So I understand the practicle application of measuring voltage drops accross a resistor.

But why does it work that way? If I measure from a point before a resistor, and then a point after the resistor, how is my multimeter measuring that voltage?

You're essentially creating an alternate path for the voltage to travel... wouldn't that alter the current accross the circuit and therefor change the voltage drop of the resistor?


If I tried to measure two points on a circuit without going through a component, shouldn't that still give me a voltage value?

I've learned a lot of this a long time ago, but haven't utilized the knowledge and much of it's lost. Google hasn't helped me much, so I wondered if I could get help here.

 

Different meters work different ways. The old analog meters works by putting a meter movement along with a series resistor in parallel with the load between the test leads. And, yes, this results in some of the current that was flowing in the load resistor now flowing through the meter. But as long as the difference is sufficiently small, you still get a useful reading. Also, if you wanted to, you could calibrate the meter to indicate what the voltage would be expected to be with the meter removed from the circuit.

Most digital meters read the voltage using an ADC (analog-to-digital converter) and generally have the same impact on the circuit as you would get my placing a 10MΩ resistor across the load. The effect is usually tiny, but in some situations can be significant and whoever is making the measurements should develop a sense for when and how to deal with those situations.

Now, when you say "without going through a component", does that mean that there is a wire between the two points? If so, then the voltage measured should be very near zero. If you had something else in mind, please throw together a sketch and post it.

 

Something that would work with both types of meters is a simple buffer, such as a unity gain op amp. I believe most ADC have something similar in place, whereas analog meters try to get by without any power. Adding a buffer like this might improve a analog meters performance.

 

I posed a problem at another forum ... what would your multimeter read if the linear potentiometer was set at 50%

Afterwards, I supplied this graph with a 10 M input impedance, a 5 M, and a simpson 260 at 20k/volt.

Knowing what your meter should read would be invaluable. That is why your meter's input impedance should be greater than 10x your circuit impedance. Remember the simple problem of the total resistance when a 1M is in parallel with a 100k or maybe a 10k ...

 

Knowing what your meter should read would be invaluable. .

Personal rule: You must know what you are expecting before you make the measurement. If you don't, the results will mean nothing to you. The best you can do is write down the measurement and then check the schematic to see what it means.

 

Eons ago, when very accurate voltage measurements were needed, they used a type of DC voltmeter that drew no current from the circuit being measured. It was called a differential voltmeter and operated by applying a very precise voltage provided by an accurate bridge circuit and voltage reference to the circuit under test through a very low current null-type analog meter. You adjusted the applied voltage with calibrated potentiometer knobs until the meter read zero current. At that point the measured voltage was equal to the applied voltage, which was read from the potentiometer settings.

 

I remember using a VTVM or vacuum tube volt meter, which had a high input impedance (>10MΩ). Until recently I had a FET VOM, which was battery operated and had similar characteristics. Tube meter had to be calibrated on each scale before use, but that was relatively simple. The tube meter was more forgiving when the input was overloaded. If you've ever damaged a DVM, you know what I mean.