I was wondering how can I find internal resistance of multimeter, in both voltmeter and ammeter capabilities?
I thought it was termed as input impedance but when I looked up the specs of multimeter, in example, Agilent 34410A Digital Multimeter, it said load impedance was 1 MΩ ± 2%, in parallel with <150 pF in True RMS AC Voltage, which did not make any sense to me at all.
I experimentally measured internal resistances for both voltmeter and ammeter capabilities of multimeter but I wanted to compare this experimental value to true value of internal resistances and make percent errors. Help please
Best Answer
In volt measuring mode, you can set up a voltage divider with two known resistor values and a known voltage source. You can then very simply compute the expected voltage you'd measure across one of the resistors. First, make sure your voltmeter is zeroed out without anything connected. Second, just apply the voltmeter to the voltage source itself and take down that reading as \$V_{source}\$. (Or tweak the voltage source's voltage so that you get a desired reading value.) Now apply your voltmeter across the chosen resistor and get a reading as \$V_{meas}\$.
Assume \$R_1\$ is the resistor across which you are not measuring the voltage. Assume \$R_2\$ is the resistor across which you are measuring. Then the effective resistance of your voltmeter will then be:
$$R_{V}=\frac{R_1 R_2}{\left(\frac{V_{source}}{V_{meas}}-1\right) R_2 - R_1}$$
This will probably work better when the values of \$R_1\$ and \$R_2\$ are equal to each other and within a factor of 5, higher or lower, of your voltmeter's expected impedance.
A similar method can be developed, assuming you have a current source and some resistors similarly near the expected impedance of your ammeter mode. The difference is that you'd set up your two resistors to be in parallel, rather than series and you'd take your current measurement using the ammeter in series with \$R_2\$ rather than across it, of course. Then:
$$R_{I}=\left(\frac{I_{source}}{I_{meas}}-1\right) R_1 - R_2$$
You should be able to work out the above equations, though.
A voltmeter should have relatively high impedance and an ammeter should have rather low impedance. So do not expect similar values.
But your meter is a \$6\:\frac{1}{2}\$ digit multimeter! It's possible you won't have similarly accurate equipment by which to judge it. But you should be able to roughly check it. Start with the voltmeter mode and use a pair of \$2.2\:\textrm{M}\Omega\$ resistor values to start. Use a voltage supply value that will test out each of the voltage ranges it accepts. (There are 5 ranges, see the manual.)