Return-Path: Sender: (Marvin Kaye) To: lml@lancaironline.net Date: Thu, 06 Nov 2003 07:19:29 -0500 Message-ID: X-Original-Return-Path: Received: from fep06-app.kolumbus.fi ([193.229.0.57] verified) by logan.com (CommuniGate Pro SMTP 4.1.6) with ESMTP id 2712055 for lml@lancaironline.net; Thu, 06 Nov 2003 03:16:24 -0500 Received: from fiz1ms01 ([194.211.242.158]) by fep06-app.kolumbus.fi with SMTP id <20031106081341.MJMN2570.fep06-app.kolumbus.fi@fiz1ms01> for ; Thu, 6 Nov 2003 10:13:41 +0200 Received: from fid1ms01.fid1.root4.net ([158.233.139.214]) by fid1ms31.fid1.root4.net with Microsoft SMTPSVC(5.0.2195.5329); Thu, 6 Nov 2003 10:16:04 +0200 X-MimeOLE: Produced By Microsoft Exchange V6.0.6249.0 content-class: urn:content-classes:message MIME-Version: 1.0 Content-Type: multipart/mixed; boundary="----=_NextPartTM-000-74a2cbd8-e6d8-4515-b322-55600214e1ba" Disposition-Notification-To: Subject: RE: [LML] Re: Ohm's law X-Original-Date: Thu, 6 Nov 2003 10:16:03 +0200 X-Original-Message-ID: <5341129A9EB2444F8A8DB3058DA5006803B5E0@fid1ms01.fid1.root4.net> X-MS-Has-Attach: X-MS-TNEF-Correlator: Thread-Topic: [LML] Re: Ohm's law Thread-Index: AcOjuWY4bOltx2rESjCQSKL7IY79igAhHQpA From: X-Original-To: X-Original-Return-Path: villi.seemann@nordea.com X-OriginalArrivalTime: 06 Nov 2003 08:16:04.0170 (UTC) FILETIME=[38A3EAA0:01C3A43E] This is a multi-part message in MIME format. ------=_NextPartTM-000-74a2cbd8-e6d8-4515-b322-55600214e1ba Content-Type: multipart/alternative; boundary="----_=_NextPart_001_01C3A43E.387227BB" ------_=_NextPart_001_01C3A43E.387227BB Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Ray wrote : =20 >Forget the resistor just put in a higher voltage bulb. See if you can = find a 24 volt one. That should cut the light in half.=20 > >Ray=20 =20 To the best of my knowledge the light will only be a forth or less, As = the the lamp only gets haft of the expected voltage, it will only draw = half of the stated amperage. Power equals voltage multiplied by amps, = and when they both are halved, power and hence light will be a fourth or = less due to increased filament resistance. =20 Best regards=20 Villi H. Seemann=20 Sen. Eng. BSEE=20 Telephony Team=20 Nordic Processor (Nordea/IBM)=20 Phone (+45) 3333 2101=20 Cell ph. (+45) 2220 7690=20 FAX (+45) 3333 1130=20 ------_=_NextPart_001_01C3A43E.387227BB Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable
Ray=20 wrote :
 
>Forget the=20 resistor just put in a higher voltage bulb.  See if you can find a = 24 volt=20 one.  That should cut the light in = half. 
>
>Ray=20
 
To the = best of my=20 knowledge the light will only be a forth or less, As the the lamp only = gets haft=20 of the expected voltage, it will only draw half of the stated amperage. = Power=20 equals voltage multiplied by amps, and when they both are halved, power = and=20 hence light will be a fourth or less due to increased filament=20 resistance.
 


Best regards
Villi H. Seemann
Sen. Eng. BSEE
Telephony = Team=20
Nordic Processor (Nordea/IBM) =
Phone   (+45) 3333 2101
Cell ph. (+45) 2220 7690
FAX      (+45) 3333 1130=20

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