The short answer is, no. Current is the same throughout in a series circuit. The resistor in series changes the current, however. You knew that; and we'll come back to that later, below.
The tl; dr answer is, we use the LED's forward-voltage characteristic to set its current with the dropping resistor.
Let's talk about that.
LED Forward Voltage
The LED is a semiconductor device (a diode), and like all diodes has a conduction threshold voltage that is more-or-less constant. This is the forward voltage, or Vf.
Because of this Vf theshold, the way the LED response to applied voltage is highly non-linear:
- below Vf, LED current will be next to nothing;
- above Vf the current rises rapidly.
Keep going much further above Vf and the LED will burn up. This is why your LED died when connected directly to 5V.
This in mind, we know that we need to regulate the LED current to keep it in its operating sweet spot: making light, but not torching itself in the process.
But let's talk about that threshold voltage Vf for a bit.
What Determines Vf?
The Vf threshold depends on the color of the LED and its exact material formulation. In general, the shorter the emitted wavelength, the higher the LED Vf threshold.
From here: http://lednique.com/test-equipment/testing-unknown-leds/
The datasheet for the specific LED you are using will include this I-V curve data. This, in turn, helps solve the rest of the current-setting dilemma.
How The Dropping Resistor Sets LED Current
We know that we need to regulate the LED current to get the brightness we want, yet not burn the LED in the process.
The simplest way to do that is to use a fixed supply (like 5V you've shown) and a series resistor. We take advantage of the LED’s known voltage drop (its threshold voltage Vf), and that fixed supply (5V), to set the LED current as follows:
Not quite. There’s a few things going on here that allow us to make that equation.
Kirchhoff Current and Voltage
The series resistor and LED currents are the same based on Kirchhoff’s Current Law for series circuits, which tells us that current in a loop circuit is the same at any point in the loop.
So we know:
Likewise, Kirchhoff’s Voltage Law for loop circuits says that the sum of all the voltage drops must equal the supply voltage.
And so we also know:
Finally, Ohm’s Law and some algebraic substitution tell us the resistor and LED current.
That is, I = E/R, so we have:
- I(resistor) = V(resistor) / R1
After substituting (5V - Vf) for V(resistor), we are left with:
- I(resistor) = (5V - Vf) / R1
And because of Kirchhoff's Current Law we know:
- I(led) = I(resistor) = (5V - Vf) / R1
How To Pick The Right Dropping Resistor
We know the supply (5V), and we know Vf (from the LED data sheet, about 2-3V depending on color), and we know the current we want (about 10-20mA), so we can solve for R1:
Example: White LED with Vf of 3V, driven at 20mA:
- R1 = (5V - 3V) / 20mA = 100 Ohms