Inductor calculations.

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I did an exercise today on my breadboard with an inductor that I made from a coil of 22 gauge stranded hookup wire wrapped in a helix coil around a 60d nail I found in my garage. The power source was a 9V battery. I took a measurement of just the battery fully open with my digital volt meter and found that I got a reading of .68A or 680 milliamps were being produced by the battery between its terminals in a circuit with little to no resistance (I hope my digital is not throwing me off here.) Upon connecting the 9V battery to my inductor I found that I was now measuring a fairly constant .04A or 40ma (from the same device so if my reading is off it will be proportionally so for all the readings). So I would have to say if I did the exercise correctly then my inductor is putting a serious load on the circuit.

That leads me to my question which is if I know how much current and voltage is in a circuit before introducing the inductor into the circuit how am I to calculate how much current and voltage is being eaten by an inductor assuming that I am talking about a commercially bought inductor with defined specifications like for example a 100uh radial coil rated at 5A (this maybe to much inductor for a small 9V circuit maybe one smaller).

How is the forward load found out if there is any change upon introducing a new inductor?
 

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  • #2
berkeman
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10,896
I did an exercise today on my breadboard with an inductor that I made from a coil of 22 gauge stranded hookup wire wrapped in a helix coil around a 60d nail I found in my garage. The power source was a 9V battery. I took a measurement of just the battery fully open with my digital volt meter and found that I got a reading of .68A or 680 milliamps were being produced by the battery between its terminals in a circuit with little to no resistance (I hope my digital is not throwing me off here.) Upon connecting the 9V battery to my inductor I found that I was now measuring a fairly constant .04A or 40ma (from the same device so if my reading is off it will be proportionally so for all the readings). So I would have to say if I did the exercise correctly then my inductor is putting a serious load on the circuit.

That leads me to my question which is if I know how much current and voltage is in a circuit before introducing the inductor into the circuit how am I to calculate how much current and voltage is being eaten by an inductor assuming that I am talking about a commercially bought inductor with defined specifications like for example a 100uh radial coil rated at 5A (this maybe to much inductor for a small 9V circuit maybe one smaller).

How is the forward load found out if there is any change upon introducing a new inductor?

Inductors are not often used in DC circuits (there are some exceptions). What is the resistance of your inductor? Also, when measuring currents with a DVM, keep in mind that the DVM introduces a resistance of its own, which is how it measures the current.
 
  • #3
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Alright so I'm looking at the specifications of an inductor right now and it says 100uh 3A. Does that mean I shouldn't run more than 3A through it?

http://ezoneda.en.alibaba.com/product/610258563-213323747/100uH_100UH_3A_font_b_coil_b_font_wire_wrap_toroid_inductor_font_b_choke_b_font_new.html [Broken]

Would a coil like this one have some measure of resistance? I would expect it to have little to no resistance by the look of things but I could be wrong I think. My application is dealing with power that comes out of a bridge rectifier from an AC power source. Supposing 1200ma is coming out of the bridge rectifer alternating between 0V and 30V @ 60Hz and then I introduce the inductor like one in the link above. How then should I know by math what is happening in the circuit? I only have the 100uh (Henrys) and 3A which I take to be the max rating of the coil.
 
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  • #4
vk6kro
Science Advisor
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When you use a DC supply, the inductance of the inductor is only important for a second or so when you first connect it.

After that, the resistance is the main concern.

If you don't consider the meter resistance for a moment, you put a short circuit across the 9 volt battery and got a current of 680 mA.
So, you can calculate the internal resistance of the battery.

Internal resistance (+ meter resistance ) = 9 volts / 0.68 amps or 13.2 ohms.

Now you put the coil you wound on a nail across the battery and got a current of 40 mA

The total resistance (including the 13.2 ohms above) would be:
Total resistance = 9 volts / 0.04 amps or 225 ohms.

Subtracting the 13.2 ohms this gives a resistance of 211.8 ohms for the coil.
This seems an unlikely value as the wire you used should have quite a low resistance per foot and you can only wrap a few feet of wire on a nail

I measured the resistance of a coil like the one you showed. It was less than 0.1 ohm.
So, it would be close to a short circuit for DC

For 60 Hz AC it would have a reactance of 2 * ∏ * F * L
So for a 100 μH coil, the reactance would be (2 * 3.14159 * 60 * 0.0001) or 0.037 ohms.

So, it wouldn't have much effect at 60 Hz.
 

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