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. 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 about 6.814 amps.

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Peter Bennett VE7CEI
email: peterbb4 (at) interchange.ubc.ca

Please quote some of the message you are reply>No. The program is telling the truth. The ammeter's resistance will

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.

--
Peter Bennett VE7CEI
email: peterbb4 (at) interchange.ubc.ca

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.

--
Peter Bennett VE7CEI
email: peterbb4 (at) interchange.ubc.ca

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:

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.

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.

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There's always one of you stupid mother f****rs 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?

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..

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.........

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