author: Zeranin
author_rank: Regular Contributor
timestamp: June 01, 2016, 04:54:08 am
content: Quote from: pulse on May 31, 2016, 09:16:42 pmI see what’s puzzling you, but it’s simple once you get your mind around it.As someone pointed out, when you load a secondary winding, the secondary voltage droops for more reasons than just the secondary resistance, namely also because of primary resistance, and leakage inductance. But let’s not let that confuse the issue. Let’s consider and understand your hypothetical example where the primary resistance is zero, the leakage inductance is zero, so only the secondary resistance causes the secondary voltage to drop when loaded.In this case, you can regard the secondary as a ‘black box’, containing a ‘perfect’ 1V source with zero internal resistance, in series with the winding resistance, and the short circuit current is correctly calculated as V(open-circuit)/R.But as Ian eloquently described, that is not really the right model, for if it was, then you could break open the black box, and measure your 1V even when the winding is short circuited.The correct model, is that the induced secondary voltage, and the secondary resistance, are ‘distributed’ all the way around the winding loop.As an approximation to this, consider 100 voltage sources, each 0.01V, and 100 series resistances each 0.017/100 ohms, all in series and evenly distributed around the loop. So as you go around the loop, you alternately see a 0.01V voltage source, followed by a 0.17 mohm resistor, followed by another voltage source, followed by a resistor, and so on.When short circuited, no matter where or how you place your voltmeter probes, you will never be able to measure more than 0.01V, because the source voltages are constantly being cancelled by the drops across the resistors.Of course, a better approximation is 1000 sources and 1000 resistors, in which case you will never be able to measure more than 0.001V, though it is easy to see that there is still a total of 1V in the circuit, namely 0.001V x 1000. In the limiting case, of an infinite number of such voltage sources and resistors, the true distributed model, then you can never measure a voltage anywhere when the secondary loop is shorted, but that does not mean the voltage is not there. It’s just distributed with the winding resistance such that you can’t detect it.Clear?

author: pulse
author_rank: Contributor
timestamp: June 01, 2016, 01:31:38 pm
content: @ZeraninBut I assume if I'd measure voltage across two most distant points of the winding, I'd get some voltage, wouldn't I? Would that be 0.5V? Also, what if I shorted that winding through another transformer core, which would have another turn of secondary winding (not shorted); what would be the voltage across that winding? As far as I understand that correctly, there's always full voltage spreaded out in the winding, with only exception being winding with no resistance (so, no voltage across it, when shorted)Actually, the main reason I'm asking about voltage is to calculate power. So, if we take our transformer with one turn of 1mm2 wire as an example, with the current of 57A and the voltage of X volts, what would the power be? If I say that voltage is 1 volt and thus the power is 57W, is that correct? As we already know, the "real life" transformer itself is the reason for voltage drop on secondary winding. If transformer would be perfect there would be no voltage drop, right? So, if my transformer is capable of providing more than 57W, there should be no voltage drop, or are there some other factors?So, are there any ways to calculate that voltage drop?Thanks!

author: Siwastaja
author_rank: Super Contributor
timestamp: June 01, 2016, 01:40:25 pm
content: Quote from: pulse on June 01, 2016, 01:31:38 pmA transformer is not a binary step function of either working with 100% efficiency, or being completely overloaded. The voltage on the secondary drops with increasing current, and this is not perfectly linear either. Also, resistances of both the primary and the secondary are equally important, because any voltage drop in primary (due to current) is naturally directly reflected into secondary.

author: IanB
author_rank: Super Contributor
timestamp: June 01, 2016, 03:00:25 pm
content: Quote from: pulse on June 01, 2016, 01:31:38 pmIf your wire has a resistance of 0.0175 ohms and a current flowing through it if 57 A, then the power dissipated in the wire is given by the formula P = I²R, so:P = 57² x 0.0175 = 57 WHowever, this is not indicative of the sustained output power of the transformer. If you short the secondary winding of a power transformer it is usually the case that the transformer will overheat. The safe, continuous load on a transformer is typically much less than what you can obtain by shorting the secondary.

author: rsjsouza
author_rank: Super Contributor
timestamp: June 01, 2016, 04:20:20 pm
content: The way I see it is very similar tothis exercise: you see the electromagnetic force inducing the current, thus IanB and Zeranin's comments about using a finite elements method instead of a traditional equivalent circuit approach using a simple VOM.In more practical terms perhaps you would be interested to take a look at some references regarding methods to obtain transformer parameters?Hereandhere?

author: Zeranin
author_rank: Regular Contributor
timestamp: June 02, 2016, 03:10:10 am
content: Quote from: pulse on June 01, 2016, 01:31:38 pmNO!You are not paying attention. For a short circuited single-turn secondary winding, you will measure zero volts between any 2 points on the loop, including two diametrically opposite points.I explained why this is so, but to understand, you need to get a sheet of paper and a pen, and actually draw out the equivalent circuit, for example assuming 10 voltage sources and 10 resistances around the loop. For simplicity, let’s assume the open circuit voltage of this single turn loop is 10V, and the total winding resistance is 10 ohms. Therefore, each voltage source is 1V, and each resistor is 1 ohm. Go on, draw it!The current is 10V/10ohms = 1.0 Amps, so the voltage across each resistor is 1V. Draw each voltage source as a battery, and label the + and – terminals on each of these ‘batteries’. Each resistor will also have 1 volt across it, except the polarity will be opposite, so now label the + and – across each of the 10 resistors. Note that the 1V across each resistor exactly cancels the 1V of the adjacent voltage source, with the result that no matter where you put your DVM probes, at any 2 points on the loop, you will either measure 0V or 1V, but never any more than 1V. I’m happy to draw the circuit out and attach it to a posting, but I’m sure you can do it yourself and see what is going on.I’ll send another posting discussing how much power you can get from the secondary of an xformer, which apparently is what you really want to know about. For this, a simple lumped circuit is all you need, but it’s worth thinking about and understanding the ‘real’ equivalent circuit when the secondary is shorted, hence the previous discussion.