Melting a nail with high currents

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greypilgrim
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Hi.
I melted a nail using a transformer with 500 windings on the primary and 5 on the secondary and 220 V input voltage, but I can't make the numbers work out. I measured the output voltage to be around 2 V, which makes sense.

Trying to measure the resistance of the nail directly with a multimeter yields around 6 Ohms, which appears be way to high. With a different power supply I can get 10 A across the nail at around 1 V, which leads to 0.1 Ohms, but this would still only lead to 20 A with the transformer, which still seems way too low to make it melt.

What's wrong here? And why is the resistance measured with the multimeter so far off?
 
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greypilgrim said:
What's wrong here? And why is the resistance measured with the multimeter so far off?
Was the multimeter calibrated? Did it have lead resistance and contact resistance? Measure it by shorting the test probes, and check the zero calibration.

What is the length and diameter of the iron? nail ?
What is the length and diameter of the copper? wire, used for the secondary winding ?
 
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The diameter is 4 mm and the distance between the contacts of the secondary is 5.4 cm. Using the resisitivity of iron, I get ##4\cdot 10^{-4}\enspace\Omega## , which seems about right.
I also found a different isolation transformer with an input ammeter that showed 3 A at 200 V, so the output current should be in the hundreds A, which makes much more sense.

First I just pushed the probes of the multimeter against the nail, which gave very unstable values. Then I used crocodile clips. Could it be that most of the resistance is due to the contacts? Should I maybe file down the surface of the nail?
 
greypilgrim said:
Could it be that most of the resistance is due to the contacts?
Yes.

Most DMMs just aren't designed to resolve very low resistance. You would probably be better off using it to measure current, at a high current level, and then voltage and calculating the resistance.

Read about "4-wire ohmmeters" or "kelvin connections".