Voltmeter and its resistance

[Arnav]

A voltmeter has a very high resistance so that it doesn't draw a considerable amount of current from the circuit when connected in parallel. Lets say we have a circuit consisting of a single cell and a load of resistance R. Now a voltmeter is connected in parallel with a load of resistance R. If the resistance of the voltmeter will be low, then it will draw some current from the circuit , leading to an increased magnitude of current in the circuit than before( or we can say equivalent resistance of the combination would be less than before, leading to increased current).

If the cell has an internal resistance r, then the voltage drop across r would be more than before hence the voltage across R, which the voltmeter is to measure, will be less. Even if we take another resistance in series with R, and ignore r, then the measurements will be same, i.e. wrong.

But, now imagine that we take an ideal cell with internal resistance=0, and connect the resistance R with the voltmeter in parallel with it. Now, the measurement of the voltmeter will not be wrong even if its resistance is low, am I right ? 

Sorry if I couldn't explain my question, if anyone wants, then I can send some calculations to put up my statement.

[Sensei]

The box should state the internal resistance of the voltmeter. Typically this is 1 M ohms or 10 M ohms. Its effect is well below the resolution of the LCD display panel.

[studiot]

1 hour ago, Arnav said:

A voltmeter has a very high resistance so that it doesn't draw a considerable amount of current from the circuit when connected in parallel. Lets say we have a circuit consisting of a single cell and a load of resistance R. Now a voltmeter is connected in parallel with a load of resistance R. If the resistance of the voltmeter will be low, then it will draw some current from the circuit , leading to an increased magnitude of current in the circuit than before( or we can say equivalent resistance of the combination would be less than before, leading to increased current).

If the cell has an internal resistance r, then the voltage drop across r would be more than before hence the voltage across R, which the voltmeter is to measure, will be less. Even if we take another resistance in series with R, and ignore r, then the measurements will be same, i.e. wrong.

But, now imagine that we take an ideal cell with internal resistance=0, and connect the resistance R with the voltmeter in parallel with it. Now, the measurement of the voltmeter will not be wrong even if its resistance is low, am I right ? 

Sorry if I couldn't explain my question, if anyone wants, then I can send some calculations to put up my statement.

 

1 hour ago, Sensei said:

The box should state the internal resistance of the voltmeter. Typically this is 1 M ohms or 10 M ohms. Its effect is well below the resolution of the LCD display panel.

 

Firstly the internal resistance of a voltmeter is usually stated as ohms per volt.

The per volt refers to the volts at full scale and is fixed for any scale on the meter.

Some meters have a fixed value for all scales and therefore a fixed resistance some have only one scale and again therefore a fixed resistance.

Quote

But, now imagine that we take an ideal cell with internal resistance=0, and connect the resistance R with the voltmeter in parallel with it. Now, the measurement of the voltmeter will not be wrong even if its resistance is low, am I right ? 

 

Everything you have said is correct, including this.

However we would normally take the cell in this condition in circuit analysis it is called an ideal voltage source and also applies to alternating voltage sources.

The point is that although sometimes we have to consider the internal resistance of the supply (this might also include the circuit board track resistance in high current situations) we usually don't.

 

There is another reason for saying that voltage measurement can introduce errors.
You have used only a single resistor, but in a circuit there may be many resistors.
Say you have a simple potential divider say 10MΩ to 100kΩ that is a100 to 1 division ratio.

It is instructive to calculate the measured voltage drop across each resistor with say 100 volts applied.

Can you do this ?

Post your attempt and sensei or I will reply.

[Arnav]

13 minutes ago, studiot said:

Say you have a simple potential divider say 10MΩ to 100kΩ that is a100 to 1 division ratio.

Sorry, but I haven't been taught about potential dividers yet.

 

13 minutes ago, studiot said:

You have used only a single resistor, but in a circuit there may be many resistors.

Yeah, I reckon in an actual circuit the number of resistors would be more than 1, as my case of "cell with 0 internal resistance and only one load" is very very ideal. So in a non-ideal case, even if we ignore the internal resistance of the source, the voltage measured by voltmeter would again come out to be less if its resistance is not very very high.

I have uploaded 3 scenarios, one which is the actual case, one with low resistance voltmeter, and one with very high resistance voltmeter. I have ignored internal resistance of the cell. 

Couldn't upload the third case, but I know you guys get what I am trying to say

[John Cuthber]

3 hours ago, Arnav said:

A voltmeter has a very high resistance so that it doesn't draw a considerable amount of current from the circuit when connected in parallel. Lets say we have a circuit consisting of a single cell and a load of resistance R. Now a voltmeter is connected in parallel with a load of resistance R. If the resistance of the voltmeter will be low, then it will draw some current from the circuit , leading to an increased magnitude of current in the circuit than before( or we can say equivalent resistance of the combination would be less than before, leading to increased current).

