Cfrgtkky

I have this setup.

It consists of a battery of no internal resistance with voltage $V$ and a resistor with resistance $R$. It also consists of a voltmeter of some (not so large) resistance as good ones should have.

Now my question is, will the resistance of the voltmeter matter for this particular circuit?

I think that the voltmeters need to be high resistance only when there are multiple resistances in the circuit... and as the internal resistance of the battery always contributes to the total resistance in real cases, a voltmeter is always expected to be very high resistance.
(Please correct me if I'm wrong)

I have seen other questions related to voltmeters on this site and googled up but couldn't find an answer.

If you have a battery with no internal resistance, or with an internal resistance that is negligibly small compared to your resistor $R$ (and the voltmeter resistance), then yes a relatively low resistance of your voltmeter won't change the measured voltage.

You are correct that the voltmeter can affect the voltage across $R$ only when there are multiple resistances in the circuit because if there are no other resistances the voltage drop across the (non-existant) other resistances has to be zero.

Now my question is will the resistance of the voltmeter matter for this particular circuit?

No.

Since the battery has no internal resistance it is an ideal voltage source, meaning the battery terminal voltage, $V$ and thus voltage across the load $V_{R}$, will always be the same regardless of how low the resistance $R$ is. Although the resistance of the voltmeter in parallel with $R$ lowers the equivalent resistance and increases the current from the battery, it has no influence on the voltage across R being measured.

A real battery has internal resistance $r_{b}$. Which means any increase in current due to current drawn by the voltmeter will cause a voltage drop internal to the battery and reduce the voltage across R

The voltmeter must have an input resistance sufficiently greater than the resistance across which it is connected so that the current drawn by the parallel combination does not result in an appreciable drop in battery voltage, which is the voltage being measured.

I have shown your circuit below but with internal battery resistance to show the effect of it on the voltage measured.

Hope this helps

The internal resistance ($R_V$) of the voltmeter is in parallel with $R$. Hence, the total load resistance on the battery ($R_L$) is slightly lower, as given by$$R_L=frac{RR_V}{R+R_V}$$

As a result a slightly larger current will flow from the battery. This in turn will cause the battery voltage to drop slightly, due to the battery's internal resistance.

In this case the voltmeter will measure a slightly lower voltage than it would if its internal resistance were infinite. A voltmeter with non-infinite internal resistance (i.e. a real one) always puts some additional load on the circuit and hence it will always affect the reading.

You are right that a low resistance in the voltmeter will not affect its measurement. The rule that a voltmeter resistance should be high compared with all resistors in the circuit is so that you can use the voltmeter in series resistance/impedance circuits without have to ask your question.

I would modify your claim to include impedance and not just resistance.

Having an extremely high resistance in a voltmeter is crucial to measuring the voltage drop across a component.

The formula for finding the equivalent resistance of 2 resistors in parallel is $frac{R_{1}R_{2}}{R_{1}+R_{2}}$.

Now, $$lim_{R_{2}toinfty} frac{R_{1}R_{2}}{R_{1}+R_{2}} = R_1$$ if you go through some algebra. Therefore, we can see that the higher the resistance in our Voltmeter is, the less relevant it becomes in the equivalent resistance in the system.

In your specific circuit, we're only trying to measure the voltage drop across 1 resistor, so the resistance of the voltmeter matters immensely. According to your problem statement, the voltmeter does not have infinite resistance so we cannot ignore its resistance in this specific case.

You have it backwards. The resistance of virtually every digital voltmeter is standardized at 10 megohms. The idea is the circuit should not "see" the meter. Almost no current flows through the meter. Conversely anmeters should have the lowest resistance possible so there is almost no voltage drop across them. The reason DMMs make lousey battery testers is because they do not load the battery. The resistor in your drawing would provide a load for a test. The problem with failing batteries is their internal resistance keeps going up as much as voltage going down..When the battery is fresh the internal resistance is very low compared to the load resistor. As it goes up, more of the battery's voltage drops across it's internal resistance and less apoears across the load resistor. A basic ohms law voltage divider.