Power consumption of resistors

[cbseattle]

This is a seriously n00b question:

If I have a simple circuit with a battery, a resistor and an LED... if I increase the value of the resistor to reduce (and dim) the LED will I be saving battery power or will the power just be consumed by the resistor instead?

I know that P=VI and so a lower current (provided by the resistor) should mean less power consumption, but does that just apply to the LED or to the whole circuit?

Thanks so much for any help - simple answer, explanation, etc.

[alm]

The voltage across an LED is (almost) constant for currents close to the nominal value. Assuming it's driven from a constant voltage source (eg. wall wart, battery), this means that the voltage across the series resistor will also be constant. Power dissipated in the resistor is V²/R. V is constant and R increases with a higher value resistor, so power dissipation will go down. In addition, power dissipated by the LED also goes down. This is a simplification that may not be accurate if you vary the current over a large range (say more than an order of magnitude), in that case you'd have to take the exponential IV curve of an LED into account, but the total power would still go down.

[cbseattle]

Brilliant! Thanks a lot. Especially for the simple explanation.

[IanB]

I think alm over-complicated the analysis. As engineers we look for the simple answer.

In this case the answer is dramatically simple. The power delivered by the power supply (the battery) is equal to the battery voltage times the circuit current. If you assume for a small LED that the battery voltage is a constant, then the power consumption of the circuit is directly proportional to the current flowing. In other words, if you increase the resistor value to reduce the current and dim the LED, then the total power consumption will go down proportionally to the reduced current.

This even works if the LED is a high power device drawing lots of current. Even though the battery voltage may decrease under heavy load the result is the same. All that happens with a heavy load is that the internal resistance of the battery becomes important and causes more power to be dissipated inside the battery instead of in the external resistor or LED.

So in summary,the total power dissipation of the circuit is proportional to the current flowing in the circuit to a very high degree of approximation. If you reduce the current by whatever means, you reduce the power consumption accordingly.

[cbseattle]

Yes that is a great and simple explanation. Fortunately I understood Alm's response the same way.
The thing that really helps from your answer is that I think I'll remember it the next time I have a similar issue.
Thanks so much for taking the time to answer such a basic question.