Interpreting Physics Results Regarding Circuitry Analysis (Highschool Level Physics) I conducted an experiment which aimed to detail and record the voltage across series resistors in series circuits.
I noticed that:


*

*The Final Voltage, V3, stayed the same.

*The resistors with the highest resistance had the highest voltage drop.


How do I put my observation in words?
 A: The total voltage difference across the resistors ($V_3$) is, by design, a constant 5.4 volts1 (because it's being supplied by a power supply that pretty well approximates a constant voltage source).
This total voltage difference must be dropped across the two resistors $R_1$ and $R_2$ in series: that is, $$V_1 + V_2 = V_3.$$  As the resistors are in series, and there are no other paths the current might take,2 the same current must flow across both resistors: $$I_1 = I_2.$$
By Ohm's law, the current across a resistor equals the voltage divided by the resistance: $$I = \frac{V}{R}.$$  Combining these equations, we see that $$\frac{V_1}{R_1} = I_1 = I_2 = \frac{V_2}{R_2},$$ which we can rearrange to get $$\frac{V_1}{V_2} = \frac{R_1}{R_2}.$$
That is to say, the ratio $V_1 / V_2$ of the voltage drops across the resistors is equal to the ratio $R_1 / R_2$ of their resistance.  Since the total voltage $V_1 + V_2$ dropped across the resistors is fixed, this means that, when $V_1$ goes up, $V_2$ must go down, and vice versa.
In particular, this means that, when you keep $R_1$ fixed and decrease $R_2$, you're increasing the ratio $R_1 / R_2$, and thus $V_1 / V_2$.  Since the sum of $V_1$ and $V_2$ is constant, increasing the ratio means that $V_1$ must increase and $V_2$ must decrease by the same amount.
For example, if, as in your first experiment, $R_1$ equals $R_2$, then $V_1$ also equals $V_2$, and thus both must be half of the total voltage drop $V_3 = V_1 + V_2$.3
Similarly, if $R_1$ is twice $R_2$, as in your second experiment, then $V_1$ will also have to be twice $V_2$.  Thus, $V_1$ will be two thirds of the total voltage drop $V_3$, while $V_2$ will be one third of $V_3$.
Note that the individual values of $R_1$ and $R_2$ don't matter, only their ratio does: you'd get the same results with $R_1 = 100\;\Omega$ and $R_2 = 50\;\Omega$ as with $R_1 = 10\;\Omega$ and $R_2 = 5\;\Omega$.  Of course, the current passing through the resistors (and the heat dissipated by them) would be ten times greater for the second case as for the first, but this would not affect the voltages in any way.4

1) You say in the text that it's supposed to be 6 volts, but the measurements say 5.4, so I'm going with that.  See note 3 below.
2) The voltage sensors do leak some tiny bit of current, but they're designed to make this leakage current as small as possible, so we can normally safely neglect it.
3) The difference between the calculated values and your experimental results is presumably due to experimental inaccuracies, e.g. in the readings of your voltmeters or in the actual resistance values of your resistors.
4) That is, it would not affect the voltages as long as the experimental assumptions remained valid.  If you reduced the resistances too far, either the power supply voltage would start to drop, or the increased current would cause your components to overheat. Conversely, if you increased the resistor values too far, at some point the minuscule amounts of current leaking though the voltage meters would start to affect the results measurably.
