First find the initial conditions, VC(0-) = VC(0+) and IL(0-) = IL(0+)
Assuming the switch has been in the position shown for a long time prior to t = 0, the circuit to the right of the switch has no voltages or currents (all are equal to zero) since there are no independent sources in this section and it is isolated from the part of the circuit to the left of the switch. Thus
IL(0-) = IL(0+) = 0
The circuit to the left of the switch (including the capacitor) looks like a simple voltage divider (since the capacitor is acting like an open circuit)
Note that VC(0-) is also the voltage across the 6 K resistor.
By voltage division
VC(0-) = VC(0+) = 75 (6K/10K) = 45 V
After the switch moves to the right, the 75 V source and the 4 K and 6 K resistors can no longer affect the rest of the circuit. The remaining circuit looks like this:
Thus, we have a parallel RLC circuit (the inductor and capacitor are in parallel). Before we can calculate a and w0, we need to find the equivalent resistance connected to the inductor and capacitor. Since the circuit connected to the two energy storage elements contains a dependent source but no independent sources, we must connect a test source to the circuit and determine its effect to find the equivalent resistance.
We need to find the voltage VS developed across the circuit by the 1 A test source. By current division (be sure you understand why the dependent source doesn't affect the current division here!) we get
if = 1 (150K/210 K) = 0.71429 A
and by KVL around the perimeter of the circuit we get
VS = 0.71429 X 104 + 0.71429 (60 K) = 50 KV
Thus
Req = 50 KV / 1A = 50 K W
a = 1 / 2RC = 1 / [(2) (50K) (1.25n)] = 8 K
Since a < w0, the circuit is underdamped and we need to calculate the damped frequency, wd, before proceeding.
The form of the solution for the parameter in common to the two energy storage elements (VC for a parallel circuit) in an underdamped RLC circuit is
V(t) = e- a t(B1cos wdt + B2sin wdt)
where
VC(0+) = B1
and
IC(0+) / C = - a B1 + wd B2
Thus
B1 = VC(0+) = 45
but to obtain B2 we must first determine IC(0+).
We know that at t = 0+ the voltage across the capacitor is 45 V and the current through the inductor is 0 A. To facilitate determination of IC(0+) we can simply replace the capacitor and inductor with sources corresponding to these values as shown below.
Note that IC(0+) was chosen so that the arrow enters the positive reference for VC.
Using mesh analysis we get
if = i3
i2 - i1 = 0 (eq. 1)
KVL around mesh 3:
60 K i3 + 150 K (i3 - i2) = 0 (eq. 2)
KVL around supermesh 1/2:
10 K i3 + 150 K (i2 - i3) = 45 (eq. 3)
Solving the three equations yields
i1 = 900 mA = - IC(0+)
or
IC(0+) = - 900 mA
IC(0+) / C = - a B1 + wd B2
- 900 m / 1.25 n = - 8K (45) + 6 K B2
B2 = [- 900 m / 1.25 n + 8K (45)] / 6 K = - 60
Thus since V0 is the voltage common to both the capacitor and inductor,
V0(t) = e- 8000 t(45 cos 6000 t - 60 sin 6000 t); t > 0
