# Does this look right?

S

#### [email protected]

Jan 1, 1970
0
I made a simple circuit in "Electronics Workbench" and the amp meter
reports over 8A across the resistor. I thought the amps would be very
low across a resistor in this kind of circuit. See link to image
below.

J

#### James Douglas

Jan 1, 1970
0
I made a simple circuit in "Electronics Workbench" and the amp meter
reports over 8A across the resistor. I thought the amps would be very
low across a resistor in this kind of circuit. See link to image
below.

What are the size of the resistors?

S

#### [email protected]

Jan 1, 1970
0
1.44ohm and .321ohm left to right.

L

#### Lacy

Jan 1, 1970
0
I made a simple circuit in "Electronics Workbench" and the amp meter
reports over 8A across the resistor. I thought the amps would be very
low across a resistor in this kind of circuit. See link to image
below.

No where near correct.. .You can not measure current across the resistor.
You need to put the ammeter in series with the circuit. The current will be
the total current flow in the circuit since it is a series circuit. Current
is the same in all parts of a series circuit.

S

#### [email protected]

Jan 1, 1970
0
That's what I thought. So the program is lying to me. He he and they
want over $2000 to license it. D #### Dave Jan 1, 1970 0 That's what I thought. So the program is lying to me. He he and they want over$2000 to license it.

Actually your circuit is designed wrong to check current here... take the
amp meter off of where you have it and but it between the positive lead of
the battery and the first resistor .... you will actually "break" the path
(wire) between the battery + and resistor and put the amp meter in its
place... amp + to the positive side of the battery and the Amp - goes to the
resistor... in a series circuit the current remains the same throughout the
whole circuit (both resistors will have the same current flow...

You have your meter is Parallel... needs to be in series..

Adding your resistors together ( 1.44 + .321) <-- I take this is point 321
ohms... this will give you an RT of 1.761 ohms

Ohms law states V/R=I ... in this case it would be 12v divided by 1.761 =
6.814 amps

Dave

L

#### Lacy

Jan 1, 1970
0
That's what I thought. So the program is lying to me. He he and they
want over $2000 to license it. No the program is not lying. Your configuration is incorrect. The ammeter needs to be in series with your circuit. Check below and try it. +----------------------------------------| | - -----1.44 ------ 321 --- Ammeter ----| P #### Peter Bennett Jan 1, 1970 0 That's what I thought. So the program is lying to me. He he and they want over$2000 to license it.

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.

What you need to do when measuring current is break the circuit, and
insert the meter in the break. If you do this, the meter should read

S

#### [email protected]

Jan 1, 1970
0
Yeah but shouldn't the ammeter read 0 amps or near zero when connected
the way shown?

S

#### [email protected]

Jan 1, 1970
0
So the ammeter becomes a 1.44 resistor?

P

#### Peter Bennett

Jan 1, 1970
0
So the ammeter becomes a 1.44 resistor?

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 said:
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

P

#### Peter Bennett

Jan 1, 1970
0
Yeah but shouldn't the ammeter read 0 amps or near zero when connected
the way shown?

No. Since it is in parallel with the .321 ohm resistor, the total
current in the circuit will divide between the .321 resistor and the
meter, in inverse proportion to their resistances. The simulator
meter probably has a simulated resistance of zero, so all the current
will flow through it, and none through the .321 ohm resistor.

In Real Life, the meter will have some resistance (my Fluke meter is
0.04 ohms on the 10 amp range). In that case, some current will flow
through the .321 ohm resistor, but most will still flow through the
meter - by my calculations, you would have about 6.6 amps through the
meter, and 1.7 amps through the .321 ohm resistor.

S

#### [email protected]

Jan 1, 1970
0
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 said:
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

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.

D

#### DSallee

Jan 1, 1970
0
So the ammeter becomes a 1.44 resistor?

nope... it just checks the current..... has no effect on the circuit
itself....

Here is how the circuit and amps SHOULD look for the circuit and I also
added 2 more meters hooked up like you have (across the resistors).. they
are to check voltage drop across the resistors... not current.... like
this......... www.audiopulse.net/amptest.jpg

Dave

S

#### [email protected]

Jan 1, 1970
0
By the way, Linear Technology has a very nice free simulator called

I went to their web site but didn't find LT Spice. Just some other
programs that didn't look like sims.

P

#### Pooh Bear

Jan 1, 1970
0
That's what I thought. So the program is lying to me.

It's teling you what the current is through a 1.44 ohm reistor with 12V
applied.

The answer is correct. The ammeter 'shunts out' the second resistor.

You can't wire up a circuit.

You don't get amps *across* a resistor.

You're an idiot. Go learn something about Ohms Law for starters.

Graham

P

#### Pooh Bear

Jan 1, 1970
0
Yeah but shouldn't the ammeter read 0 amps or near zero when connected
the way shown?

No !

Stop making a fool of yourself and learn something about current flow.

Graham

S

#### [email protected]

Jan 1, 1970
0
There's always one of you stupid mother fuckers on these NG huh?
Calling someone an idiot without any real knowlege of their own. My
guess is you don't know anymore about electronics than I do. In fact
you are probably not intelligent enough to learn so you just spew
insults. Typical huh?

S

#### [email protected]

Jan 1, 1970
0
Yep, pooh is what you are full of so I guess at least your name fits.

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