This is a question that I'm sure will seem straightforward to someone
well versed in electricity and/or electronics.  I'd like to think the
answer is simple, but as  I've observed it doesn't seem to be. 
Anyway, to get to the question:

I'm wondering about DC power adapters (like the power bricks that come
with laptops, cell phones, MP3 players, etc) that convert AC wall
current to DC current for whatever device they're designed to
power/charge, and whether they still use power (as in cost money on
the electrical bill) even when the device is not plugged into them. 
For example, if I leave my laptop's power brick plugged into the wall
without the other end connected to the laptop, I can observe 2 things
that lead me to believe that it is still drawing current (and perhaps
as much as when the laptop is plugged in):

1) The green light that indicates power is flowing is on (and I would
assume that this would not light unless the laptop was plugged in to
complete the circuit if the circuit was not complete inside of the
adapter)
2) The adapter gets warm, just as it does when the laptop is actually
plugged in.  I assume that there would be no heat generated if it was
not drawing current.

This leads me to believe that the adapter is still converting whatever
amount of power it is designed to whether the device is plugged in or
not.  As a result, I generally unplug the adapters from the wall when
they aren't in use, but that's inconvenient, and I don't want to look
like an idiot if they are in fact using current (which I'm sure the
light and warmth indicate) but far less current than when the device
is plugged in.

I hope this question is worded well enough to answer, I have limited
knowledge of electrical circuits (far less, probably, than a
programmer like myself should have), and I tried to make it
straightforward enough to understand even if I wasn't using proper
terminology.  Thanks!

Hi swrobel-ga,

   DC adapters do consume power even if they do not supply actively to a
connected electronic appliance.In a simple dc power adapter, it is used up
by the eddy current discharge that happens in the iron core of the stepdown
transformer, also supply power to a no load resistance, leakage of the
rectifying diodes etc. Eventhough the transformer's coils are inductive(thereby 
the average power in each cycle is zero), there is also the resistance of the
coil that consume power during each cycle(make the coil a non ideal inductor).
Incase of the switching power supplies used in the laptop power adapter bricks,
there is still a loss in the ferrite core, apart from the idle no load
resistance, battery trickle charge. Most power adapters use a no load
resistor
to improve the power supply regulation, so that there is no wide swing in
voltage output, as the current fluctuates between noload and full load.

Hope this answers your question.

Regards
ldavinci-ga

I appreciate the explanation, and while it does clarify what is going
on inside the adapter, I'm still wondering if it does consume as much
power as when a device is actually connected?

The power supply will use more power when there's a device connected.  

Think of an electric motor - a coil sits in a magnetic field, a
current's applied to the coil and the coil moves.  Now think of a
generator - a coil sits in a magnetic field, the coil's moved and a
current's generated.  It's the same system.  So, when an electric
motor's running it also acts as a generator! But, the current
generated flows in the opposite direction to the one you're applying,
so it limits the speed at which the motor can turn. The generated
current is known as a back Electro Motive Force (EMF).

In a stepdown transformer, such as you have in a power supply, there
are 2 coils - a primary which is connected to the mains and a
secondary which is connected to a low voltage device.  The primary
coil induces a current in the secondary coil, which is fed to the
device.  It also induces a current in itself, another back emf , which
helps to limit the current it can draw.  With nothing connected to the
secondary coil, the back emf in the primary is higher, so the current
drawn is lower.

Ian G.

swrobel,

I'm not here to impress you and make you think I'm smart like the
previous comments tried to do.  Anyone can read a website and
summarize what they read to make themselves come off as smart.  Yes, I
understand completely what the previous comments said, but swrobel
didnt ask for a physics lesson, just a common sense reply.  (Side
note: I also noticed that both of the previous replies completely
ignored magnetic flux.  The flow (and direction of flow) of magnetic
flux is absolutely crucial in a transformer.  Without it, there is no
conversion of voltage.)  I had to throw that in there, sorry.

You honestly seem to be a pretty sharp person for making this
observation and asking the question.  The answer IS simple: yes it
consumes power when not powering a device (no load).  However, it is
negligible in regards to any actual cost you would incur on an
electric bill.

Let me explain:
In the case of a laptop, the DC power supply (simply a transformer
with a rectifier and filter in most cases) will consume the highest
amount of power when charging the battery of the device.  It will
consume a lesser amount of power when powering a device with a full
battery.  And finally, it will consume a negligible amount of power
while plugged into a receptacle and not connected to your device.

To clarify what has been said before (and quite confusingly) in the
previous posts:  A transformer will always have losses.  From these
small ones in power supplies to the 500kV transformers in switchyards,
they all have losses.  However, the losses are due to the resistance
of the copper (or whatever conductive material makes up the windings
of the transformer).  Eddy currents, iron core, blah blah whatever -
the electric company can only charge you for REAL power consumed, ie,
power that is consumed by a resistive load (wattage).  Notice on your
bill that you pay for kWh (kilowatt-hours) NOT kVAh
(kilo-volt-ampere-hours).  Motors, solenoids and other devices DO
require imaginary (volt-amp reactive (VAR)) power, but there's no
point in exploring that here.  (I just brought it up in case you had
heard the term.)

Here's the point:
What the previous posts arent telling you is (maybe they dont
understand themselves), these small transactions between capacitive
and inductive elements in this rectifier circuit arent costing you a
dime (believe it or not, its true).  However, what DOES cost you is
the copper loss caused by the resistance of the wire itself that makes
up the windings.  This resistance is very small in a power supply (of
course, the loss gets bigger in power transformers, and their losses
are much greater).

As far as power supplies go, here's my take:  If they stay unused for
quite awhile (weeks, months) go ahead and unplug them.  Not for cost
reasons, simply for safety reasons, ie, if something was to fail
inside of one of these (and it does happen on occasion) it can build
up a decent amount of heat and cause a fire (again, pretty uncommon,
but it does happen).  However, if it powers a device that is used
daily or every other day, go ahead and leave it in.  You are spending
more on the inrush of current to turn on your lightbulb when you enter
the room than when you leave your DC power supply hooked up all day.

However, let common sense be your guide.  If the thing stays hot while
its not loaded, it probably has a high resistance in the windings (ie,
cheaply made) and is consuming a small amount of power (more of a heat
concern than cost concern still).  If the DC supply is room temp when
not being used, its probably OK.  The light on the thing is a LED
anyways, which are very, very low power and quite efficient.

Let me know if this answers your question.

TK

Thanks TK, that's exactly what I've been waiting for.  I really
appreciate the fact that you gave a common-sense answer to my question
and provided the scientific background (I'm interested in that as
well, of course, but I was hoping that I wouldn't have to decipher it
to pry out the answer to my question as with the previous comments). 
Thanks again!

Hah that's pretty funny -- as I read through the first two answers,
they sounded straight from an encyclopedia and never really answered
the question... the english post is definitely appreciated, as I
seeked an answer to this as well!

Rich