AS CIRCUITS HELP please

Yes the pd across V₂ and the resistor R is 2V and yes the resistance of the combination of voltmeter and resistor R in parallel is 5Mohms.
But that 5Mohm is the combined resistance of R and the 10Mohm voltmeter in parallel, not just R.
This gives a value of 10Mohm for R, which when combined in parallel with the 10Mohm meter equates to 5Mohm.

Why doesn't the value of the resistance change in the Voltmeter in parallel with the resistor?

Yes the pd across V₂ and the resistor R is 2V and yes the resistance of the combination of voltmeter and resistor R in parallel is 5Mohms.
But that 5Mohm is the combined resistance of R and the 10Mohm voltmeter in parallel, not just R.
This gives a value of 10Mohm for R, which when combined in parallel with the 10Mohm meter equates to 5Mohm.

I fully understand that and fully understand why you are correct. However referring to Kirchhoff's Second Law specifically (please refer to last diagram on this website http://www.regentsprep.org/Regents/physics/phys03/bkirchof2/) it says that in any closed loop in the circuit, the sum of emfs = sum of pds. Taking the closed loop with R and V1 we have 6 V dissipated in each component, therefore IN THIS LOOP we have R having 2V dissipated.

Then it all goes wrong. Is it because you can't take a potential divider circuit in a loop, it has to be in the whole circuit?

Correct.


It goes "wrong" because you have two loops in this problem and you can't solve it using Kirchhoff's Law on only one of the loops.

Now apply it to the lower loop.

So applying K2 on the loop with R, we deduce that R dissipates 2V. Applying K2 on the loop with V2, we deduce that R dissipates 2V.

Can we now say that since they are in parallel, there must be a total of 2V coming into the 'junction' and then the result follows?

Yes but
If you decide to use K's Laws you don't talk about "parallel" in that way.
The reason for the confusion is that there is absolutely no need to do this problem using Kirchhoff's Laws on the system of two loops. This is a simple series circuit with one of the parts being a parallel system of two resistors. The answer I gave previously is all you need. The two pds add to 6V, we agree.
There is 2V across the parallel bit. The equivalent resistance of the parallel bit must therefore be half that of the component with the 4V across it.

K's Laws are normally applied when you have more complex circuits with more than one source of emf.

This circuit can be solved using basic series and parallel ideas.