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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.

enter image description here

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    \$\begingroup\$ Remove the meter, set it to voltage, hook up your circuit and measure the exact voltage across each LED, and the exact battery voltage. Then using the actual numbers report back on what the current should be. \$\endgroup\$
    – John D
    Sep 30 '20 at 18:57
  • \$\begingroup\$ Probe is not at the correct position. Red probe should be inserted to the leftmost socket. Also be suret that the crocodile clips do not touch each other. \$\endgroup\$ Sep 30 '20 at 18:59
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    \$\begingroup\$ @Rohat, that's the 10 A socket. Usually the right socket is V\$\Omega\$mA. \$\endgroup\$
    – Transistor
    Sep 30 '20 at 19:03
  • \$\begingroup\$ @Transistor I've never seen those. Interesting. \$\endgroup\$ Sep 30 '20 at 19:05
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    \$\begingroup\$ @Mac_79: You have a yellow, a red, and what appear to be two green LEDs. The forward voltage of each color is different. \$\endgroup\$
    – JRE
    Sep 30 '20 at 19:05
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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.

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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.

enter image description here

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.

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  • \$\begingroup\$ Oh, nifty chart! It's nice to see the color vs. voltage all gathered into one chart like that. \$\endgroup\$
    – TimWescott
    Sep 30 '20 at 20:55
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If you use in your calculations 22 ohm + internal resistance of the battery you will get 1.3mA

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  • \$\begingroup\$ A warm welcome to the site. Your Answer needs to contain some explanations rather than just firing off one line. It should be written clearly enough to teach and future readers. This is little more than a comment. Please can you edit it and greatly improve it, making it informative rather than just factual. Note that the OP is a "total beginner". Thanks. \$\endgroup\$
    – TonyM
    Oct 1 '20 at 6:28

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