Date:         Mon, 19 Mar 2007 23:24:20 -0700
Reply-To:     Michael Elliott <camping.elliott@GMAIL.COM>
Sender:       Vanagon Mailing List <vanagon@gerry.vanagon.com>
From:         Michael Elliott <camping.elliott@GMAIL.COM>
Subject:      Re: ammeter question
Comments: To: David Etter <detter@MAIL.AURACOM.COM>
In-Reply-To:  <f06240800c224eec9bb18@[192.168.1.100]>
Content-Type: text/plain; charset=ISO-8859-1; format=flowed

No current loss at all. Current going in would be exactly equal to
what's coming out. The current goes through a coil and the resultant
magnetic field moves the pointer. This does take power, which is caused
by passing that current through a shunt resistor of a few milliohms,
causing an eentsy weentsie voltage drop. It's common to see analog
ammeters set up with shunt resistors such that there is a 50 millivolt
(.05 volt) drop across the resistor at the desired full-scale pointer
reading. So for a 10A meter, the shunt would be R = .05/10 = .005 ohm.
Power consumption at 10 amps is .05 x 10 = 1/2 watt. Nothing to worry
about. And when there is no current flowing (from the panel) the meter
isn't drawing anything. It just reflects what's going through it.

--
Mike "Rocket J Squirrel" Elliott
71 Type 2: the Wonderbus
84 Westfalia: Mellow Yellow ("The Electrical Banana")
74 Utility Trailer. Ladybug Trailer, Inc., San Juan Capistrano
KG6RCR



David Etter typed:
> Can anyone tell me if using an analog ammeter (located in-line from
> my solar panel) would result in current loss?  It would seem to me
> that by creating a field in a coil to operate the ammeter it would
> perhaps act much like a choke in a normal electrical circuit.
>        Any field offers resistance to current flow; but perhaps the
> loss is so small....  ???????
>
> Thanks!
>                                        David
>