# RC circuit analysis with parallel capacitances (using Thevenin equivalent)

I am trying to determine the voltage across capacitor C3 (voltmeter VM1) with respect to time. I believe that simplifying the circuit to its Thevenin equivalent is the correct way to proceed but I'm not sure how to.

Specifically, I am unsure how two parallel capacitances can be combined if only one has a resistor connected in series. I know C2 and C3 can be be simplified into a single capacitor with capacitance C2+C3. However, I am unsure how this can then be combined with C1, as resistor R2 is connected in series with C2 and C3 but is parallel to C1.

• how two parallel capacitances can be added ... unclear what you are asking .... be very clear .... specify components Commented Jun 5, 2018 at 2:51
• I think first you should remove the c3 and find the voltage across the c2 and then by short circuit the bat1 find the impedance by seeing the two terminals of c2...you have vth,zth,c3 connect them in series....find the voltage by using voltage division rule Commented Jun 5, 2018 at 3:33
• Have you omitted a switch, and initial conditions?
– Chu
Commented Jun 5, 2018 at 7:38

I believe that simplifying the circuit to its Thevenin equivalent is the correct way to proceed but I'm not sure how to

An alternative way is to work in the frequency domain, use an on-line solver that calculates the transfer function and then use "inverse Laplace" to solve the time domain step function. In the frequency domain we have: -

If we plug the values into this solver we get: -

Then go to this inverse Laplace solver, plug in the formula from above: -

Noting that I've multiplied the denominator by "s" because the input is a step function

And the answer is: -

$$-e^{-11000t}\cosh \left(\sqrt{101000000}t\right)-\frac{11000}{\sqrt{101000000}}e^{-11000t}\sinh \left(\sqrt{101000000}t\right)+1$$

This needs to be multiplied by 9 because the input step voltage is 9 volts and not 1 volt.

I am trying to determine the voltage across capacitor C3 (voltmeter VM1) with respect to time.

If you need a formulaic answer then use the one above. Or, just draw the circuit in your favourite sim tool to get something like this: -

And just in case the formula derived doesn't stack up, here's a double plot of Vout and the derived formula (with the derived formula being made 1 volt higher so the two can be compared): -

Note the "10" in the formula indicated by the arrow is "9+1", just to shift it a tad higher than Vout.

• Can I use Elmore delay to approximate the time constant? Commented Jun 6, 2018 at 18:58
• I have no idea what that is or why you prefer an approximation to the real unambiguous answer. Commented Jun 6, 2018 at 21:10