If the cell has an internal resistance r, then the voltage drop across r would be more than before hence the voltage across R, which the voltmeter is to measure, will be less. Even if we take another resistance in series with R, and ignore r, then the measurements will be same, i.e. wrong.

But, now imagine that we take an ideal cell with internal resistance=0, and connect the resistance R with the voltmeter in parallel with it. Now, the measurement of the voltmeter will not be wrong even if its resistance is low, am I right ? 

Sorry if I couldn't explain my question, if anyone wants, then I can send some calculations to put up my statement.

You are right but, in that instance, why would you bother to measure the voltage?
The voltage would be written on the side of the cell.

 

3 hours ago, Sensei said:

The box should state the internal resistance of the voltmeter. Typically this is 1 M ohms or 10 M ohms. Its effect is well below the resolution of the LCD display panel.

Unless it isn't.
It would, for example, be wrong for systems with effective source resistances of more than 10K or 100K ohms.

 

1 hour ago, studiot said:

Firstly the internal resistance of a voltmeter is usually stated as ohms per volt.

That depends.
It's broadly speaking true for analogue meters but not for digital ones.

 

[studiot]

On 9/18/2020 at 11:11 AM, John Cuthber said:

That depends.
It's broadly speaking true for analogue meters but not for digital ones.

There are many learning points available in this thread.

The OP is trying to learn, not quibble.

[John Cuthber]

3 minutes ago, studiot said:

There are many learning points available in this thread.

The OP is trying to learn, not quibble.

And it's not going to help them learn if people say things that are simply not true any more, is it?

In the "good old days" when a voltmeter was essentially a galvanometer with a resistor in series, then the resistor could be calculated from the "ohms per volt" figure for the galvanometer- the figure is the reciprocal of the full scale current. (You needed to subtract the resistance of the meter itself)

The OPV was typically just about halved when using AC.

But if you got to buy a meter today, you are much more likely to get a digital one where the  input resistance is more or less fixed.

 

It's actually quite a useful feature- it means you can use the meter (on the volts range) as a very sensitive current meter.

[studiot]

1 hour ago, John Cuthber said:

And it's not going to help them learn if people say things that are simply not true any more, is it?

In the "good old days" when a voltmeter was essentially a galvanometer with a resistor in series, then the resistor could be calculated from the "ohms per volt" figure for the galvanometer- the figure is the reciprocal of the full scale current. (You needed to subtract the resistance of the meter itself)

The OPV was typically just about halved when using AC.

But if you got to buy a meter today, you are much more likely to get a digital one where the  input resistance is more or less fixed.

 

It's actually quite a useful feature- it means you can use the meter (on the volts range) as a very sensitive current meter.

 

So why don't you explain these things, and in particular answer the OP's questions,  instead of just shooting at me all the time ?

BTW just humour me and type into google 'analog meters for sale'  I just got over 15 million instant results.

There is considerably more to this than your rather glib and superficial comments make out.
In particular your omissions about the effect of a voltmeter in a more complicated arangement stand out.
This was where I was heading with my introduction via potential dividers when you intervened.

I will leave it to you to explain what happens when you measure the voltages across a 100k resistor in series with a 10 M resistor using a bog standard digital meters of input resistance 1M and 10M that Sensei mentioned.

Arnav has come here to learn, so to tell him that the applied voltage is 'written on the side of the cell' is about the worst piece of advice you could give.

@Arnav  I don't know if you are still interested in this thread  ?

 

 

 

Edited September 20, 2020 by studiot

[John Cuthber]

1 hour ago, studiot said:

he applied voltage is 'written on the side of the cell' is about the worst piece of advice you could give.

It's perfectly true in the circumstance he was referring to.

 

On 9/18/2020 at 7:31 AM, Arnav said:

imagine that we take an ideal cell with internal resistance=0

and that's the reason for his confusion.
In that case the resistance of the voltmeter is irrelevant.

Yet he has been told it is important.

1 hour ago, studiot said:

So why don't you explain these things, and in particular answer the OP's questions,  instead of just shooting at me all the time ?

Why didn't you explain it- rather than saying something that's not true?

 

1 hour ago, studiot said:

in particular answer the OP's questions,

Here's the question he asked
 

 

On 9/18/2020 at 7:31 AM, Arnav said:

am I right ? 

And here's the answer I gave
 

 

On 9/18/2020 at 11:11 AM, John Cuthber said:

You are right

 

 

1 hour ago, studiot said:

BTW just humour me and type into google 'analog meters for sale'  I just got over 15 million instant results.

OK, I tried and then I tried the important second part of the comparison.; I searched for 

 'analog  digital meters for sale' 

And guess what... there are more of them.

When you have finished moaning about me, the fact will remain.
What you said was wrong. Did you expect  that not to get picked up on a science site?

[Arnav]

Thanks guys for taking out your time and answering my doubt. Sorry I've been a little too inactive

Edited October 27, 2020 by Arnav