Re: Multimeter woes



Circa 4 May 2007 03:11:26 -0700 recorded as
<1178273486.828798.201530@xxxxxxxxxxxxxxxxxxxxxxxxxxx> looks like Noodle
<dean@xxxxxxxxxxxxxx> sounds like:

It looks like a decent meter, but it has no current ranges. It looks like
the usual current functions were traded off for the capacitance meter and
frequency counter functions. You have a 200 mV range, so you could insert
a .1-ohm precision resistor into the circuit and measure the voltage
across it, reading .199V as 1.99 amps.

Thanks for that Stephen, wish I understood what you was going on about
hehe. I'm gonna bookmark this thread then read this in a few months
time when I have learn't a bit more and will probably understand what
you mean then hehe. :P

In your first post, it seemed clear that you know how to find amp values by
combining voltage and resistance:

"...cannot for the life of me find how to do it unless I measure the
resistance and voltage separately."

What Stephen is saying is the same thing, and many high-power circuits that
have a need to know current conditions real-time, use these small,
precision resistors. They are called shunt resistors.

The value is chosen small enough so not to interfere in more than a
negligible way with the current in the circuit. So, you build a circuit,
get it wired up, and oh by the way put a 0.1 ohm resistor in the common
path. Don't forget to put test terminals on each side. Get the thing
running, and measure the voltage across the resistor. Multiply the voltage
by ten (which is the same as division by 0.1: I = E / R), and you have the
current.

You can use this technique as a way to measure current in any circuit.
Interrupt the output path, and insert the precision resistor in the path
(if you put it in the return/common/ground path, exposure to high voltage
will be avoided). Reconnect everything, and voltage measurements across
the resistor will allow you to simply convert the measurement to amps.
This is precisely how ammeters work, when you buy one that has the
function. The big giveaway hint is that in addition to AMPS range on the
meter selector, it will have a separate input jack for measuring amps.

.



Relevant Pages

  • Re: Einsteins Doppler equation wrong?
    ... the voltage is zero, i.e dE/dt is at its greatest, and ceases when the ... short circuit. ... :>: consisted of a connector with a small resistor tying the core ... :>: Without the terminator, the thinnet cable spits back echoes into ...
    (sci.physics.relativity)
  • Re: Need help with a MOSFET electronic load
    ... questions about circuit 1 that I copied from an internet source. ... To behave like a fixed resistor I=V/R. ... let's look at the voltage across and current through the MOSFET ...
    (sci.electronics.basics)
  • Re: 18 volt charger
    ... and I am trying to build a charger circuit for ... I am starting with this basic circuit (attached -- I ... That resistor will also need to be a beefy one to carry the ... How do I calculate Ry to give the proper voltage output? ...
    (sci.electronics.basics)
  • Re: Understanding circuits
    ... I'm going to breadboard this circuit up and perform some ... correct voltage to the respective areas. ... If I have a 2k and a 1k resistor in series supplied by 9V, ... the opamp is powered by a constant voltage some electronics ...
    (sci.electronics.basics)
  • Re: Thevenin-equivalence
    ... he's doing things like this (Thevenin and Norton equivalences). ... Consider this circuit: ... U1 is calculated using the formula for voltage drop over R1: ... calculation of Rth (resistor for the Thevenin-equivalent) ...
    (sci.electronics.basics)