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A simple voltage divider designed with standard resistors from the E12/E24 series (For E12 with 10% and for E24 with 5% tolerance typically).
Try to press the switch and see what happens with the voltages in the circuit! 🙂
If you for example have a 6V battery as shown here and you need to supply something running only 2V as the LED, then it can be practical to make a voltage divider with two resistors in series, where you can calculate the resulting voltage U2 after the division from the supplied voltage U1 as follows:
U2 = U1 x R2 / (R1 + R2) e.g.
3V = 6V x 22 / (22 + 22) = 6V x 1 / 2.
You could also have chosen to divide with 3 to supply the 2V for the LED directly, but you usually include a resistor R3 in series with the LED to limit the current going through it, here dropping the voltage about 1V more.
The final thing to remember with voltage dividers is that they also need to be able to supply the needed current for the intended load at their output. A good rule of thumb is to set the current going through R2 the output resistor about 9 to 10 times higher than the needed load current, here through the LED!
If you forget this, then you risk not being able to maintain the required voltage U2 at the output of the voltage divider!
NB: This circuit has partly been inspired by Simply Electronics' series Basic Electronics part 12. (Or Electronics Basics).
Check also my other circuits about basic electronics and if you like them, please give me a thumbs up! 🙂
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