Electronic – Recommended Layout for Parallel Current Sensing Resistors

Power on the resistor will be \$I^2R\$. Using an higher resistance will lower the wattage as in the formula above, but your led will not shine too much. You can solve the problem by using two 1/4 w resistors in parallel of 160 Ohm to have a single resistance value of 80 Ohm with 1/2 W power, but of course you really won't do that, because making your circuit so ugly just to save a resistor it is not a good idea.

Your calculations for both the resistor value and wattage are off.

For the resistor, you show the formula R = V / I, but then calculate using the entire current for all three LEDs (1 A) instead of the current for just one LED.

So it should be

$$R = \frac{V}{I} = \frac{(12 - 3.1)}{.333} = 26.7\space Ω$$

For the wattage, you are trying to use the formula P = VI, but then you substitute R for the V and use the current for all three resistors (1 A) instead of just one.

So the correct calculation would be

$$(12 - 3.1) * .333 = 2.96\space W$$

or you could use:

$$P = I^{2} R = (.333)^{2} * 26.7 = 2.96\space W$$

So you could use a 27 ohm, 5 W resistor.