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I have seen optocoupler series with common cathode and resistor on every anode. My question is whether or not one resistor on the cathode side will be sufficient, with the right wattage, of course, and what is the reason behind these many resistors.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Please post a schematic. A schematic is better than words. You can add one in using the CircuitLab button on the editor toolbar. Double-click a component to edit its properties. 'R' = rotate, 'H' = horizontal flip. 'V' = vertical flip. Note that when you use the CircuitLab button on the editor toolbar an editable schematic is saved in your post. That makes it easy for us to copy and edit in our answers. You don't need a CircuitLab account, no screengrabs, no image uploads, no background grid. \$\endgroup\$ – Transistor Jul 2 at 19:25
  • \$\begingroup\$ If you're driving the LED with a current source, it doesn't matter what resistor you use. You could use no resistor at all and have no trouble. \$\endgroup\$ – The Photon Jul 2 at 19:49
  • \$\begingroup\$ Well, I just didn't see the 24 volt voltage, so the input is 24 volts. \$\endgroup\$ – user6740407 Jul 2 at 19:52
  • \$\begingroup\$ I've added the 24 V sources for you. If you don't have constant current sources then remove them. Double-click the resistors to set their values. \$\endgroup\$ – Transistor Jul 2 at 20:01
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    \$\begingroup\$ If only one LED or optocoupler will be on at a time, you may use a single resistor. If more than one LED or optocoupler will be on at the same time, the current will be divided between the LEDs, which may make them too dim to operate the phototransistor in the optocoupler. \$\endgroup\$ – Peter Bennett Jul 2 at 20:08
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. OP's schematic redrawn.

In Figure 1a a typical LED wired as shown would receive a current of about \$ I = \frac {24 - V_f}{480} = 45 \ \text {mA} \$ if the forward voltage, Vf is about 2 V.

In the Figure 1b the same current will result if one LED is switched on. If two are switched on they will receive about half the current, three -> one third, etc.

A further problem will occur due to variation in the Vf of the various LEDs. The one with the lowest Vf will hog the current, get hotter, its Vf will decrease, causing it to hog more current, etc., and may fail. For more on the topic see what I've written in LED variations and binning.

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