Starter magnet coil

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Jpflex

Jpflex

Electrician big leagues
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Electrician commercial and residential
Hi I’m working with older contact starters for 3 phase AC motors. I was wandering that if the coil from series Connected start and stop buttons was 17.4 ohms (just the coil) and the coil was rated for 120 volts

And assuming power is delivered from lines 1 and 2 of A and B line phases (ONLY USING TWO PHASES OF THREE) then if I wanted to know how much current was traveling through coil

would I divide 120 volts / by 17.4 ohms resistance

Or 120 volts / 17.4 ohms x assumed power factor between 80 to 90%

Or 120 volts / 17.4 x 1.732 (3 phase although only 2 phases used)

Or 120 volts / 17.4 x 2 (2 phase)

Or A combination of these figures? I also don’t know what power factor would be if it matters?
 
Correction I meant multiplying voltage by 1.732 or 2 for three phase or two phase respectively but without watts or amperes known I have too many variables
 
Not sure if you can figure the math on it being purely resistive load since it’s alternating current.

And the coil is only using single phase of the three phase system so no sqrt of 3 to figure for current through the coil.

(If I understand your set up) control circuit being two tapped from two hots of your 3 phase circuit.)
 
The resistance of the wire making up a coil is almost useless if you are trying determine the current draw of the magnet coil.

As the magnetic force is created, the resistance becomes less of a component of the total impedance. Once the air gap in the magnetic circuit is closed the total impedance changes again, but the resistance is still a minor component.
 
Dsg319 said:
Not sure if you can figure the math on it being purely resistive load since it’s alternating current.
Click to expand...

Yes, in fact the inductive reactance of the coil is typically significantly higher than its DC resistance, particularly when the contactor has closed and the magnetic circuit has a lower "reluctance" than when its open.
 
Dsg319 said:
Not sure if you can figure the math on it being purely resistive load since it’s alternating current.

And the coil is only using single phase of the three phase system so no sqrt of 3 to figure for current through the coil.

(If I understand your set up) control circuit being two tapped from two hots of your 3 phase circuit.)
Click to expand...
No this is how circuit was ran prior.
 
jim dungar said:
The resistance of the wire making up a coil is almost useless if you are trying determine the current draw of the magnet coil.

As the magnetic force is created, the resistance becomes less of a component of the total impedance. Once the air gap in the magnetic circuit is closed the total impedance changes again, but the resistance is still a minor component.
Click to expand...
Yes as I recall now that the formula for impedance / inductive reactance Or equivalent inductive resistance for an inductor or coil is

2 x pie x frequency Henry or inductance L

XL = 2x pie x frequency x L (Henry)


so how would I find out what the Henry (WHAT THE HENRY!!) inductive reactance of a coil is so I can plug in value to find amperes?
 
Jpflex said:
Yes as I recall now that the formula for impedance / inductive reactance Or equivalent inductive resistance for an inductor or coil is

2 x pie x frequency Henry or inductance L

XL = 2x pie x frequency x L (Henry)


so how would I find out what the Henry (WHAT THE HENRY!!) inductive reactance of a coil is so I can plug in value to find amperes?
Click to expand...
The value changes based on the reluctance of the magnetic path. At what point of the closing/operating cycle are you trying to calculate?

But more importantly, why are you trying to do this?
 
Jpflex said:
Yes as I recall now that the formula for impedance / inductive reactance Or equivalent inductive resistance for an inductor or coil is

2 x pie x frequency Henry or inductance L

XL = 2x pie x frequency x L (Henry)


so how would I find out what the Henry (WHAT THE HENRY!!) inductive reactance of a coil is so I can plug in value to find amperes?
Click to expand...
You generally will never see the inductance of the coil specified anywhere. In addition, the inductance will be lower with the armature at rest with the magnetic circuit partly open and will increase when the armature gap is closed (the holding condition).
What will instead be given, as mentioned earlier, will be the VA drawn by the coil at the rated voltage. Divide the VA by the rated voltage too get the steady state current. Using the related impedance you can calculate the steady state current at different applied voltages, but you should not extrapolate to excessively different applied voltages.
Applying a much higher voltage may cause core saturation with greatly increased current, and a much lower voltage may not actually close the contactor, leading again to a higher than normal current.
If you are concerned about fusing rather than supply transformer rating, you need to allow for inrush current which is several times the steady state current.
 
The counter emf is what gives increased impedance with increased frequency or opposition to a change in current flow (offered by an inductor). That is why as you said with an armature slowing down there is less impedance (true).

This is because there is less back emf counter voltage being induced into the same wire inductor to oppose the change in current for each altering cycle and therefore impedance goes down while current jumps to the point of coil saturation you mentioned.

However, in my engineering book why give out these formulas if we can never use them?

Also if I divide VA of coil by line voltage, wouldn’t it be better to divide watts by voltage and power factor for amperes?
 
Look up the VA rating if the coil from the manufacturer. There will be two values: Inrush and Sealed. Inrush is the peak value when you initially energize it, Sealed is the steady state value to keep the contactor closed.

If a contactor is so old that the coil data is not available, it might be time to consider replacing it. They don’t last forever.
 
Ok no prob. Will have to search manufacturer for specs I guess
 
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