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Hey everyone. I'm fairly new to the hobby electronics world, and today I decided I'd build an electromagnet. I wound my coil around a steel bolt, and I used 3 Duracell AA (1.5V) batteries for a power supply. When I turned on the switch, I found that my beautifully wound coil took a little bit of prying to get it off of the metal base I was working on. I thought "Let's try 9 volts!" So I grabbed a Rayovac 9V battery that I had around and hooked that up in place of the 4.4V (measured) bank. The Rayovac measured to be ~9V. The electromagnet was noticeably weaker. When I tested my magnet with just one of my 1.5V batteries, it felt like it had similar strength to the 9V battery.

Can anyone explain what might be going on?

Short of that, could someone tell me how to measure the current coming off of each power source?

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## 2 Replies

Orngrimm (author)2012-12-27

Magnetic force is proportional to the current flowing. Thats the law. :)

First, let my (guess and) explain why the 9V is weaker than the 3x1.5V.
See, often inside a 9V-Block you find 6 AAAA-Batteries in series. They are really small and therefore the surface where the chemical reaction can go on, is quite small.
Now another law in batteries is that if you have the same chemicals but smaller surface, you can only deliver smaller currents.
And thats exactly whats going on here! Your little tiny AAAA-Style batteries simply fail to hold the voltage if your (i think qiute low-ohmic) coil is connected.
Low voltage --> Low current (Since the resistance of your coil doesnt change. see Ohms law).
You bigger AA-Cells can hold the voltage better than the smaller AAAA-ones and therefore deliver more current.
Now what will happen (by your observations) is something like the following:
The 4.4V may drop lets say to 4V if connected to the coil. Your 9V drop below that. Lets say 2.5V if connected to the coil. But for sure below what the voltage was with the 3x1.5V!

Edit: "Drop"? Yes. measure the open-voltage (Nothing attached to the batterie(s). That gives you Voltage_open
Now attach the coil and measure at the contacts of the coil. This gives you Voltage_load
The drop is simply Voltage_open - Voltage_load
This drop is caused by the internal resistance a battery has. And since your 9V has 6 tiny batteries in serial (often small bats have a high resistance too!) and the 3x1.5-Thing only has 3 in serial (the bigger the cells, normally the lower the internal resistance), you have at LEAST 2-3 times the resistance in your 9V-block than in the 3x1.5V-thing.

Back to the calc! :)
Lets assume your coil has a resistance of 1.5Ohm.
And since I=U/R you have:
3x1.5V: I = 4.0V/1.5Ohm = 2.666 Ampere. Quite some juice there!
9V: I = 2.5V/1.5Ohm = 1.666 Ampere. Lower current than the 3x1.5V-configration...

Now if i take a look at normal specifications, a 9V-Cell is often rated to around 200-500mA. So i think the 2.5V made up by me are quite generous... make that something in the neighborhood of 1.5V. That would give you 1 Amp which is still a LOT for a 9V-block.

Now i hope you understand why the 9V may give worse results than 3x1.5V
Now to the second part of the question: "How to measure the current?"
If you have a hand-multimeter, normally they also have a current-meter.
Simply Come from the +Side of the battery(s) to the Red port where "I" is labeled. From the black (often labeled "GND") you go to the coil. The other side of the coil goes directly to the -Side of the battery(s).
Be sure you switch the multimeter to Current-measure beforehand.
Now you can measure away! :)

However, if your meter does NOT have a current-meter, you may go another way:
If you can measure the resistance of your coil, do that. We use that resistance as measuring-device! :)
Now connect the battery(s) and the coil as usual and measure the voltage across the coil (So from one end of the coil to the other). If you get negative values, ignore that and - and think nothing of it. You simply had the battery or the voltmeter the wrong way... Nothing bad there. :)
Now with the known resistance of the coil and the (now) known voltage accross the coil, you can get the current by Ohm's Law:
Remember? I=U/R

I hope i gave some backgroundinfos and helped you to understand why more (unloaded) voltage is not always better if it drops lower than a lower (unloaded) other battery.

And here is something you can do, but should only do with proper savety-measures in place (like goggles, Outdoors and (thick) Cables at least 10 feet long, Thermal-insulating leather-gloves, ...):
get yourself a lead-battery. They are 12V and rated for enormous currents. 100 Amps? Easy bake! So lets say, your current is only limited by your resistance in the coil. with this kind of power, your coil will probably melt/vaporize! But if your resistance is enough to limit the current enough (lets say to around 4-5Amps) than you should get a few dozens of secs to play with an increasingly in temperature increasing, but quite strong magnet).

Also the Lantern-Batteries (Quite common in the US, but scarse here in Europe) noted by fellow verence should be able to deliver some current.

verence (author)2012-12-27

In theory, you are right, the magnet should stronger with 9V instead of 4.5V. In the ideal world of theory...
But you entered the real, non-ideal world of electrical engineering. Be welcome.

In the ideal world, your circuit is modelled by your wire with given resistance (defined by material, length and gauge) and a power source that supplies a fixed voltage. According to Ohm' law I=U/R, the current should double if the voltage is doubled. And as it's the current that produces the magnetic field, your magnet should be twice as strong. So far, so good.

In the real world, your battery is the problematic, non-ideal part (for your circuit, the wire can be almost seen as ideal). Any battery has something called internal impedance. It is more or less the resistance of the internal parts of the battery. It's like having an ideal voltage source and a resistor in series inside the battery housing.

Given the same technology (chemistry) of the batteries, as a rule of thumb you can say the bigger the battery cell, the lower the internal impedance, the higher the current it can supply.
I'd guess, you used a 4.5V lantern battery (3R12) made of 3 single cells and a PP3 9V block battery, made of 6 cells inside. Just by looking at those batteries and imagining the size of the cells inside, it becomes obvious the the internal resistance of the tiny cells in a 9V block is much higher compared with the bigger cells inside the 4.5V battery.

The effect? Just what you saw. The 9V battery is less strong. You can check that by measuring the voltage over the electromagnet. Where does the voltage go? It is lost inside the battery. If you leave the magnet connected longer, the 9V battery will warm up.