LEDs like to be powered with a constant source of current-ie. a fixed current regardless of the voltage it takes to achieve this. In practice for simple applications we assume a fixed forward voltage drop, and use a resistor to achieve the correct current.
However, with changes such as process variation, temperature etc. the forward voltage, and hence the current, will change. For simple applications this is not an issue, but for high power application such as you mention, this does become a problem, and so resistors are not used.
The solution is to include feedback in the circuit. As part of the driver circuitry, the current will be measured and the voltage across the LED controlled to always keep the current at the desired value; as a useful bonus, this also give you the ability to dim the LED by reducing the current.
As you point out, if we turn the excess voltage into heat it ends up being pretty inefficient (this is a form of
linear regulator)
The solution is use a switching regulator, which turns the voltage either fully on, or fully off. A capacitor is used to "average" this voltage, and by changing the ratio of the time turned on to the time turned off, we control the average voltage. All with an efficiency of 90%+.
If you're interested, then a commonly used circuit is a
buck converter
And if you'd like to get in-depth, then these two videos with Howard Johnson and Bob Pease are extremely good,
Driving High Power LEDs Without Getting Burned - Part 1
Driving High Power LEDs Without Getting Burned - Part 2
You have to start with a closed circuit, so that there can flow current. Do you have the resistor connected between the + and the - of the power supply? Then the voltage difference is 3.3 V. And you use Ohm to calculate the current.
The LED. Did you place that in series with the resistor? Which is how a LED circuit is built: the resistor makes sure that there's not too much current through the LED. Always use one.
If the LED's voltage would be 3 V then the difference between your supply voltage and the LED's voltage would be across the resistor. Kirchhoff is to blame for that. Kirchhoff's Voltage Law (KVL) says that the total of the voltages in a closed loop is zero. So we'll have 1.1 mA through LED and resistor. The 3 V was specified at 20 mA, so we're an end below that. It's normal for a LED to have a lower voltage at low currents.
But note that the 1.1 mA was true for 3 V LED voltage. We're apparently at 2.4 V, so the difference is now 0.9 V, and the current 3.3 mA. If you decrease the resistor value so that the current increases, you'll notice that the LED's voltage will increase as well.
How do you calculate the value? (Here we go again)
\$ R = \dfrac{\Delta V}{I} = \dfrac{3.3 V - 3 V}{20 mA} = 15 \Omega \$
edit re your update of the question
Welcome to the real, imperfect world. What you have at hand is measurement error. This is an important issue in engineering, and handling it properly can be a painstaking process.
You're giving your numbers in three significant digits, that's probably what the multimeter gives you. A multimeter's precision is most of the time expressed as a percentage (relative error) + a "count" (absolute error). A hobby quality meter may for instance have 2 % precision +/- 1 count. The 2 % should be clear: a 100 V reading may actually represent anything between 98 V and 102 V. The 1 count is an error in the last digit. A 5 may actually be a 4 or 6. That's an absolute error and doesn't depend on the value the meter gives you. If you measure 100 V then 1 count represents 1 %, if you read 900 V (same number of digits!) then 1 count is 0.11 %.
Let's presume you have a decent multimeter with 1 % +/- 1 digit precision. Then worst case your values may become
3.28 V - 1 count = 3.27 V, - 1 % = 3.237 V
267 Ω + 1 count = 268 Ω, + 1 % = 270.7 Ω
11.7 mA + 1 count = 11.8 mA, +1 % = 11.92 mA.
3.237 V / 270.7 Ω = 11.96 mA, which agrees well with the 11.92 mA we calculated for worst case. If your multimeter has a 1.5 % precision the calculated current will fall perfectly within the measured value's error range.
Best Answer
Your resister choices are reasonable so there will not be overcurrent on the PSU .This does waste lots of power .The resister wastes more power than the electromagnet .You could consider a lower voltage power supply or rewind the electromagnet with more turns of smaller diameter wire.