# Inductor calculations.

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?

berkeman
Mentor
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.

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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vk6kro
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.