If you had only one LED which dropped 3.1 volts with 333 milliamperes through it and you were driving it with a 12 volt supply, you'd need a series ballast resistor to drop the remaining 8.9 volts.

Then, since in a series circuit the current is everywhere the same, that 333 milliamperes would pass through both the LED and the ballast and the value of the resistor would be:

$$ R = \frac{Vs-Vled}{Iled} = \frac{12V - 3.1V}{333mA}\approx 27\Omega $$

Since your supply is a voltage source it'll maintain its output voltage at 12 volts as long as you don't exceed its 1 ampere rating, the effect being that you can connect any number of LEDs and their associated ballsts across it and each pair will think its connected to its own independent supply.

Consequently, each of your LEDs would be ballasted with a 27 ohm resistor which would limit the current through each LED to 333 milliamperes, for a total of one ampere for the array.

Each resistor would dissipate about:

$$ P = IE = 333mA \times 8.9V \approx 3\text{ watts,} $$

so it would be prudent to use 27 ohm, 5 watt resistors for your ballasts.

If you had only one LED which dropped 3.1 volts with 333 milliamperes through it and you were driving it with a 12 volt supply, you'd need a series ballast resistor to drop the remaining 8.9 volts.

Then, since in a series circuit the current is everywhere the same, that 333 milliamperes would pass through both the LED and the ballast and the value of the resistor would be:

$$ R = \frac{Vs-Vled}{Iled} = \frac{12V - 3.1V}{333mA}\approx 27\Omega $$

Since your supply is a voltage source it'll maintain its output voltage at 12 volts as long as you don't exceed its 1 ampere rating, the effect being that you can connect any number of LEDs and their associated ballasts across the one supply and each pair will think it's connected to its own independent supply.

Consequently, each of your LEDs would be ballasted with a 27 ohm resistor which would limit the current through each LED to 333 milliamperes, for a total of one ampere for the array.

Each resistor would dissipate about:

$$ P = IE = 333mA \times 8.9V \approx 3\text{ watts,} $$

so it would be prudent to use 27 ohm, 5 watt resistors for your ballasts.

If you had only one LED which dropped 3.1 volts with 333 milliamperes through it and you were driving it with a 12 volt supply, you'd need a series ballast resistor to drop the remaining 8.9 volts.

Then, since in a series circuit the current is everywhere the same, that 333 milliamperes would pass through both the LED and the ballast and the value of the resistor would be:

$$ R = \frac{Vs-Vled}{Iled} = \frac{12V - 3.1V}{333mA}\approx 27\Omega $$

Since your supply is a voltage source it'll maintain its output voltage at 12 volts as long as you don't exceed its 1 ampere rating, the effect being that you can connect any number of LEDs and their associated ballsts across it and each pair will think its connected to its own independent supply.

Consequently, each of your LEDs would be ballasted with a 27 ohm resistor which would limit the current through each LED to 333 milliamperes, for a total of one ampere for the array.

Each resistor would dissipate about:

$$ P = IE = 333mA \times 8.9V \approx 9\text{ watts,} $$

so it would be prudent to use 27 ohm, 20 watt resistors for your ballasts.

If you had only one LED which dropped 3.1 volts with 333 milliamperes through it and you were driving it with a 12 volt supply, you'd need a series ballast resistor to drop the remaining 8.9 volts.

Then, since in a series circuit the current is everywhere the same, that 333 milliamperes would pass through both the LED and the ballast and the value of the resistor would be:

$$ R = \frac{Vs-Vled}{Iled} = \frac{12V - 3.1V}{333mA}\approx 27\Omega $$

Since your supply is a voltage source it'll maintain its output voltage at 12 volts as long as you don't exceed its 1 ampere rating, the effect being that you can connect any number of LEDs and their associated ballsts across it and each pair will think its connected to its own independent supply.

Consequently, each of your LEDs would be ballasted with a 27 ohm resistor which would limit the current through each LED to 333 milliamperes, for a total of one ampere for the array.

Each resistor would dissipate about:

$$ P = IE = 333mA \times 8.9V \approx 3\text{ watts,} $$

so it would be prudent to use 27 ohm, 5 watt resistors for your ballasts.