Why does the potential difference across a type of parallel circuit not act like a potential divider? 
Image credit (Q3)
In this attached circuit, when $R_1=0\Omega$, I am failing to understand how the two cells affect the potential difference across the central resistor R3. I understand that potential difference is constant across different strands in parallel, and that so 12 volts should be distributed between R2 and R3 as a potential divider, and so 9V should be across R3 from V2. Likewise, from V1, 10V will cross R3 from V1, and so the total potential difference across R3 should be 19V. 
However, according to the answer sheet, the potential difference across R3 is 10V. Is my misunderstanding here conceptual, or something more basic, and why does potential difference act in this way?
 A: The pd across the terminals of the left hand battery is always 10 V. But because $R_1=0$, so is the pd across $R_3$! [The zero resistance conductors that run between the left hand battery terminals and the black blobs above and below $R_3$ effectively extend the battery terminals as far as these black blobs (and beyond)!]
I'm afraid that's all there is to it! When $R_1=0,$ the right hand battery has no effect on the pd across $R_3.$ [The excess 2 V (difference in battery voltages) is 'dropped' across $R_2$.] 
A: So this is how I've solved it generally, where the resistance for R1 is equal to k volts:
P.D. across R3 from V1: Equal to the P.D. across the parallel resistors of R3 and R2, as P.D. is identical across parallel components. Resistance of R3 and R2 branch equal to $\frac{1}{\frac{1}{30} + \frac{1}{10}}$, or 7.5. Therefore P.D. across this branch is equal to $\frac{10(7.5)}{k+7.5}$. 
P.D. across R3 from V2: Equal to the P.D. across the parallel resistors of R1 and R3, this branch has a resistance of $\frac{30k}{30+k}$, and so the P.D. across this branch is equal to $\frac{\frac{12(30k)}{30+k}}{\frac{30k}{30+k} + 10}$, or $\frac{9k}{7.5+k}$. 
Therefore P.D. across r3: $\frac{10(7.5)}{k+7.5} + \frac{9k}{7.5+k}$. As k-> 0, VR3->$\frac{75}{7.5}$ so in this circuit VR3 = 10v. 
A: If you are trying to view the system as a linear superposition of the two voltage sources, you must "kill" each source separately. To kill a defined voltage source (like the batteries), you should replace them with a short circuit.
Killing the 12 V source leaves $R_3$ and $R_2$ in parallel with the $10~V$ battery: $10~V$ contribution to each.
Killing the $10~V$ source leaves $R_3$ in parallel with a short circuit, attached to the $R_2$ and $12~V$ source, so there is a $0~V$ contribution to $R_3$.
$10+0=10~V$
You can't throw away the short circuit on the left.
If you are trying to view $R_3$ as part of a voltage divider, you can't ignore that it's in parallel with something else. You must evaluate that effective resistance.
A: 
so 12 volts should be distributed between R2 and R3 as a potential divider

If you're trying to solve this with superposition, you've forgotten a step or two.
When solving for the effect of the 12-V source, you need to zero out the other source. So you set $V_1$ to 0 V. Then R1 is in parallel with R3, and you need to use this parallel combination as the lower half of the voltage divider, not just R3. 
If R1 is 0, this means that the voltage across R3 due to $V_2$ is zero.
A: First, when $R_1 = 0\Omega$, resistor $R_3$ is in parallel with the source $V_1$ (a zero ohm resistor is identical to an ideal wire).  Parallel connected circuit elements have identical voltage across and so $V_{R_3} = V_1 = 10\,\mbox{V}$.
But, as an exercise, you should work out the general solution for $V_{R_3}$ when $R_1 \gt 0\Omega$ and then look at that solution as $R_1 \rightarrow 0\Omega$.  In fact, it would be good form for you to derive that solution and post your work as an answer to your own question.
