Finding time constant in terms of R and C

I got a different expression for the time constant than the actual solution. Can anybody tell me where I went wrong? Why is it that in the solution all three resistors were put in parallel? What do you think is the correct way to go about doing this? • Look at $R_2$ and $R_3$. They are obviously in parallel. Convert them to their parallel equivalent and replace them with the new resistor value. Now, you have a resistor divider formed by this new resistor and $R_1$. Convert this to its Thevenin source+resistance equivalent. To do that, you must treat $R_1$ as being in parallel with the newly created resistor (not by adding it) which itself was $R_2$ in parallel with $R_3$. That's why.
– jonk
May 17 '19 at 18:04

Since this is effectively an AC analysis with time constants, you short out the Voltage source since it is 0 Ohms then you can see R1//R2//R3=Req gives you the correct resul for $$\\tau=Req*C\$$

While a DC voltage divider gives you the steady state voltage for Vc.

• Thanks! It seems I did not go all the way through with this problem by turning off all the independent sources. May 17 '19 at 19:15

The following steps are the transformations that are probably easiest to see, in order: simulate this circuit – Schematic created using CircuitLab

You did the first step, correctly. That was just two resistors obviously in parallel with each other. That's Step 1 shown above.

However, for the following step you don't add $$\R_1\$$. The reason is that once you've reached the upper-right corner schematic, you are left with a voltage divider pair of resistors. And this means you need to perform a "source transformation."

To help make the transformation a little clearer to discuss, I decided to ground the (+) rail. The reason is that $$\R_1\$$ is in parallel with $$\C_1\$$ and grounding the top rail allows me to flip the schematic over (upside down, from the voltage perspective) to make the resistor divider analysis a little more obvious. That's Step 2, above.

A Thevenin source transformation of a voltage source and a resistor divider pair is pretty simple. In your case, the new equivalent voltage source is $$\V_\text{TH}=V_\text{source}\cdot\frac{R_1}{R_1+\left(R_2\mid\mid R_3\right)}\$$, where $$\V_\text{source}=-V_1\$$, and its source impedance is $$\R_\text{TH}=\frac{R_1\cdot\left(R_2\mid\mid R_3\right)}{R_1+\left(R_2\mid\mid R_3\right)}\$$; this latter calculation simply being $$\R_1\$$ in parallel with $$\R_2\mid\mid R_3\$$.

If you haven't been exposed to the Thevenin source transformation for a voltage source and a resistor divider pair, as yet, then you should immediately work on that problem and make sure you understand why it works as it does.

• Thank you! I messed up on finding R Thevenin because I did not turn off the voltage source. May 17 '19 at 19:15