Amp Meter

 

Hi guys
I have a circuit with a 24 volt DC power supply 2 amps that feeds about 20
small relays and I have an amp meter that just says 1 to 5 on it.  How would
I put this in the circuit I think it would go in series but to draw 2 amps
through the meter would that not damage it
Thanks
Nikki

 

Nikki wrote:


most likely the meter is already designed to be used inline directly.
other wise you would need to know the R of the meter coil and voltage
of the meter to calculate for a shunt resistor.

 

On closer inspection of the meter I have taken the cover off and the needle
will not move so I think the meter is no good.  I have a second I would like
to use this one goes from 0 to 60 On the back it has a + on one of the posts
and printed on the back is 120  on the front of the meter is says (its very
small print) FS-50mvDC  and it also says D C Amperes. Now I know this came
from the Phone company and it was on a 48 volt circuit.
Nikki

20

would

amps

 

Nikki wrote:

Amp meters have a sensitive movement connected in parallel with a low
resistance current carrying shunt that produces the small voltage
required to move the meter (50 mV in this case).  The open question
with any amp meter is whether that low resistance shunt is sealed
inside the meter or is assumed to be external to the meter.  Measure
the resistance of your amp meter.  If it is less than an ohm, the
shunt is almost certainly inside the meter, and it is safe to pass
amperes through the meter terminals.  If higher than an ohm, the meter
is supposed to have an external shunt resistor connected to it, and
the load current is supposed to be passed through that resistor
through an extra pair of terminals.  Here is what an external shunt
resistor might look like:
http://www.carrel.co.nz/shunt/main.htm
Note the big terminals for the load current and the smaller ones for
the meter connection.

A more detailed review of ammeters:
http://www.allaboutcircuits.com/vol_1/chpt_8/4.html

--
John Popelish

 

Nikki wrote:


  you will need to determine the coil R (resistance), 120 maybe it.
  lets assume for the moment that it is.
  your scale is 0..60, we will use that for amp scale.

  Imeter = 0.050/120 = 416 uA ( does not seem likely etc).

  Rshunt = 0.050/(60.0 - 0.000416) = 0.00083 ohms

   so this would mean you need a shunt resistor on the
meter terminals of 0.00083 which is more like  a bar
of copper across the terminals of the meter.
   if the 120 number you are reading is actually indicating
I of the meter then you use 120 uA or 120 mA  instead of the
  416 uA in the first calculation.

      in either case, you will need to refer to a wire&meteral
chart to get teh ohms per 1K' (per 1k feet) to determine what
you need to create a shunt .

    these are only my thoughts off the top of my head going by the
info your passing.