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Not true, all circuits are essentially the same, the equipment for medical equipment is just more sensitive to the capacitance and inductance that is inherent in the circuit. Even semiconductors have resistance, capacitance and inductance.
Also, where you were talking about Ohm's law above is somewhat correct, the added resistance of the wire will drop the current draw. However, 12 volts is not constant. The wire is in series with the lamp, this causes a voltage drop across the wire, therefore, at the lamp the voltage is not 12 volts, this is why the lights are dimmer. The lower voltage at the light causes the current through the light to be less, which generates less light.
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Absolutely correct. All electrical circuits operate on the same principle. The capacitance of the wire (as minimal as it is) doesn't affect this circuit significantly.
In my example the 12 volts is a constant for the entire series circuit. If you break it down to individual load devices (pretending that there was a 1 ohm resistor and the light bulb) then yes, you're correct, 6 volts to the resistor, and 6 volts to the lamp is the reason that the current flow is so low, if you prefer to calculate it that way.
However, Electrical law states that in a series circuit, the total resistance of the circuit equals the sum of the resistances of the individual load devices.
So, in my example, the wire was offering 1 ohm, and the lamp 1 ohm. Combined, the resistance of those two items in a series circuit is 2 ohms, which if you calculate it that way, 12/2 = 6 amps.
Either way, no matter how you look at it (6 volts divided by 1 ohm is still 6 amps, and current is the same no matter where it is measured in a series circuit so in this case you arrive at the same answer).
So now that I've typed out all of this to say that 6 / 1 = 12 /2 which are both equal to the exact same number, 6, I hope I am now making sense to you.
Your point about individual voltage drops across the wire and the lamp is noted, however, if you re-read my above post you'll see that I simply added the resistances and figured current that way which is perfectly acceptable and correct.
To whoever said I stand corrected about the capacitance of a wire, as I said above, whether or not wire has capacitance or not is irrelevent to this circuit anyway containing only wire and a light bulb.
Damn guys, it doesn't get any simpler than a circuit with one light bulb, a source, and conductors. This is like first week of electronics class stuff and you guys are bringing in stuff that your average student doesn't learn until late in the first semester or so. /forums/images/graemlins/rotfl.gif
Now if we were arguing about something more complex than this it'd be one thing, but having a bitch fest about the capicitance of a wire in a circuit containing nothing but a light bulb for a load device is absolutely insane. Looks like someone is simply reaching to tell someone else that they're wrong considering this is an insignificant factor in this circuit.
