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How would you go about finding V across the leftmost capacitor after reaching a steady state and having the voltage source is set to 0V? Is it possible find the time constant for the leftmost capacitor [and then apply Vc0 * e^(-t/RC)]?

And how will the discharge times scale with two loops (as shown here) or a third loop compared to having just one?

There's no need for a complete solution, but I'm genuinely curious and would love your thoughts on a good way to approach the problem.

What we know so far (Thanks JensenR30):

  • The Voltages across both capacitors is 4 V
  • After deactivating the source, Vc of the right loop alone would be 4e^(-t/0.0001) V

Would it is possible from there to find the Thevenin equivalent for the leftmost capacitor?

  • 3
    \$\begingroup\$ When you say "having the voltage source shut off", do you mean that it is disconnected from the circuit (open circuit)? Or do you mean that its voltage is set to 0 V (closed circuit)? \$\endgroup\$ Commented Oct 29, 2015 at 21:08
  • 2
    \$\begingroup\$ LTSpice can be your friend \$\endgroup\$
    – Andy aka
    Commented Oct 29, 2015 at 21:11
  • \$\begingroup\$ @Andyaka, LTSpice is indeed our friend here, but I'm curious to find an analytical solution. \$\endgroup\$
    – Meridian
    Commented Oct 29, 2015 at 22:04
  • \$\begingroup\$ Google "second order rc circuit". \$\endgroup\$ Commented Oct 30, 2015 at 0:36
  • \$\begingroup\$ In fact "2nd order lpf" is even better; sim.okawa-denshi.jp/en/CRCRkeisan.htm \$\endgroup\$ Commented Oct 30, 2015 at 2:02

2 Answers 2


You could try to use Laplace transform.

Consider the following equivalent circuit:


simulate this circuit – Schematic created using CircuitLab

This is the situation in the Laplace domain when your circuit has the generator off, i.e. shorted, and the caps were charged at 4V. Note the two 'initial condition' generators, that's \$\frac{4V}{s}\$, while for the capacitance value is \$\frac{1}{sC}=\frac{1}{s\cdot 10n}\$. The unit is not Farad, as the schematic tool suggests.

Now you can solve the circuit with the usual linear circuits tools, a good idea might be superimposition, then find \$V_x(s)\$. If you're lucky enough you may even be able to anti transform it and come up with the time domain funcion.

If you don't know what the Laplace transform is you can just write all node voltages as a function of the currents, keeping in mind that in a cap \$i=C\dot{v}\$ and throwing in the initial conditions when you integrate. Or you could just learn to use the L-transform that is basically syntactic sugar for differential equations.


In steady state, the voltage on all of the capacitors (regardless of how many iterations of this pattern your have) will be 4 V.

If you disconnect the voltage source (open circuit), You should not expect to see any change in the voltage across the capacitors. The voltage across each capacitor will stay at 4 V forever unless you provide a path through which they can discharge.

If you talking about replacing your voltage source with a short circuit, then you will have a path through which all of the capacitors will discharge. The rate at which they all discharge should be able to be determined by Kirchhoff's Laws, Ohm's Law, the capacitor equation ( i = C*(dv/dt) ), and some formula manipulation (i.e. circuit analysis).


Very cool question!


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