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Does the power dissipation in an LED and resistor change depending on the resistor placement?

Let's assume the following:

  • Supply Voltage: 3.3v (DC)
  • LED Forward Voltage: 2.1v
  • Resistor: 330 Ohm

I think my problem is that I'm thinking that the circuit "starts" at 3.3v, then goes through the resistor which restricts current and then goes through the LED which drops voltage. Alternatively, if I place the LED before the resistor, then the LED will drop the voltage (3.3v - 2.1v = 1.2v) before it reaches the resistor so the resistor is dissipating less heat?

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The order of the components does not make any difference. The voltage is diveded on the two components in the same way, in your case 2.1V on the LED and 1.2V on the resistor. This results in 1.2V/330Ohm = 3.6mA through your series circuit. So the power dissipation is also the same in both configurations.

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What matters is the voltage across the terminals of the resistor (as you would measure with a meter connected directly across the resistor). It doesn't matter how the voltage gets there.

So, no it doesn't matter which order they're connected when they're in series, because the magnitude of the voltage across the resistor will always be the supply voltage minus the LED forward voltage.

Whether you measure the voltage across the resistor as positive or negative doesn't matter either, since the power dissipation of the resistor is the square of that voltage divided by the resistance.

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