So in the case cited, an improvised dropping resistor could be the stated amount of number 10 wire.
2.
Even if your MHD cell were shorted, with the stated dropping resistor, the current would not increase much.
3.
You could measure the current with an ammeter in series, or with a voltmeter across the dropping resistor and use Ohm's law.
4.
If the 9 volt load is a constant resistance a single dropping resistor might work, at the cost of the energy dissipated in the dropping resistor.
5.
If the 9 volt load is a constant resistance a single dropping resistor might work, at the cost of the energy dissipated in the dropping resistor.
6.
If the battery produces 12 volts exactly and you wanted to reduce the current to 30 amps, you must have a dropping resistor in series with your . 001 ohm resistor, to drop the voltage from 12 volts to . 03 volts.
7.
The power dissipated in the dropping resistor will be 30 squared times the resistance or 359 watts, so such a resistor would be very hard to find, and the wire will heat up, so the current should be only briefly applied.
8.
A linear regulator, or an op amp or any other active semiconductor solution will be more satisfactory than a simple dropping resistor if the demand of the device varies, since the active device can keep the voltage supplied to the device from varying.
9.
It is true that with the proper combination of forward biased voltage drop in the LED and proper battery voltage, the circuit can be operated without a dropping resistor ( and with higher efficiency due to that absence of power wasted heating up the resistor rather than producing light ).
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