From owner-freebsd-questions@FreeBSD.ORG Tue Jun 8 21:33:00 2004 Return-Path: Delivered-To: freebsd-questions@freebsd.org Received: from mx1.FreeBSD.org (mx1.freebsd.org [216.136.204.125]) by hub.freebsd.org (Postfix) with ESMTP id EA55616A4CE for ; Tue, 8 Jun 2004 21:33:00 +0000 (GMT) Received: from smtpout.mac.com (smtpout.mac.com [17.250.248.97]) by mx1.FreeBSD.org (Postfix) with ESMTP id BDB8B43D46 for ; Tue, 8 Jun 2004 21:33:00 +0000 (GMT) (envelope-from cswiger@mac.com) Received: from mac.com (smtpin07-en2 [10.13.10.152]) by smtpout.mac.com (Xserve/MantshX 2.0) with ESMTP id i58LW0Ou021084; Tue, 8 Jun 2004 14:32:00 -0700 (PDT) Received: from [10.1.1.193] (nfw2.codefab.com [199.103.21.225] (may be forged)) (authenticated bits=0)i58LVmjq001011; Tue, 8 Jun 2004 14:31:58 -0700 (PDT) In-Reply-To: <40C62A70.8060102@bah.homeip.net> References: <20040608122101.GA68204@dogma.freebsd-uk.eu.org> <87zn7ednwg.fsf@pele.r.caley.org.uk> <20040608172756.GA70798@dogma.freebsd-uk.eu.org> <20040608135903.024729b8.wmoran@potentialtech.com> <6CCB3AEC-B97C-11D8-8148-003065ABFD92@mac.com> <40C62A70.8060102@bah.homeip.net> Mime-Version: 1.0 (Apple Message framework v618) Content-Type: text/plain; charset=US-ASCII; format=flowed Message-Id: <3F7DF4CB-B993-11D8-8148-003065ABFD92@mac.com> Content-Transfer-Encoding: 7bit From: Charles Swiger Date: Tue, 8 Jun 2004 17:31:48 -0400 To: "Bernt. H" X-Mailer: Apple Mail (2.618) cc: freebsd questions Subject: [OT] Re: Leaving a server on all day X-BeenThere: freebsd-questions@freebsd.org X-Mailman-Version: 2.1.1 Precedence: list List-Id: User questions List-Unsubscribe: , List-Archive: List-Post: List-Help: List-Subscribe: , X-List-Received-Date: Tue, 08 Jun 2004 21:33:01 -0000 On Jun 8, 2004, at 5:06 PM, Bernt. H wrote: >> No need to guess, use an amp-meter. :-) > > Well If it measure trueRMS then you could use it, otherwise no. You are correct that one needs to measure the voltage and use the RMS value, or DC series equivalent if you like that phrase, in order to figure out the power consumption accurately, but an {ammeter, amp-meter, DMM} which can deal with AC will do the right thing. >> Radio Shack and the like will sell something with male and female >> plugs that will measure both voltage and current, and give you the >> current power load in Watts. Smart UPSes may also have a similar >> capability. > > Yes but it will only show you the correct value if the load is a pure > resistans, not if it's reactiv, as all switching psu's are. The ratio between the actual load and a purely resistive load is known as the power factor, and is why UPS are rated in terms of kVA rather than in terms of the wattage of the load. For computer equipment [1], the power factor is lagging, representing an inductive rather than capacitive load, and the PF is typically about 0.9. However, the electric company bills you for the power you draw from them, they don't give you a refund for the power "wasted" because your load is not purely resistive, so the notion of measuring the kVA rather than the "useful wattage" is not really "incorrect". -- -Chuck [1]: And almost everything else, too. Most things use a transformer to convert line voltage into whatever voltage the device wants, which is inductive, or consist of a motor, also inductive. Motors which draw a lot of current when starting (which is most of them) tend to have a "starting capacitor" to help manage the surge current and also help adjust the power factor back towards 1.0 to improve their efficiency. The so-called "ballast" in fluorescent lights serves much the same purpose. We thank you for tuning in to basic electronics, and return you to your regularly scheduled FreeBSD programming. :-)