Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

[SOLVED] Voltmeter Internal Resistance

Status
Not open for further replies.

~analoger~

Banned
Member level 1
Joined
Jan 4, 2013
Messages
37
Helped
1
Reputation
2
Reaction score
1
Trophy points
1,288
Location
Mars
Activity points
0
Hello,

Based on Boylestad's Introductory Circuit Analysis, a voltmeter places an additional resistance across the element of interest. As I understand this is voltmeter internal resistance, which should be very high. However, in another book, Nilsson & Riedel Electric Circuits, it shows that a voltmeter internal resistance is connected in series with the voltmeter itself. I'm lost here, which configuration is correct?

Thanks.
 

In essence all old fashioned voltmeters were actually current meters. A good one used a 50 microamp movement, so needed an external resistance of 20,000 ohms per volt measured, hence a voltmeter (meter calibrated in volts) would have a high resistance in series with it, although normally internally. Modern DVMs or old fashioned VVTMs, normally have a constant impedance on their voltage range of 10 M ohms, so can be disregarded for most measurement.
Frank
 

So far as the other book and resistance shown in series is concerned, this is in fact what is shown in post #2 by Chuckey.
The voltmeter conventionally is a current meter (sensitive galvanometer) connected in series with a high value resistance. So many times a voltmeter is shown as a meter with a series resistor if the internal resistance of the voltmeter is to be included in the circuit analysis ( if it contributes substantial load).
 

An ideal voltmeter has infinite resistance. It does not take any current from the circuit. A real voltmeter will take current from the circuit under measurement. So to model a real voltmeter you can use an ideal voltmeter with a resistor in parallel.
A resistor in series with an ideal voltmeter does not make any sense. Because the ideal voltmeter is an open circuit.

An ideal ammeter has zero resistance. To model a real ammeter you place an ideal ammeter in series with a resistance.

In practice as chuckey said, a voltmeter can be made with a resistance in series with an ammeter. Probably that is what you saw at the 2nd book.
 

So when asked about the reading of voltmeter of a given internal resistance without specifying what type of voltmeter (could be digital or an old analog), we add the internal resistance in parallel to the element/points/branch across which we want to find the voltage, and then find the voltage across this internal resistor?
 

That is right. Suppose that you want to measure the voltage of a 12v battery which has a 100k resistor in series.
An ideal voltmeter will measure exactly 12v.
If you place a voltmeter with 20000 ohm per volt sensitivity in the 15v scale it will have 20K x 15v = 300k internal resistance.
With that voltmeter you will measure V= 12v x 300K /( 100K + 300K) = 9 volts . Because the internal resistance of the voltmeter forms a voltage divider.
In the same way. A modern DVM with 10M internal resistance will measure V= 12 x 10M/(10M+100K)= 11.88v
 

Status
Not open for further replies.

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top