Solving a Multimeter Problem with 10k Internal Resistance

[scientifico]

Hello, I'm trying to learn some electronics by myself but it's quite difficult, now I'm studying the multimeters and a problem says

What will a 20000Ω/V meter read on its 1 volt scale, when attached to a 1 volt source with an internal resistance of 10k?

Can you set up this problem for me because I didn't yet understood well the argument. Thank you!

 

[Averagesupernova]

Analog meters have an ohms per volt spec. This means that if the meter a 20,000 ohms per volt spec and has a 10 volt range that means that the load the probes will put on a circuit while in the 10 volt range would be 200,000 ohms. So do the math figuring a series circuit of 10K ohms with 200K ohms and a 1 volt source.

 

[grzz]

It is equivalent to having the voltmeter in series with a resistance R1 of 10k and another R2 of 20k.

Use V = IR to find V2 which is the reading of the voltmeter.

 

[scientifico]

But doesn't an high resistance mean a lower sensibility (because it "resist" to more electrons)? Instead why it's the opposite here?

 

[phinds]

But doesn't an high resistance mean a lower sensibility (because it "resist" to more electrons)? Instead why it's the opposite here?

I don't understand what you just said, but here's the thing: high resistance in a meter means the meter will have less effect on the circuit that it is measuring because it will draw less current out of it. That's a good thing.