What is the load of a multi-meter when used as a voltmeter.?

Question

Also, how does the load of a multi-meter (voltmeter) affect the resultant voltage across two resistors in series in a simple circuit. What would be the affect of a load resistance on the voltage divider ( the two resistors in series) performance.

Answer 1

You need to know the internal impedance of the voltmeter, and that will depend on the voltage scale you are using. The meter should have a specification of "ohms per volt". Multiply this by the full-scale voltage setting of the meter to get it's internal resistance. Then apply parallel resistance formula to find the restance with the meter connected. For example, if you have a series resistance circuit with two resistors R1 and R2, and you want to measure the voltage across R2, you connect the meter across R2. If the meter has a specificaton of 1000 ohm per volt, and you are using the 10-volt scale, the internal impedance is 10,000 ohms. The value of R2 will be changed to (R2*10,000)/(R2+10,000).

This applies to passive multimeters; electronic meters (used to be called VTVM, but now most are digital) have such a high internal impedance that it can be ignored.

Answer 2

Ideally a voltmeter should have infinite resistance. Practically, depending on the make of the devide, the resistance varies in terms of Mega Ohms. The main aim here would be to reduce the amount of current that is tapped from the main circuit into the voltmeter so as to no to affect the conditions in that circuit.

An average voltmeter shouldn't have any effect on a potential divider (pot) having considerably low potential (maybe less than 50V). Above this voltage, the effect of the voltmeter might manifest itself as unwanted load across the pot. this is where the quality of the voltmeter comes in to the picture.

As far as the effect of a load on a pot is considered, this is basically dependent on the resistances of the load and the ones in the pot. This can be calculated by a simple formula available in popular texts.