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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?enter image description here

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  • \$\begingroup\$ 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. \$\endgroup\$
    – jonk
    May 17 '19 at 18:04
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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.

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  • \$\begingroup\$ Thanks! It seems I did not go all the way through with this problem by turning off all the independent sources. \$\endgroup\$
    – Hector
    May 17 '19 at 19:15
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The following steps are the transformations that are probably easiest to see, in order:

schematic

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.

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

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