See the attached image.
Sticking with the water analogy. The LED is a one way check valve. It prevents current going the wrong direction. If you hook an LED up backwards to a battery, it won’t work just like a check value only works in one direction.
Current is like the water flowing through the pipe. The amount of water flowing at one end is the same as through the other end.
Resistors are narrower pipes. The narrower the pipe, the higher the restriction on water flow.
VB - Battery Voltage - 5V
VF - LED Forward Voltage - 3V
VR - Resistor Voltage - 5V - 3V = 2V
CrossRoads provided a pretty good description of the maths.
You’re not really using the right equation.
The LED will have some voltage across it, Vf, when the 30mA of current is flowing.
so (5V - Vf) = voltage across the resistor, Vr.
Now knowing Vr, you can calculate R for the current using Ohms law: V = IR
Rearranging, V/I = R
So (5V - Vf)/I = R
If the LED has 3V across when current flows and you want 30mA:
(5V - 3V)/.03A = 67 ohms. Use a 68 ohm standard resistor.
Now, P=IV can come into play.
You have I, 30mA and you know Vr, so you can calculate the power dissipated in the resistor.
Taking Vr=IR, and subbing that in for V above,
P = IIR, so .03A *.03A * 68 ohm = 62mW, so a 1/4W rated resistor would be fine.
These calculations are for continuous operation, not pulsed. For pulsed usage, the amount of current can be higher but until you are used to the maths, then it is best to use these values.
There are various pdf’s and web sites that offer an introduction to electrics and electronics.
Hope this helps