# what determines the max current that can be drawn from a transformer's secondary coil? Answered

how could i figure out what the max current i can get out of a transformers secondary coil? example: a welding transformer

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here are my calculations:

winding ratio:

winding ratio = voltage secondary / voltage primary

voltage primary = 120v
voltage secondary = about 20 or less(i can rewind the transformer if i need a lower voltage)

ratio = 20 / 120

winding ratio = about 0.17

input current:

when i measured the resistance of the primary coil it came out as .5 ohms could this be right?

I = V / R

I = 120 / 0.5

I = 240 - this can't be right, there is no way this transformer is drawing 240 amps from the main. it has to be less than 20 amps cause its only on a 20 amp breaker

i am going to guess it is maybe 5 ohms because

I = V / R

I = 120 / 5

I = 24 which is alot more reasonable

so about 24 on the primary of the transformer

secondary current:

current secondary = current primary / winding ratio

current secondary = 24 / 0.17

current secondary = about 141 amps

the most the 12 gauge windings can handle is about 40 amps

is 40 amps all it will be able to draw? or will it try and draw 140 and overheat melting the secondary windings?

.  Figure out what gauge/gage wire you have. Strip the (enamel) insulation and measure with a dial caliper or similar. Compare measurement to an AWG table.
.  Look up ampacity for that gauge wire. May get lucky and find that figure on same page as above.
.  This will give you a ballpark figure.