measurementtemperature
A four terminal sensing is used to compensate the resistances of the conductions. Why is that as I even use more conductions than in a two terminal sensing?
The current from the current source will flow through conductions which have a resistance and the voltage metering is also connected via connections.
So why are these conduction resistances negligible?
Indeed, if you hold only one probe of your meter to the circuit, then the meter's circuit isn't closed, so there's no voltage difference to be measured.
Absolute voltage does not exist, it's always relative to some other point. So if you only use one probe of the meter it will say "compared to what?".
Ground is your reference for all voltages in your circuit. If a voltage is +12 V that will be relative to ground, unless otherwise noted. So you place the meter's reference probe, the black one, on the circuit's ground, and then that same +12 V will also seen by the meter as +12 V, because circuit and meter share the same reference.
OK, problem solved. The bridge is connected like this. Only one resistance in the load cells is variable, the other is fixed.
Why the confusion above? I was measuring resistance of a load cell which came from a different scale. The cells looked pretty similar, therefore I thought they were the same. But they were not! Eureka!
simulate this circuit – Schematic created using CircuitLab
Best Answer
Because a voltmeter ideally has infinite resistance, and in practice is typically 10 megohms or more.
In the case of the current meter that means the current it measures is very nearly the same as the current through the device under test (DUT).
And in the case of the voltmeter it means the current through the leads that connect it to the DUT is very small, and therefore any resistive voltage drops in those leads is also very small.
Edit
In comments you said
Let's work an example. Suppose you're trying to measure the resistance of a 1 ohm resistor, and your interconnect wires have 10 milliohms resistance each. We'll use a 10 mA probe current. So the two measurement scenarios look like this:
simulate this circuit – Schematic created using CircuitLab
Let's assume our voltmeter has 1 megohm burden resistance. Now in the 2-wire measurement, the ammeter reads 10 mA and the voltmeter reads .010199 V. So we estimate the resistance of our DUT to be about 1.02 ohms.
In the 4-wire measurement, the ammeter again reads 10 mA (being in series with the 10 mA source), but our voltmeter reads 0.00999999 V. So we estimate the resistance of our DUT to be about 0.999999 ohms.
The error in the 2-wire case is about 2%, dominated by the series resistance of the interconnect wires. The error in the 4-wire case is only about 0.0001%, from the effect of the 1 megohm voltmeter in parallel with the load.