# Why the potential across the voltmeter is not affected by changing metals? How can we prove it?

A voltmeter has both its leads made of metal Cu. When connected across a battery, it measures a voltage V. One of its leads is replaced by metal Au. But the potential across the voltmeter is not affected. I understand the sentence conceptually I am thinking that driving force will be the same the electric field will be the same that's why. Is this reason correct or there is any other reason with proof

A voltmeter is designed to have a high input impedance in order to draw as little current as possible so as not to affect the voltage being measured. The resistance of the leads is so small compared to the internal impedance that changing from Cu to Au will have a negligible affect on the current it draw and therefore negligible affect on the voltage being measured.

Hope this helps

The two most common voltmeters are the moving coil meter and the digital voltmeter (usually as part of a digital multimeter).

On the 25V range, a moving coil meter with $$50 \mu A$$ sensitivity will have an effective internal resistance of 500$$k \Omega$$ and is the range I will use for this discussion. A typical digital meter will have an effective input resistance of 10 M$$\Omega$$.

I am ignoring circuit loading effect (where the source being measured is a significant fraction of the input resistance of the measurement device) as it doesn't really apply here as we are looking at what difference will be seen in the reading due to the change in wire type (there will be very small differences in this situation but they can be ignored for the purpose of this discussion).

Now let's look at the resistance of the wires:

Assume that each set of wires is 1metre long, with a radius of 1mm (for the conductor).

As $$R= \frac {\rho L} {A}$$ for a metallic conductor where R is the resistance, $$\rho$$ is the resistivity of the material and A the cross sectional area, we can calculate the resistance of the wires.

For Cu with a resistivity of $$1.7 x 10^{-8} \Omega m$$ then each wire will have a resistance of approximately 2.7 $$m \Omega$$ for a total resistance (2 leads) of $$5.4\ m \Omega$$.

The result for Al is $$8.6\ m\Omega$$ for both wires.

Compared to the resistance of the moving coil, this will induce a reading error of such a small magnitude (about 10 parts per billion for Cu and 18 ppb for Al with a difference of 8 ppb assuming a very low source impedance but even at twice these numbers it is still incredibly small) as to be indistinguishable to the display and indistinguishable from noise for the DMM as the digitiser could not resolve such a small difference (except perhaps at NIST)