BEGINNER - Basic LED circuit, power dissipation question

I'm beginning the Arduino for Dummies book and and it is talking about power dissipation in a circuit which has a 5v supply with an LED and a 100 ohm resistor. (page 80)

It states the LED has a forward voltage of 2v and requires 30mA to light. The author calculates the power as:

P=(5v-2v) x .03A = 0.09W

or

(0.03A x 0.03A) x 100R = 0.09W

Can I confirm this is the power dissipated by the resistor?

BUT what about the power dissipated by the LED? Why is this not considered?

Yes, that is the power dissipated by the resistor.

The power dissipated by the LED is .03A x 2V = .06W

The forward voltage drop is essentially fixed by physics. And if your circuit keeps the current well within the limits of the LED (documented in the data sheet), we don't really care what the power dissipation is.

Paulcet:
Yes, that is the power dissipated by the resistor.

The power dissipated by the LED is .03A x 2V = .06W

The forward voltage drop is essentially fixed by physics. And if your circuit keeps the current well within the limits of the LED (documented in the data sheet), we don't really care what the power dissipation is.

Thank you for clearing that up. I was puzzled since the author was discussing the power consumption of the circuit and completely omitted the power consumption of the LED??

A led is made for a certain current, mostly 20mA.
A resistor can be bought for almost any Watt you like. It is therefor often important to know how much the power dissipation is for a resistor.

The led has 2V and 30mA, that is 60mW, and maybe 20% into light, the other 80% into heat.
The resistor gets 3V and 30mA, that is 90mW, 100% of that is turned into heat.

About the forward voltage : Light-emitting diode - Wikipedia (it is the "voltage drop" in the table).

An ideal LED would convert all the power to light, but we don't have anything that efficient
quite yet - but we are getting close - LEDs are among the most efficient light sources.

When the first red LEDs appeared way back in the 70's they were about 1/10000 times
as efficient, its worth noting...

BUT what about the power dissipated by the LED? Why is this not considered?

Maybe he made a mistake! Maybe he didn't explain it very well. Maybe you didn't understand what he was doing.

But in any case, it looks like you do understand the electronics and the calculations!

P=(5v-2v) x .03A = 0.09W

Obviously he did because he calculated the power for a led with VF of 3V !

P = V x I (Normally expressed as P = I x V)

(5V-2V=3V)