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Does resistor placement change power dissipation in simple LED circuit?

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3

$begingroup$

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?

power resistors basic

share|improve this question

asked May 31 at 13:18

SofaKngSofaKng

182128

$endgroup$

add a comment | 

3

$begingroup$

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?

power resistors basic

share|improve this question

asked May 31 at 13:18

SofaKngSofaKng

182128

$endgroup$

add a comment | 

$begingroup$

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?

power resistors basic

share|improve this question

asked May 31 at 13:18

SofaKngSofaKng

182128

$endgroup$

share|improve this question

asked May 31 at 13:18

SofaKngSofaKng

182128

share|improve this question

asked May 31 at 13:18

SofaKngSofaKng

182128

asked May 31 at 13:18

SofaKngSofaKng

182128

asked May 31 at 13:18

SofaKngSofaKng

182128

2 Answers
2

active

oldest

votes

6

$begingroup$

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.

share|improve this answer

edited May 31 at 13:51

answered May 31 at 13:25

jusacajusaca

1,546625

$endgroup$

add a comment | 

6

$begingroup$

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.

share|improve this answer

edited May 31 at 13:54

answered May 31 at 13:25

Spehro PefhanySpehro Pefhany

217k5166447

$endgroup$

add a comment | 

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6

$begingroup$

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.

share|improve this answer

edited May 31 at 13:51

answered May 31 at 13:25

jusacajusaca

1,546625

$endgroup$

add a comment | 

6

$begingroup$

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.

share|improve this answer

edited May 31 at 13:51

answered May 31 at 13:25

jusacajusaca

1,546625

$endgroup$

add a comment | 

$begingroup$

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.

share|improve this answer

edited May 31 at 13:51

answered May 31 at 13:25

jusacajusaca

1,546625

$endgroup$

share|improve this answer

edited May 31 at 13:51

answered May 31 at 13:25

jusacajusaca

1,546625

answered May 31 at 13:25

jusacajusaca

1,546625

answered May 31 at 13:25

jusacajusaca

1,546625

6

$begingroup$

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.

share|improve this answer

edited May 31 at 13:54

answered May 31 at 13:25

Spehro PefhanySpehro Pefhany

217k5166447

$endgroup$

add a comment | 

6

$begingroup$

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.

share|improve this answer

edited May 31 at 13:54

answered May 31 at 13:25

Spehro PefhanySpehro Pefhany

217k5166447

$endgroup$

add a comment | 

$begingroup$

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.

share|improve this answer

edited May 31 at 13:54

answered May 31 at 13:25

Spehro PefhanySpehro Pefhany

217k5166447

$endgroup$

share|improve this answer

edited May 31 at 13:54

answered May 31 at 13:25

Spehro PefhanySpehro Pefhany

217k5166447

answered May 31 at 13:25

Spehro PefhanySpehro Pefhany

217k5166447

answered May 31 at 13:25

Spehro PefhanySpehro Pefhany

217k5166447