Specific Mike Holt video example

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DerrickM

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Billings, Montana, USA
Hi guys,

I have purchased some of Mike's DVD packages in the past and I cannot find one of the theory disks that has the clip I am trying to find.
The clip shows Mike doing a demo using spools of 12 AWG wire, checking resistance and ohmic values, attaching the wire to a circuit, and calculating ohm's law to prove why it does not trip the breaker.

He also uses the spools of wire to show inductance and and attaches the second spool to a light bulb to show the induced voltage onto the light bulb.

THEN, he takes a large screw driver and sticks it into the spools to show the magnetic effects of the material of the shaft of the screw driver and how shutting off the electricity while it's still inside the spool magnetizes the screw driver.

I am teaching a class to the Boy Scouts of America in February and would really like to do this demo for them/ with them as it really brings together so many parts of basic electricity.

Any help tracking this down would be appreciated.
Happy New Year.

Derrick MacAskill
 
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161228-2515 EST

I looked at a roll of #16 copper about 500 ft long. My room temperature resistance was 1.82 ohms and measured inductance 9.2 mH. Thus, at 60 Hz the inductive reactance is relatively small, about 0.35 ohms.

With 28 V 60 Hz applied the current was 15 A. This is a calculated impedance of 1.87 ohms.

If you went to a 1000 ft roll of #16 copper and applied 120 V you will be pushing around 64 A, but not for long because the wire will heat quite a bit. I would not want to run the wire that hot.

You could run the experiment at a lower voltage. #18 wire is 6.4 ohms per 1000 ft. This operated at 60 V would be about 10 A or 600 W. Still a lot of power into a small volume.

I would not use a screwdriver for an iron core. It will get too hot from eddie current losses. Rather get a roll of small diameter iron wire. Possibly 0.05" diameter. Cut into lengths you want for your core length. Bundle together to make a larger diameter core.

You can do your own measurements and make logical judgements on what is feasible. Do your own study and research. You will learn more from that and provide a better demonstration to your kids (really students).

Since you are at a University there should be equipment and people to help you.

.
 
161229-1929 EST

In my sleepy state last night I made some mistakes.

Redoing some measurements today I get 9.4 mH for inductance. Close to my previous measurement of 9.2 mH. My big mistake was not using a magnifying glass to read my Shure reactance slide rule. 9.4 mH is close to 3.4 ohms at 60 Hz ( not 0.3 ohms ). 1.82 and 3.4 combine to form an impedance of 3.86 ohms.

Second mistake was my clamp-on Amprobe was too close to the the 500 ft coil. The stray field from the 500 ft coil affected the current reading.

I have switched to a resistive shunt for current measurement. Now adjusting for 10 A current my voltage reading is about 38 V. Good correlation with measured impedance from the DC resistance and inductance.

I found an old thread that referenced Mike's experiment. He apparently used a 1000 foot roll of #14. This is a resistance of about 2.53 ohms per 1000 ft from a table at 20 C.

I can only very approximately estimate the inductance as being 4*9.4 = 37.6 mH or a reactance of 13.6 ohms for the 1000 ft roll of #14.

Calculating an estimated impedance we get 13.9 ohms. At 120 applied to this coil the current is approximately 8.6 A. Internal power dissipation is about 200 W.

You should be OK if power is not applied for too long a time.

Find a fresh 1000 ft roll of #14 copper, and run some tests.

.
 
yikes. stay away from line v.
1k ft roll, sounds heavy to tote around.
variac --> step down xfmr (120 to 12v as ex) --> 12v incadescent light bulb (landscape bulb sounds good, etc).
turn up variac to get some light.
use a strong neodymium magnet to swipe around the step down --> watch the bulb
 
170102-1230 EST

I think you are fine with line voltage. But since you will have what at this point may be an unknown load (a 1000 ft spool of #14 copper wire) I would first get some parameters.

A 250 W incandescent bulb is a sufficient current limiter for a spool of #14 wire.

Connect a 250 W bulb in series with your your roll of #14.

Measured characteristics of a random sample of a 120 V 250 W incandescent vs voltage is:
Volts-Amps
10 -- 0.63
20 -- 0.81
30 -- 0.98
40 -- 1.14
50 -- 1.27
60 -- 1.40
70 -- 1.52
80 -- 1.64
90 -- 1.75
100 -- 1.86
110 -- 1.95
120 -- 2.05
A tugnsten incandescent in combination with a constant voltage source behaves part way between being a linear resistance and a constant current source. No matter what load you add in series with the bulb and the AC 120 V line voltage the maximum current will be limited to 2.05 A with this particular bulb. This assumes no energy source in said load. 2.05 A in a #14 copper 1000 ft roll will not produce enough heat at room ambient to cause any thermal problem.

With my 500 ft roll of #16 as the load the voltage and current of the coil in series with said bulb is 7.92 V and 2.02 A. This calculates to an impedancre of 3.92 ohms. Close to my previous calculations and measurements.

When you find a 1000 ft roll of #14 and run the experiment you can guess on about 4 times the voltage and and a somewhat lower current in the above experiment. Based on guesses from virtually no real data on the 1000 ft roll.

If the measured impedance is in the range of 10 ohms, then current when the coil is connected to 120 V is about 12 A. Safe for #14 in a coil for a short time.

Putting a 3/4" dia steel rod in the coil hole raised the coil voltage to 12.6 V and current dropped to 2.01 A. With your coil with more turns than my coil (higher inductance) the insertion of the rod should produce a visible change in the brightness of a 250 W bulb.

