Are these resistors connected in parallel? Thus, when I calculate the overall resistance to get the time constant ($RC$), I should be using $$R_{tot}=\left(\frac{1}{R_1}+\frac{1}{R_2}\right)^{-1} \;.$$ Right? In my situation, $R_2$ is an oscilloscope that has an internal resistance which is the same magnitude as $R_1$. The time constant expected from using only $R_1$ is twice the one I measured, which would work out perfectly if they are in parallel. 
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 A: The time constant of a first-order $RC$ circuit like that one can be calculated as the product of the capacitance $C$ and the equivalent resistance $R$ seen across the capacitor's terminals, when all the independent sources are nulled (that is, ideal voltage sources become short circuits and ideal current sources become open circuits). The reason for this is that to analyse the transient response of a first-order circuit, you can transform the electrical network connected to a capacitor or an inductor into its Thévenin or Norton equivalents (see [1, chapter 6] for more on this).
In the case of that circuit, the equivalent resistance, calculated as above, after the closure of the switch, is exactly the parallel of the two resistances, and the time constant is $\tau = C(R_1||R_2)$.
When the switch is open, instead, the equivalent resistance is just $R_2$ and the time constant is $CR_2$.
Note that calculating the equivalent resistance by direct inspection can be done only if there aren't any controlled sources, otherwise one should use the technique of applying a test voltage or current.
[1] L. Chua, C. A. Desoer, and E. S. Kuh, Linear and nonlinear circuits, McGraw-Hill, 1987.
A: That's a tricky one! When you do AC calculations, you can pretend that there is a short circuit around your power supply. And what happens? With the switch closed, and the power supply shorted, suddenly your two resistors are indeed "in parallel".
But as Samuel Weir mentioned in his comment, that will only be true when the switch is closed. When you open the switch, the only resistor that is "in the circuit" is $R_2$. If you don't get a different time constant for "opening" and "closing", you are definitely doing something wrong.
