How To Design Solonoid Coil for AC Voltage

AI Thread Summary
To design a solenoid coil for 240 VAC, it is essential to calculate the coil's resistance (R) and inductance (L) based on the desired impedance of 10k ohms. The reactance (XL) is determined by the formula XL = 2 * π * fHz * L, where frequency plays a crucial role. Wire diameter affects both resistance and inductance, with a smaller diameter allowing for longer wire but increasing resistance and inductance significantly. An example calculation is requested to clarify these concepts, along with references for further understanding. Understanding these principles is vital for effective solenoid coil design.
bhavu21fri
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Hello Friends,

Can you help to solve the problem of my solenoid coil as before i created 12VDC coil & its working fine thanks for that help guys. but now i want 240 VAC solenoid coil. so can you please help me which SWG I have to select. how many turns have to select..i want the equitation how to calculate it.

Voltage is 240VAC
Ohms 10000

Thank You Friends
 
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The current through an AC coil is limited by the coil resistance, R, and the coil inductance, L, in series.
Zcoil = R + j XL The reactance, XL, is proportional to power frequency, XL = 2 * π * fHz * L.

If you halve the cross section area of the wire, you will be able to wind twice the length on the former, the resistance will rise by four times, (half the area and twice the length), the inductance will also rise by four times since the inductance is proportional to the square of the number of turns.

You have specified “Ohms 10000” so I assume you want Zcoil = 10k
You must calculate the resistance and inductance of the coil for different wire diameters and solve for the diameter that gives the minimum cost of wire. Then √(R2 + XL2) = 10k
 
Baluncore said:
The current through an AC coil is limited by the coil resistance, R, and the coil inductance, L, in series.
Zcoil = R + j XL The reactance, XL, is proportional to power frequency, XL = 2 * π * fHz * L.

If you halve the cross section area of the wire, you will be able to wind twice the length on the former, the resistance will rise by four times, (half the area and twice the length), the inductance will also rise by four times since the inductance is proportional to the square of the number of turns.

You have specified “Ohms 10000” so I assume you want Zcoil = 10k
You must calculate the resistance and inductance of the coil for different wire diameters and solve for the diameter that gives the minimum cost of wire. Then √(R2 + XL2) = 10k
Thank you for your replay,

I would like to appreciate if you explain with an example.. because i m new & its very hard to understood. or give some link or document for reference.

My bobbin length is 18mm & radius is 4 mm

Again Thank you very much sir
 
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