# steady state voltage in capacitor network

simulate this circuit – Schematic created using CircuitLab

I have this capacitance network in my circuit and I can't figure out the math behind why the final steady-state value of Vx=0.542v when SW2 closes to the other position. Coincidentally I found out this is Vx - 1.(c2 /(c1+c2+c3)) but I would like an intuitive approach. Please ignore the absence of resistance in the circuit to make it practical.

• Maybe this helps to get you started: capacitors in series all have the same charge Q. Each voltage is then V = Q/C. electronics-tutorials.ws/capacitor/cap_7.html Commented Jul 16, 2020 at 19:07

C1 and C2 are initially in parallel for an equivalent capacitance of 31fF. So, the voltage across the parallel combination of C1 and C2 will be V = Q/31fF. The voltage across C3 will be Vx = Q/31fF. From these 2 equations you can see that V = Vx in this state, which leads to Vx = 0.5V. You can reach this same conclusion by applying Laplace transform circuit theory where you would end up with $$Vx = \frac{1}{s}\frac{C3}{Ce+C3}$$ where Ce is the parallel combination of C1 and C2. This is clearly = 0.5 for initial case. The charge on C2 will be Q2 = C2V2 = 8f Coulombs. The charge on C3 will be Q3 = C3Vx = 15.5f Coulombs.