Modelers Corner

Winding Coils

| November 2005

  • 11-05-023-Model-Corner.jpg

  • 11-05-023-Ohm-1.jpg
    Applying Ohm’s Law: V = voltage, I = current in amps, R = resistance in ohms.

  • 11-05-023-Model-Corner.jpg
  • 11-05-023-Ohm-1.jpg

Hello again, everyone. Well, my experiment building a coil, which I discussed in the September 2005 issue, caught the attention of model engine fan David Metz, Muscatine, Iowa. David, who seems to have a lot more experience on the subject of wiring, was nice enough to send me a long letter detailing what I need to know. This is good, basic information I know we could all benefit from. So here's the start of David's letter, and we'll finish his thoughts next issue. Is this hobby great, or what?

Ohm's Law

David Metz writes: Rusty, my opinion is that you did everything right. You had the right wire, the right number of turns, the right core and the right insulation between layers of windings. My guess is that your only problems are in your measurements. Read on:

You made the comment that you were attempting to reach a value of 2 volts on the contacts of the igniter using 12 volts on the coil. I think what you wanted to say is that you wanted to have a current of two amperes flowing through the coil and battery in a simple series circuit. The relationship between the voltage and current in this circuit can be defined with Ohm's Law (P = power in watts, E = voltage, I = current in amps, R = resistance in ohms):

R = E/I: Voltage divided by amperage = resistance in ohms
I = E/R: Voltage divided by resistance (ohms) = amperage
E = I x R: Amperage times resistance (ohms) = voltage

And for determining power:
P = I x E:
Amperage times voltage = power dissipated in Watts

So, with 12 volts across the coil (contacts closed) and a 6-ohm winding resistance, you will have a current flow of 2 amps.


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