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Here's a circuit connecting 4 LEDs with 22 Ohm resistor. According to my calculation, I'm supposed to get 45 mA flowing.
Battery is 9 V. Voltage drop of each LED is 2 V. Resistor is 22 Ohm.
\$(9 - 4 \cdot 2)\ /\ 22 = 0.045 \$ or 45 mA.
However, if you look at the multimeter, it says 1 mA? I'm really confused.
I'm a total beginner, so I appreciate any help.
Nominal voltages can vary a lot from actual, and both batteries and diodes show this tendency strongly.
The 1.5V nominal voltage on a dry cell is for a fresh cell that's not being asked to deliver much current. Ask for more current, and you'll get less voltage out of the cell. 45mA is a lot to ask from a 9V battery, so it's undoubtedly delivering less.
Similarly, the 2V drop across a diode will vary with the individual diode characteristics and with temperature.
The rule of thumb for dry cells is that you haven't gotten all the energy out of them until they're running at around 1V to 0.9V per cell -- that works out to designing your circuit to work with a battery voltage from 9V down to 6V, or even 5.4V. That, in turn, means that you should design for two LEDs in series, with an appropriate current-limiting resistor -- and expect that your intensity will drop considerably from a fresh battery down to a dead one.
You seem to be expecting 2 V for each of the four LEDs but you appear to have four different colours and that means you will have four different forward voltage drops.
Figure 1. Current and voltage relationship for various LED colours. Source: LEDnique.
If you draw a horizontal line the graph above at the current you are reading you can add up the Vf of each of your LEDs and see if it matches your readings. Notice that blue has a much higher Vf than red.
Note on use of multimeter: You are reading 1.3 mA on the 200 mA range. Your precision will not be good. It's the correct way to start but once you've established the magnitude of the current you can step down to the next range (20 mA) and get more precision, etc.
If you use in your calculations 22 ohm + internal resistance of the battery you will get 1.3mA
