Circuit - Why a more accurate value of the resistance?

[vadevalor]

Circuit -- Why a more accurate value of the resistance?

Note: CURRENT CAN FLOW THROUGH THE VOLTMETER IN THESE CASES.
Why does the second setup as shown by the arrow in the picture give a more accurate value of the resistance of that wire?

It says the current in setup 1 on the left measured by the ammeter includes both the current through the wire and current through the voltmeter,
but for setup 2 on the right, it says this setup the ammeter will measure the current in the wire only.

They are both series circuits, so i can't get why there is a difference in currents for the 2?

 

[lewando]

The image on the left does not have the wire and ammeter in series. They are in series on the right.

In case you are not confused enough, the ammeter has a tiny internal series resistance. The voltmeter has a large shunt (parallel) internal resistance (this is why a tiny amount of current can flow through the voltmeter). Their presence in your circuit affects the circuit. For high-quality meters, the effect is negligible (measurement error is within the meter's specified uncertainty).

 

[vadevalor]

Why is the wire and ammeter not in series? Well i see the voltmeter as a separate entity, and its a complete rectangle circuit to me :0 i understand you :P

Parallel should be like on different lines right? ( in layman's term) one on a line in a small rectangle, one on a line in a big rectangle (where rectangle is the circuit)

 

[lewando]

Two elements are in series if the node (wire) that connects them has no branches. Series elements will have the same current going through them.

Two elements are in parallel if they are both connected across 2 nodes (wires). These nodes may have branches (unlike the node joining the two series elements). Parallel elements will have the same voltage across them.