Please quote some of the message you are replying to, so we can see
what you are talking about. (most of us don't use Google Groups, and
our newsreaders will only show one message at a time.)
I previously wrote:
>No. The program is telling the truth. The ammeter's resistance will
>be very low (probably zero in the simulator), so you effectively have
>1.44 ohm resistor across a 12 volt battery. By Ohm's law, the current
>through the resistor should be 8.333 amps.
In Real Life, the resistance of an ammeter is very low - probably
under 0.05 ohms, however, I would expect the simulator's ammeter to
have a resistance of zero.
If the meter resistance is zero, and you have it connected across your
.321 ohm resistor, the only remaining resistance in the circuit is the
1.44 ohm, so the current in the circuit is 12/1.44 = 8.33 amps.
By the way, Linear Technology has a very nice free simulator called
"LT Spice" available for download.
Yes the meter is set to 1nm. I forgot that electricity will take the
path of least resistance so with the ammeter resistance being near 0 it
simply allows the current to bypasses the .321 resistor leaving only
the 1.44 in the circuit.
Thanks for clearing that up.