[SOLVED] What is the response of the given circuit

[iVenky]

Hello. Atlast I got it. I don't know why it says as 100 nV in the Schematic but in the Probe window I got it as 2 V.

 

[albbg]

Sorry, I usually don't use pspice I know better microcap. I think you just need one switch (between the capacitors) and two .ic statements to set the initial conditions of C1 and C2.
However the charge is moving from the charged capacitor to the other one and at the final state the charge must be conserved while the excess energy is dissipated through a pulse. Since the not initially charge capacitor act as a short circuit you will have (ideally) the charge flowing istantaneously. This means you will have a the complete redistribution of the charge in a time interval dt --> 0. Since the current is given by:

I = dq/dt and dt --> 0 (dq is the charge of the charged capacitor thus is a finite number)

Then I --> ∞ for a time duration --> 0 (the dirac pulse)

As I said the pulse will be a current pulse not a voltage pulse. In your you must plot the current flowing between C1 and C2 to see the pulse, not the voltage across them.

The area of such a current pulse is finite and represent the dissipated energy.

In pspice where the resistance of the switch can be very small, but not zero, you will see a current pulse when the switch between the two capacitors is closed. The amplitude increases as the on resistance of the switch decreses while the duration of the pulse will decrese.

Here below you can find a simulation did with microcap evaluation version. The first graph shows the behaviour of the two voltage (across C1 and across C2), the second one (lower side) the current.



The on resistance of the switch is 1mohm. I used a voltage controlled switch that closes at t = 1 second

 

[FvM]

I don't know why it says as 100 nV in the Schematic but in the Probe window I got it as 2 V.

Good. The voltage display in schematic only shows the initial solution. But you apparently managed to perform a transient analysis, that shows the final state.