Why does a drop in resistance lead to a drop in voltage? 
I was working on this question which I know the answer to but I am still confused. I get that resistance is due to collisions on the atomic scale where the electrons' kinetic energy is transferred to the metal lattice and that a lower resistance means a low voltage because less energy is transferred as less collisions.
However, in this case, lowering the resistance of the variable resistor will lower the total resistance of the circuit which increases the current. So since $V=IR$, surely the voltage would remain the same because, yes the resistance is smaller but, the current is bigger?
 A: Ohm's Law (the equation, $V=IR$), like any other physical law, only is meaningful in the proper context. $V=IR$ is not a universal law that applies to just any Voltage $V$, any current $I$, and any resistance $R$.  If the $R$ is the resistance of a single, two-terminal, resistive/conductive component in a circuit, and $I$ is the current through the component at some particular instant in time, then the $V$ is the Voltage between the two terminals of that component at the same instant in time.
Your circuit diagram shows two resistors, but the Volt meter only shows the voltage across one of the two. According to Kirchoff's Voltage Law (KVL), the sum of the Voltages across the two resistors must always be the same as the Voltage supplied by the battery.*
The current through the two resistors must be the same. And it will be limited to some finite value by the other resistor. Ohm's Law tells us that as the resistance $R$ approaches zero, the current $I$ will approach the limit set by the other resistor, and the Voltage $V$ must go toward zero. Meanwhile, the Voltage across the terminals of the other resistor will approach the battery Voltage.

* Whenever you see a battery symbol in a circuit diagram, you should assume that it represents a constant voltage source unless something in the drawing tells you otherwise.
A: I'll explain it through the water analogy to get a mental picture which makes it easier to make predictions in the future. In my analogy the two resistors are represented by two dams which let a certain amount of water through via a small pipe. A smaller the diameter pipe means the flow is more restricted which corresponds to a bigger resistance. The flow rate is the current and the height of the water is the voltage which is natural because water height and voltage are both measures of energy density. We assume the flow rate through a pipe is directly proportional to the pressure difference between the two sides. Since  the pressure increase linearly with depth this means the flow rate is proportional to difference in water height between the two sides.
If we call the variable resistor $R_1$ and the second resistor $R_2$ then the picture looks like this:

If we decrease $R_1$ the pipe becomes larger. Initially the flow rate through the first pipe increases and the water level in the middle starts to rise. As this level rises the flow rate through the second pipe increases because the pressure increases and similarly the flow rate through the first pipe starts to decrease. When the two flow rates match exactly the water level stops changing and equilibrium is achieved. In this new state the middle water level is higher and the new flow rate is higher than it was before.

A: The way to think about it is as follows...
The battery provides a total voltage VT, so that must be the total voltage drop across the two resistors in series. Since the same current passes through both of the resistors, the overall voltage drop VT is shared across the two resistors in proportion to their individual resistance. So as the variable resistance increases and decreases, its share of the overall voltage drop increases and decreases.
A: There are three voltages in this circuit: the voltage, $V_{batt}$ across the battery terminals, the voltage, $V_r$ across the fixed resistor, $r$, and the voltage, $V_R$, across the variable resistor. By definition of voltage (aka potential difference),
$$V_{batt}= V_r+V_R$$
To a fair approximation (provided that $r+R$ isn't too small), $V_{batt}$ is constant.
The current in the circuit is
$$I=\frac{V_{batt}}{R+r}$$
The pd across $R$ is therefore
$$V_R=IR=\frac{V_{batt}R}{R+r}$$
It's easy to show that, for a given value of $r$, as $R$ is increased from zero, the value of $V_R$ increases towards a maximum value of $V_{batt}$.
It's the presence of $r$ that spoils your argument!
A: The variable resistor R forms a voltage divider with the other resistor.
The voltage divider is proportional: R gets a "slice" of its voltage which is proportional to its "share" of the total resistance. For instance if the total resistance is 100Ω, and R is 15Ω, then 15% of the voltage drops across it.
Suppose you then decrease R to 1Ω, the total resistance drops by 14Ω to 86Ω. But now, R is only $\frac{1}{86}$ of the voltage divider; it gets a much smaller proportion of the voltage.
Whenever we decrease the lower leg of a voltage divider, it will get less of the total voltage.
Whenever we have a formula of this shape:
$\frac{R_1}{R_1 + R_2}$
representing $R_1$'s fraction of the total, where all $R$s are positive, real values, whenever $R_1$ decreases, the value of the formula drops.
If you have a share in something, and the share's value decreases, you cannot become a larger shareholer, only smaller!
