How can I measure power draw?

Is there a simple way to measure how much power my components are drawing from the wall? My interest is prompted by experimenting with the PS Audio Power Plant and various conflicting info about how much it draws. (The answer I'd love to get is that there is a Radio Shack box I can stick in between the DUT and the wall, but you tell me...)
Get an Amprobe metering device. It clamps around the AC wire, usually in your fusebox(clamps on a single conductor)either the hot or neutral line for the circuit you want to measure. Or you can rig an AC extension cord with an opened jacket, & measure with that in series with the load in question. Use caution; don't do this if you don't understand the risks, because you could be electrocuted. Also don't do any critical listening when rigged this way for measuring; it will probably sound awful. Quiescent current measurement will be lower than with signal applied (to a power amp) outputing any appreciable amount of power.
Here is a simple solution. Look on the back of the power amp, and pre-amp or in the manual. The wattage will be listed for each component in either Watts or VA. of consumption. If all else fails call the manufacturer ask them the Wattage of the peice then simply add it all up. Jan Olsen
Thank you both for your responses. I have the published specs on all of my components.The thing is, I don't believe them.The P300 will give you a readout of power consumption of the components plugged into it. My Pass amp is supposed to draw 200 watts at idle, but if I run it and my front end stuff into the P300, the reading is about 190 watts. And then what is the P300 itself drawing from the wall? Some of the reviews of the PS Audio Power Plant P300 have stated that the unit draws 300 watts all the time, but PS Audio says this isn't true. So the reason for my question is to try to get a true measure of what's going on here. Bob, your solution is too tricky for me to attempt. I'll tell you, if someone could devise an in-line meter for this, I think it would be a useful addition to my audiophile toolkit. -Dan
Bob's solution is a good one, and not all that difficult. But, if you really want to know without any trouble, Allied Electronics sells a nice digital wattmeter, just plug in the equipment and read watts used. Regret to say the cheapest model is $450. Check ebay, etc. for used analog ones, listed as Variacs, Powerstats; they have all variable output voltage, a few have the wattmeter included. Rich
Drubin, Go to and check out their "wattsup" device. This unit plugs into the wall and then your amp plugs into this unit. Turn on the "Wattsup" and then turn on your amp. This unit measures the power consumption of your amp. It can also calcuate the cost of electricity. Pretty need stuff. I saw one at Fry's electronics for $80.00. Have fun.
The Wattsup sounds like just the ticket. I'll get one and report back. Thanks. -Dan
An accurate power draw cannot be obtained without considering the power factor. In essence this means measuring the phase angle of voltage vs current (as well as measuring voltage vs current). The true RMS power figure is obtained by: P = VI cos phi. The cosine of the phase angle is called the power factor. This formula holds for all sinewave sources of power (such as AC from the wall).
As a more simple method to the above suggestions, you can measure the power consumption by watching the meter at the power distribution box for your house. I'm currently in Japan, and I'm from Ireland, but both countries use the same type of meter; so the following probably also applies in America. My meter, has a rotating disc, one full revolution is 1/100 on a kWHr. You can calculate the average power consumption, over a given period of time, by turning on your equipment, and counting the number of revolutions over that time. The power in Watts is given by the following equation P = 600 X (#Revolutions/Time in Minutes) This will directly give the power for which you will pay your Electricity Supply company. Make sure to turn off all other appliances in your house before making the measurement. During the day is the best time to do this experiment!! Brian Kearns
Sorry, I checked my meter this morning, and I have to make a correction to the formula above. One full revolution is 1/300 kWHr, so the correct value is three times smaller, ie: P = 200 X (#Revolutions/Time in Minutes). Brian Kearns.