.
 
DerrickM said:
I have purchased some of Mike's DVD packages in the past and I cannot find one of the theory disks that has the clip I am trying to find.[/FONT][/COLOR]
The clip shows Mike doing a demo using spools of 12 AWG wire, checking resistance and ohmic values, attaching the wire to a circuit, and calculating ohm's law to prove why it does not trip the breaker.
Click to expand...


Derrick, Dereck here. Perhaps you might be confused?

1000 feet of 12 AWG copper wil easily trip a 20 amp breaker with 120 volts applied, or do nothing at 12 volts. I think you might have the video confused where Mike uses 12 AWG wire to connect to a ground rod driven into dirt, then connecting the orher end to a 20 amp breaker to prove Earth cannot be used as a conductor in NEC applications.
 
170102-1535 EST

dereckbc:

I don't believe DerrickM is confused.

1000 ft #14 copper has a DC resistance of about 2.525 ohms at 20 C. And yes 2.525 ohms with 120 volts applied to it would draw about 47.5 A and that should blow a 20 A fuse or breaker. But this 1000 ft of wire is in a small coil and has a moderate amount of inductance.

Based on my measurements of a 500 ft roll of #16 wire which has an inductance of about 9 mH I grossly estimate a 1000 ft roll of #14 has an inductance around 4*9 = 36 mH, and thus, an inductive reactance of about 14 ohms at 60 Hz. This alone without considering the DC resistance will not trip a 20 A breaker from a 120 V source.

To make a good judgement on the current we need an actual measurement on an actual 1000 ft coil.

Two results from a limited search.

http://forums.mikeholt.com/showthread.php?t=149817

http://forums.mikeholt.com/showthread.php?t=108695

.
 
170102-2236 EST

I found a 1000 ft roll of #16 that visually appears to be nearly full length. DC resistance = 4.1 ohms, table for 1000 ft is 4.016 ohms. So it is near full length.

-----------------------------

With the 250 W bulb in series the coil measurements at 60 Hz are:

Just sheet metal spool
55.8 V, 1.71 A, Z = 32.6 ohms, Xl calculated = 32.3 ohms, L calculated = 85 mH
Based on this test, then at 120 V current would be about 3.68 A, but that is not what is seen in an actual test. See below.

Sheet metal spool with 3/4" dia. steel rod inserted
77.4 V, 1.39 A, Z = 55.7 ohms, Xl calculated = 55.5 ohms, L calculated = 147 mH
At 120 V current would be about 2.15 A. Also see below.

Can easily show the dimming effect of the steel rod.


-----------------------------

Next I changed to directly connecting the 1000 ft roll to 120 V.
120 V, 6.49 A, Z = 18.5 ohms,

Note the gross disparity in the calculated impedance compared to the above at 55.8 V. Implies a non-linearity. My giess is the thin steel sheetmetal of the spool is having an effect. At the higher voltage more saturation of the sheet steel.. I was not expecting the spool to have this much effect. But also note that the inductance was higher than what I might expect.

With the air core coil and the steel rod inserted and 120 V applied.
120 V, 2.85 A, Z = 42.1 ohms.
The steel rod gets too hot to touch where it was inside the coil.


With some additional tests I still saw the non-linearity. It is real and roughly repeatable.


-----------------------------

1000 ft roll on steel spool connected with the source being metered by a Kill-A-Watt EZ. The question is can I get a rough measure of the power dissipated in the steel rod.

120 V input without the steel rod.
120 V, 6.89 A, 210 W, 827 VA, PF = 0.25 . Rdc from above is 4.1 ohms. From the 4.1 ohms and 6.89 A calculated power is 194 W. Not a real good coorelation, but I don't want tp study the difference. It is possibly related to the spool.

120 V input with the steel rod.
120 V, 2.97 A, 182 W, 354 VA, PF = 0.51 . Rdc from above is 4.1 ohms. From the 4.1 ohms and 2.97 A calculated power is 36.2 W. The great difference between the calculated copper losses here and the power input is the heat being dissp[ated in the inserted steel rod. Induction heating.

About 145 W going into the rod. The answer is yes I can see the power into the rod.


-----------------------------

Going back to the spool of 500 ft on a plastic spool and raising the spool a foot above my bench I got these results:

10.4 V, 2.64 A, Z = 3.94 ohms
20.5 V, 5.22 A, Z = 3.93 ohms
31,5 V, 8.05 A, Z = 3.91 ohms.

No particular non-linearity.

.
 
I've seen the video. As I recall, it eventually did trip the breaker but it took minutes. He constructed a transformer out of two spools of THHN and a screwdriver.
 
170103-0948 EST

From the measurements and calculations I have made I doubt that in the air core state with no secondary load and for a typical spool size that a 1000 ft roll will trip a 20 A breaker with a source voltage of 120 V. But a 500 ft spool would trip the breaker.

Adding a loaded secondary contributes a shunt impedance (parallel) to the impedance of the primary coil. How much loading is added is dependent upon the amount of magnetic coupling, and the secondary load.

As soon as a magnetic core is added to the magnetic circuit, then the inductance of the primary coil goes up and its component to primary current goes down. Series leakage inductance from primary to secondary goes down. If there is a load on the secondary, then its contribution increases the primary current. How things balance out depends on the improved magnetic coupling, the improved primary inductance, and the load.

We need to know what the details are for the components used in the video presentation. Can anyone find that video?

.
 
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