Right, your real-life circuit will not need a resistor in the supply leads.
Hello Brad,
What about power rating of the ~5R resistor?
What about voltage rating of the inductor?
What about wiring and mm² concernings of wiring? For ~3Amps current drawing, thickness of wire loading the 220/50VAC should not be less than 3mm² , and interconnection between cap and RL circuit should also have a significant thickness.
I mean, soldering the total circuit using 0.8mm soldering wire wouldn't be a good idea. Right?
Solderng wire sounds like the wire that melts when you touch it with the soldering iron tip.I mean, soldering the total circuit using 0.8mm soldering wire wouldn't be a good idea. Right?
In real circuit you cannot connect across any supply directly a coil; plus how can a coil draw 3Amps without manifesting at any operation? The coil is not driving a shaft to turn, nor transforming any voltage, how is it supposed to draw 3 Amps doing nothing.
The only relevant restriction is that the choke must be operated within it's voltage and current ratings. If "supports 50volts" means that it's designed to work up to 50 VAC, you shouldn't exceed the voltage level much to avoid core saturation.
Inductors are commonly referred to as chokes. They restrict AC.
The 50mH inductor gives the right amount of choke action, so that it has a reactive impedance of 15.7 ohms at 50 Hz (from the formula XL = 2 * Pi * f * L).
An ideal inductor has only a current rating because a choke cannot stand a voltage because it has a low resistance.
It will have enough reactance at 50Hz and that is the factor that will restrict the current. It will have only around 15ohm reactance that may not be sufficient at 220V (it will go up in flame) but at 50VAC it will take 3-4A current (that is still too much for the small size).
How can I get a bigger one? What about connecting the secondary of a 220/50 transformer to a source of 50VAC? With 5Amps rating, can I see 3 amps circulation and correction of phase using pre-calculated caps?
Yes, the secondary of a transformer is a suitable load. I was experimenting with a few transformers in my parts box. Normally these are for step-down, and carry a few A. However what I did was to apply AC to the secondary thru a variac. They admitted current, which I measured as voltage across a 1 ohm resistor rated for 10W.
You need a choke/transformer which can handle 100 or 200W. It needs to be much larger than your photo in post #369. Expect it to be a few inches on each side. Perhaps it can be a twin of your step-down transformer.
Ideally you would have an assortment of chokes/transformers to try in your lab.
I've applied 220VAC on the primary of this transformer, to know its L:
I've used Ohm's law to know its impedance Z.
XL = √(Z^2 - R^2)
Then: L = XL / 2πf >>>>>>>>> I got around 8H.
There still one problem: I do not know its inductance in Henry, and my multi-meter doesn't have an inductance function.
How can one know the inductance of a coil?
What if its inductance is larger/smaller than 50mH little much or too much? What are the consequences (beside that its impedance would have a different value).
So you were saying that using a variac (autotransformer) feeding the secondary of 5A rated 220/50VAC with/without capacitor would let me see power factor behavior?
Shall I try to feed it from the 50VAC and try by trial and error the replies for adding/removing caps?
But if the secondary Henry are tens, this experiment won't work because Impedance will be so much big, that only few milli-amps will travel from source to choke and thereby I won't be able to see current reduction using caps.
Right?!
R, Z, L, C... is transferred from primary to secondary (or back) with the square of the winding ratio.So for a 220/50 > n: 4.4 Therefore secondary inductance would be around 8/4.4 = ~1.8181 <<< This is too high and results accordingly in a too high XL = 2 * pi * 50 * 1.8181 , and 50 / XL = few milli-amps.
R, Z, L, C... is transferred from primary to secondary (or back) with the square of the winding ratio.
In your case: 4.4 ^2= 19.36
Therefore the secondary (theoretically) is 8H / 19.36 = 0,41H
(A real circuit deviates from the idealvalues)
A solid state relay?Are there any solid-state techniques that functions in the same way as an open-relay switch functions but with no moving mechanical parts?
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