For below circuit, the simulation result is not matching with theoretical prediction

Question

Please note that i1(t) is the loop current (in loop analysis method) and not the inductor current.

The answer to above problem is option B, which is derived below Following principles are used:

1. Capacitor acts as an open circuit under DC steady state.
2. Inductor acts as a short circuit under DC steady state.
3. Capacitor volatge and inductor current cannot change instantaneously.

When I tried the simulation, it seems V(C1) (V(1,2) in the snap) is settling to input volatge and V(C2) (V(3) in the snap) remain as zero. I realize this is also a stable steady state solution.

My doubts are:

1. Looks like there are 2 stable solutions for this circuit.

   V(C1) = input voltage, V(C2) =0

   V(C1) =V(C2) = half of input voltage

   What is the exact condition which determines which solution the circuit will eventually settle to?

2. In the simulation result, I see the capacitor voltage rising instantaneously. Not sure how is it possible. (Note: Initial voltage is set to zero). Looks like some simulation basics I am not aware of.

Answer 1

At \$ t=0^- \$ right before the switch opens, we're at DC steady state, so inductors are shorts and caps are open. The top capacitor has charged to voltage V, which means no current flows in the battery. The right cap is shorted by the inductor, so it is charged to 0V.

There can not be an instantaneous change in current in an inductor. So, at \$ t=0^+ \$ right after the switch opens, both capacitor voltages and inductor currents are the same as previously.

This means i1 (inductor current) is 0 at \$ t=0^+ \$.

For i2 we have a loop formed by two caps, one charged to V and the other to 0V, and two resistors R. Current flows in the opposite direction relative to the i2 arrow drawn on the schematic, so i2=-V/2R.

As time passes, inductor current and capacitor voltages will change, so this is only valid at this precise time of course.

Looks like there are 2 stable solutions for this circuit. V( C1) = input volatge , V(C2) =0 V(C1) =V(C2) = half of input voltage

The second one is not possible because the inductor will short the right side of C1 to ground. So after the oscillations have settled, there is no voltage on C2.

In the simulation result, I see the capacitor voltage rising instantaneously. Not sure how is it possible. (Note: Initial voltage is set to zero)

The sim will calculate the DC operating point of the circuit, which as expected has the supply voltage across C1. So it starts at 10V. It'll just ignore your ".ic" initial condition. If it has not ignored it, the graph would have started with 0V across C1.

You say "I see the capacitor voltage rising instantaneously" but I don't see that on the plot. I just see it starting at 10V. I guess you think it rises instantaneously because you set ".ic"...

Answer 2

That's not the correct answer. I won't spoil it for you, but your approach with shorted inductances and opened capacitors is the right way to go.

In particular, L1 and C2 in parallel, and L1 is an inductor

Then your LTspice approach is not correct, either. You used an .IC card which, for voltages, it relies on specific node names. Since you haven't assigned any labels, your node names are generic: N001, N002, etc. It's best to not rely on these names and, instead, add labels, since the default names change as soon as there is a change in topology. Instead you wrote V(C1), probably meaning to specify the voltage across C1 to be zero. In order to achieve that you need to write two initial conditions, one for each node at each side of C1: .ic V(a)=<value> V(b)=<value> (considering a and b to be the labels). Alternatively, you can simply add:

ic=10

to C1's value, which will mean that many V across it.

\$\scriptsize{\text{In the meantime, }}\$ bobflux \$\scriptsize{\text{ posted his answer, so I guess spoilers are useless now, but I'll keep them, anyway.}}\$

Then you can either use ic=0 for C2 and L1, then use the uic flag for simulation, or leave them untouched, since their initial conditions are, by default, zero.

Here's the reworked schematic, in two different ways. Note that there are two .tran cards, only one circuit should be considered:

Answer 3

The right-hand C has no charge at t=0 (L is short circuiting it at t<0), therefore this C looks like a short circuit at t=0.

The upper C is charged to V, with + on its left-hand plate. This is the only voltage source in the loop.

Hence the path for the current at t=0 is anticlockwise through left-hand R then anticlockwise through the right-hand R. This path has resistance 2R, hence the current anticlockwise is V/2R. Current flow is in opposite direction to i1, so the current is i1=-V/2R.

The inductor has no effect in the loop as it retains status quo at t=0, viz, iL=0.