# Changing the resistance of a voltmeter

1. Sep 2, 2009

### pp280

The setup shows a Galvanic cell connected to a voltmeter. The question asks how the reading of the voltmeter would change if the voltmeter's resistance became finite (instead of infinite) and current started to pass through the voltmeter.

The answer given is "The reading will decrease because the voltage decreases as the current increases"

I've tried drawing circuit diagrams to show this, but I'm stuck. Here's how I approach the problem:

First, I draw a circuit with a battery and a single resistor. The galvanic cell is the voltage source and I assume there is some internal resistance in the circuit. Since the voltmeter initially has infinite resistance, I leave it out.

When the voltmeter shorts, that's like adding another resistor in parallel. However, I don't see why this would change the voltage measurement. Adding a resistor in parallel decreases the total resistance of the circuit, and causes the total current to increase. BUT isn't it the case that while the total current in the circuit is greater than before, it splits in such a way that the same voltage drop occurs across both resistors?

As far as the given answer's reference to voltage decreasing as current increases, that sounds like P= IV, but I can't see how that would enter the picture.

Would appreciate any help!

2. Sep 2, 2009

### rl.bhat

Voltmeter consists of a galvanometer with a high resistance in series with it. Galvanometer is calibrated to measure the voltage depending on the small current flowing through it. When it is connected across the galvanic cell, current flows through the small internal resistance and high external resistance.
The emf of the cell is given by E = I*r + I*R.
The voltmeter shows only I*R reading. If you reduce R, the current will be more . But I*r drop will also be more. By changing R, emf of the cell does not change. Hence the voltage reading will decrease.

3. Sep 2, 2009

### jambaugh

The main point is that the galvanic cell has internal resistance.

The equivalent circuit is an ideal battery in series with a resistor.

4. Sep 2, 2009

### pp280

jambaugh - That's what I assumed. If that's the main point perhaps I'm missing something. Could you clarify? The confusing issue is what the voltmeter does when it has a finite resistance. Does it act like a resistor in parallel? Or is it a resistor in series as rl.baht seems to be saying?

rl.bhat - are you saying that the equivalent circuit is actually 2 resistors in series? I didn't quite follow your explanation about the voltmeter involving two resistors. I can see that IF the circuit was set up in the way you suggest: E = I*r + I*R things would make sense, but I'm not convinced as to why the circuit should be set up that way. Also, why is it that the voltmeter only measures I*R?

5. Sep 2, 2009

### rl.bhat

Yes. The circuit is actually 2 resistors in series.
The voltmeter measures the voltage across the terminals of the cell. That is why it is called the terminal voltage.
Definition of emf is the amount of work done to take a unit positive round the circuit. If there is any resistance in the path, some amount of the work is lost. Ideal voltmeter measures without taking current. And ideal cell is one which has zero internal resistance.
So the terminal voltage is E - I*r

6. Sep 3, 2009

### jambaugh

Basically the voltmeter is measuring the a microcurrent through a calibrated high-value resistor. Say for example 100 mega-ohms so ten microamps corresponds to a volt.

Now say the Cell has internal voltage of 1 Ohm and terminal voltage of 1 volt. With this setup you get an actual series resistance of 100,000,001 Ohms and current of about 9.9999999 microamps which reads as 0.99999999 volts.

Recalibrate the voltmeter to a 1000 ohm load and 1 milliamp per volt and you'll get total 1001 ohm series resistance with current of 0.999 miliamps which reads as 0.999 volts.