196 Practical MATLAB® Applications for Engineers
Note that the network shown in Figure 2.95 is an ideal circuit (since there is no resis-
tance, R = 0 in the LC loop for t ≥ 0), and the solution clearly indicates that the current
shows an oscillator behavior with W = 12 rad/s = _____^1
√
___
LC
and T = π__ 6 s.
Example 2.29
Steady-state conditions exist in the network shown in Figure 2.97 at t = 0 −, when the
V 1 = 120 V source is connected to the RCL parallel circuit. At t = 0 +, the switch moves
downward, and the source V 1 = 120 V and resistor R = 10 Ω are disconnected from the
parallel (RLC) structure.
Analyze the transient response (t > 0) of the source-free parallel RLC circuit, for each
of the following values of R, R = 3, 9, and 72 Ω.
- Determine the analytical response vC(t) for each value of R, for t ≥ 0
- Create the script fi le transient_RLC_parallel that returns the MATLAB solutions of
part 1 and its corresponding voltage plots - Compare the MATLAB solutions of part 2 with the analytical solutions of part 1
FIGURE 2.97
Network of Example 2.29.
L = 9 H
R = 10 Ω
R = 3,9 and 72 Ω
V 1 = 120 V
Switch moves down at t = 0
C = 1/36 F
ANALYTICAL Solution
From the circuit diagram of Figure 2.97 for t ≤ 0, the initial conditions are vC(0) = 0 and
iL(0) = 120/10 = 12 A, and
Cdv t
dt
C()tC 0 it( 0000 ) iC L R() i() i()
then
dv t
dt
C()tC 0 36 i() 0 36 12( 0 ) 432 V/s
Recall that the node equation is
dv t
dt
dv t
dt
CCvtC t
2
2
() 11 () () 00
RC CL
for
For R = 3 Ω, the resonant frequency is
w 0 ^12
LC
rad/s