How do multimeters work?

Question

I am helping a high school student with a physics experiment on internal resistance of a cell. My background is in chemistry, not physics.

Browsing SE for lab design, Measurement of internal resistance of a battery pops up. Now the high school lab cannot generate square waves so I was looking for alternatives.

I understand that in an open circuit, emf is the same as the voltage across the cell. The school does not have a potentiometer. There are videos such as this video at 4:20 that shows an open circuit, with an emf measurement. But wouldn't the cell form a circuit with the multimeter, thereby voiding the set up because the current is non-zero?

Please kindly point out what I am missing, and provide me with tips on how to avoid shorting the circuit and/or the multimeter. Thank you!

Answer 1

The multimeter does indeed form a circuit with the cell. The multimeter is designed to have a very high resistance, such as 10 mega-ohms, when in voltage mode. So the current is nearly zero - it's nearly an open circuit. Because the current is so small we assume it makes no noticeable difference and call it an open circuit. If you have a \$2V\$ cell and a \$10M\Omega\$ meter, the current is \$0.2\mu A\$.

If the cell's resistance isn't much lower than the multimeter, then you do need to calculate a correction, or use a different kind of measurement. For example, you could measure the voltage with different amounts of current, and then calculate what the voltage would be with zero current (open circuit).

(Likewise, the multimeter is designed to have a very low resistance in current mode - an approximate short circuit)

Answer 2

You can measure the internal resistance by applying a reasonable load (the definition will vary depending on the cell construction and intended use) and measuring the change in output voltage. Since batteries are not simple ideal voltage sources with fixed resistances in series (as they are sometimes modeled) the voltage drop will not necessarily be proportional to the current drawn, it will change with the time the load is applied (and how long since a load was last applied) and with environmental variables such as temperature. Also, of course, with state of charge of the cell.

If we think of it as a simple resistance in series with and ideal voltage source, then applying a load of R1:

^{simulate this circuit – Schematic created using CircuitLab}

Open circuit voltage is 9.000V. Voltage with 100 ohm load is 8.824V, so a current of 88.24mA drops 0.176V so the internal resistance must be 2.0Ω in this case. As it happens that's the default internal resistance in Circuitlab for a 9V battery.

If the load is relatively light then the voltage change will be small so you need a good resolution meter that will be stable during the measurements.

Answer 3

To add some background clarification to the excellent answers already given. There is a conundrum under this question which bears some explanation.

What we ideally want to do is measure electric potential without drawing ANY current. In practice this is rather difficult. There are physics techniques such as using an electroscope, but these are too impractical for most electronic engineering situations.

So what we do is measure while drawing a very small and known current. That is, we use a meter with a known, very high input, resistance. "High" is of course a relatiuve term; for most multimeters it is 10M ohm, which is chosen to work well enough with general purpose circuits which will use resistances typically in the ohms or kilohms region.

However we should always remain aware of what we are doing, in order to make allowances when working with "high impedance" circuits. This might occasionally be a concern when working with sensitive FET based circuits that (for instance) work with a capacitive transducer, such as an electret microphone. These circuits are designed to have impedances in excess of 100 Mohms or more, and we would not expect them to work properly with our multimeter connected. Indeed the reading obtained might well differ from the voltage in the circuit in normal operation, due to the excessive load presented by it.

For your purposes, you expect the load resistor to be perhaps a few hundred ohms or so. Measuring with a device with 10M input has a negligible effect : yes, we are drawing current, however such a small one that it can be disregarded.