Can any one tell me the formula used to calculate the value of a single phase motor start cap, I know this involves the inductance value (L) of the coil, but if this is also not known, is there then a way to calculate the value of its inductance from its measured DC resistance ?
Motor Start Cap
- Thread starter housebasher
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Not really. The problem is some motors are wound different to others and one has timed connection of capacitors, Load dependent connection, centrifugal switch connection, permanent connection, and just winding with no capacitance and a mixture of them all. So unless one knows what system is used hard to work out. In fact many now use electronics and are three phase motors. Even the 12vdc fan in a computer is often a three phase motor with electronics to give it 3 phase. All this is hidden in the case so we as the user have little or no chance on guessing.
I have in the past tried to guess and often been short lived where winding has burn out. As to if winding starting to burn out made original capacitor fail, or capacitor failing had damaged wiring impossible to tell. But through this experience with exception of submersible pumps, if capacitor goes its whole motor to rewind guys.
I have in the past tried to guess and often been short lived where winding has burn out. As to if winding starting to burn out made original capacitor fail, or capacitor failing had damaged wiring impossible to tell. But through this experience with exception of submersible pumps, if capacitor goes its whole motor to rewind guys.
I know there different, but I thought if you knew the DC resistance there was then a method to calculate the XL at 50 Hz, and then from that you could calculate the inductance with the formula L = XL / 6.28f.
But then I am not a kitchen fitter with part P
But then I am not a kitchen fitter with part P
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As above, there's no relationship between resistance and inductance.
Here's the problem:- Consider a selection of different windings. one has ten turns of 0.5mm wire, one is 20 turns of 0.5mm wire. The third is 10 turns of 1mm wire and the fourth is 20 turns of 1mm wire. Which would you expect would have the highest resistance, and which would have the highest inductance?
Here's the problem:- Consider a selection of different windings. one has ten turns of 0.5mm wire, one is 20 turns of 0.5mm wire. The third is 10 turns of 1mm wire and the fourth is 20 turns of 1mm wire. Which would you expect would have the highest resistance, and which would have the highest inductance?
20 turns of 0.5 mm wire has the most resistance and inductance
10 turns of 1 mm has the least resistance and slightly less inductance to above
10 turns of 0.5 will have a resistance in between the other two but the least inductance
so.......... point well put ticklyt
Inductance is proportional to turns squared... Slightly counter-intuitive but in essence if you double the turns, there is twice as much magnetic flux and twice as many turns for it to act on, hence the squaring.
if 1mm and 0.5mm are diameters then
20x0.5 has four times the inductance of the other two
10x0.5 has half the resistance of the above
10x1 has one eighth of the resistance of the 20x0.5
if they are cross sectional areas then the resistances work out differently but you get the picture.
if 1mm and 0.5mm are diameters then
20x0.5 has four times the inductance of the other two
10x0.5 has half the resistance of the above
10x1 has one eighth of the resistance of the 20x0.5
if they are cross sectional areas then the resistances work out differently but you get the picture.
One of the best plain language explanations I have heard. Simple and accurate.
Thanks, I can see the point, the inductance would vary depending on the thickness of the wire and its winding method but the DC resistance would obvesualy remain the same.
I simply cannot understand why it wasn't immediately obvious to you that your idea could not possibly work - a straight piece of wire would have DC resistance, but b-all inductance.
What are you then?
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