- Yes,You only need to have L x W = R (L = inductance, W = 2 x Pi x Frequency, R = resistance), you can play with wire lengh, wire section, wire resistivity, coil radius or simply add a resistor.Laurent DAMOIS

2012/3/25 Harvey Norris <harvich@...>Can I make an AC inductor with high resistance wire so that it's current lags the voltage by 45 degrees?

http://physics.ucsd.edu/neurophysics/cou…

implies that the inductor can have any phase angle negotiated by the reactance/resistance ratio where this will equal the tan of the phase angle,(pgs. 700,701) This being true for a 45 degree phase angle all that is needed is to make X(L) = R. When I pointed this out to my stubborn friend he still insisted that the current will still be 90 degrees out of phase on just the inductor alone, and cited numerous references on this, and continued to insist that I was looking at things the wrong way, as if R and L would have different actions imposed on them according to the series diagram that typically represents this circuit of a real inductor having resistance. HOW CAN THEY POSSIBLY HAVE DIFFERENT ACTIONS BETWEEN R AND L IF THEY ARE CONTAINED IN THE SAME PART? Hasn't he somehow taken the schematic diagram literally, so that even if an additional external resistor was incorporated in the schematic, it would be at a different phase angle then the inductor

itself. He is trying to say that no phase angle change would be on the external resistor, but the phase angle change across the inductor will still be 90 degrees out of phase. Even the cited text seems to make the same distinction when they state on previous page that the voltage across the inductor will be 90 degrees out of phase with the source voltage. I just dont get it at all!!! How can this be if R (int) and L are contained as the same part, and are only separated for schematic explanation. AM I confusing voltage being out of phase with source voltage, and resultant current being out of phase with the source voltage? If this is true can it be shown by two channel oscilloscope? I would think that unity of phase angle currents exists all along the schematic series chain, but if the phase voltage across the inductor alone is different from the source voltage, again shouldnt that same voltage phase difference also exist across the resistance contained

within the inductor? Wouldnt this just be impossible to measure since again R and L are contained within the same part??? If all this is true then should it be more proper to say that the current may lag the voltage by 45 degrees, but the voltage across the inductor will still lag the source voltage by 90 degrees? Shouldnt BOTH of these phase angle differences be the SAME thing? I am truly confused here about these explanations, any clarity would help.

http://answers.yahoo.com/question/index;_ylt=At0SsnOLp0WTkJqXLSZ17o_sy6IX;_ylv=3?qid=20120319035721AAlW9Je

From Physics Q & A

Answerer 2

You are right.

The equivalent circuit representation of a real inductor as an ideal inductor and a resistor in series, treats the model as black box ie you can't look inside, you can only measure at the terminals. This will indeed result in phase angles between voltage and current anywhere between the limits of 0 and 90 degrees.

Source(s):

Electronics engineer 20 years

Answerer 1

I think I see your problem. Consider a simpler example which you are hopefully familiar with – a cell (emf=E) with internal resistance (r). This is represented as a ‘pure’ cell in series with a separate resistor. Of course, it isn’t really, because the resistance is distributed throughout the whole cell.

Yet we can think of there being 2 ‘voltages’ – the emf across the cell and the pd (in the opposite direction to E) across the (series) internal resistance. In reality there aren’t two separate voltages because the ends of the ‘pure cell’ are in fact the same as the ends of the resistor (the cells' terminals).

Treating the cell and r in series is just a ‘model’. It works because in the model:

- the cell and r have the same current;

- the terminal pd = (E–Ir) which is the algebraic sum of emf and the pd across r, sometimes called the ‘lost volts’.

This is the same as the behaviour you get for 2 physically separate series components.

So the real cell can be treated as a pure cell plus series resistance. But you can never directly measure the emf or the potential difference across r. So you mustn’t take the series representation too literally – it is a model and it works mathematically.

A very similar situation holds for the inductor:

- the voltage across the inductor corresponds to the emf of a cell;

- the wire’s resistance corresponds to the internal resistance of a cell.

A real inductor, with some resistance, can be treated mathematically as a pure inductor and a series resistor. You can get a phase angle of 45⁰ as you describe (but you wouldn’t want to do it this way, as we try to avoid wasting heat when designing circuit).

From Engineering Q&A

http://answers.yahoo.com/question/index;_ylt=Aio5svB.rhXEp2h0NaGKYwrsy6IX;_ylv=3?qid=20120319035314AADYRqP

Answerer 1

Your friend is incorrect. If what he were saying was true, then impedance analyzers and RLC bridges would not work. The Q of an inductor is measured by comparing the voltage across an inductor with the current going through it. If what your friend thought was actually true, all inductors would be ideal, with infinite Q. Just ain't so.

Answer 2 is so stupid I do not repeat it.

Answer 3

The phase angle in the inductor is due to its internal resistance and L, and while calculations (based on sine-wave) split the two components for the purpose of analysis, they are not so in reality. There is one current in a series circuit, with a certain phase difference as you say, to the source voltage. It is between 0 and 90 for a combination R and L, and 0 degrees for a pure resistor. If there is a resistor and inductor (almost ideal) in series, yes the voltage across the inductor is not in phase with the source, and the current in the inductor is 90 degrees out of phase with the voltage across it.

The equivalent series or parallel R for a non ideal inductor can be measured using a sine-wave test frequency (single frequency). Calculate the impedance |Z| using V/I, and then with the voltage to current phase angle (theta) calculate the X and R. All the losses are lumped as equivalent series R.

Rs = |Z| cos(θ)

Xs = |Z| sin(θ)

Here is a reason to try simulation. Down load LTspice, a free simulator. Draw a simple schematic with voltage source = 1V sine-wave, and f = 8000Hz. The series resistor is 740 ohms, the inductor 14mH. Do a transient analysis for 30ms. Monitor the current in the resistor and inductor, and the voltage across the inductor and the source (put the inductor on the grounded side). Note that the circuit must have a reference ground (one side of the source). If the two currents are out of phase, reverse the resistor or inductor. These are just values that will give around 45 degrees phase shift. You can demonstrate all the points in an few minutes once you are familiar with it. The display is like an oscilloscope. You can try 50Hz source and adjust the inductor and resistor to suit.

Inductor start single phase motors rely on the different resistances of the start and run windings. It is for this reason they have lower starting torque, because it is not as close to 90 degrees as a capacitor start can achieve. The start winding is switched out once it is running. These were popular in the days that capacitors were more expensive.

Answerer 4

In an ideal inductor the current lags the voltage by 90 degrees. In an RL circuit the phase lag will depend on the component values and the frequency. It's not complicated and doesn't take 250 words to explain it.

Pioneering the Applications of Interphasal Resonances http://tech.groups.yahoo.com/group/teslafy/