Consider the following secondorder circuit below The switch

Consider the following second-order circuit below. The switch opens at t = 0. Determine the initial condition, v_0(0*) Determine the initial condition, dv_0/dt |t=0+. Derive the differential equation for v_0(t) Determine the particular solution. Determine the characteristic equation, find the solutions to the characteristic equation, and determine the characteristic solution to the DE. Determine the unknown coefficients from the initial conditions. Write out the complete solution.

Solution

Vo(0+)= 10

Capacitor is fully charged just before opening the switch

Hence io= current through capacitor is zero

dVo(0+)/dt=io=0

Now switch is opened.

i = current through capacitor= C*dVo/dt

Apply kvl:

iR+ Ldi/dt + Vo=0

0.025dVo/dt +0.25d²Vo/dt²+ Vo=0

Particular solution is 0,

Homogeneous solution is Voh

d2Vo/dt2 + 0.1dVo/dt + 4Vo=0

Characteristic equation is

D^2+0.1D +4=0

D= -0.5 +- j1.99

Solution is Vo(t)= e^(-0.5t)[Acos(1.99t)+Bsin(1.99t)]

Apply initial condition, Vo(0+)=10v,

Hence A=10

At t=0, dVo/dt=0 , hence -0.5A+B1.99=0,

Hence B=0.5A/1.99=5/1.99=2.51

Vo(t)=e^(-0.5t)[10cos(1.99t)+2.51sin(1.99t)]