I'm pretty sure it doen't matter, the total resistance of the circuit is what counts, the diagrams I have seen however predominantly put the resistor on the + (anode) side of the LED.
Of course it depends entirely on the ohm's resistance of the resistor. The higher the resistance, the lower the comparison to a short circuit.
Yes. You can use a voltage divider. Say, for instance, one 1KOhm resistor in series with a 3KOhm resistor. Connect the 3k resistor to the 48 volts and connect the 1k resistor to ground. The 1k resistor will have 12 volts acress it. These resistors need to be at least 1 watt each as they are going to dissipate 0.576 watts and get warm. Now, if you attempt to pull power from the 1k resistor, note that regulation will be poor because the impedance of the load will go in parallel with the 1k resistor and change its value.
Two Hundred
If there's nothing else between the ends of the resistor and the power supply, then the voltage across the resistor is 24 volts, and the current through it is 2 amperes.
If there is nothing else in the circuit, then the voltage drop across the resistor will be the full supply voltage of 5 volts. The size of the resistor does not matter in this case - it will always be 5 volts.
9v
In its simplest use a resistor in a circuit is used to limit the amount of current flow, or to decrease the amount of voltage applied to a device. One example is you had a 12 volt battery and you need/ wanted to connect it to a device that ran on 9 volts then a resistor can be chosen to reduce the 12 volts to the 9 volts required.
If the 3-ohm resistor is the ONLY thing in the circuit, then the current flowing through it is (12 volts)/(3 ohms) = 4 amperes. If there are other things in the circuit besides the resistor, then the current depends on all of them.
600 VDC.
Depends on the led forward bias threashold, if its a typical led it will be .7 volts so, .7x6=4.2V, so pick a resistor that will drop around 7 volts. What is the current? Then just to V=IR, 7=IR.
Assuming it's 90 v dc, get a 1.5 k-ohm resistor and an 82 k-ohm resistor. Put them in series across the supply, then there will be slightly over 1.5 v across the smaller resistor.
The question is a bit ambiguous, but I will try to address it. If the 6 ohm resistance is in series with another resistance then some of the 5 volts would be dropped across the 6 ohm resistance and the remainder of the voltage would be dropped across the other resistance. To calculate the voltage, use the 'resistor voltage divider equation' (Google it). If the 5 volts is applied across only a 6 ohm resistance, then the top of the resistor is at 5 volts and the bottom of the resistor would be at 0 volts. The resistor would drop all of the voltage.