204 Practical MATLAB® Applications for Engineers
Observe that by changing the value of R (R = 75, 36, and 3 Ω), the three solutions for
the second-order differential equation—over, critical, and underdamped cases—
are obtained. Also observe that the analytical solutions completely agree with the
MATLAB solutions.
Example 2.31
Steady-state conditions exist in the circuit of Figure 2.105 for t ≤ 0, while the voltage
source V 1 = 6 V is connected to the network. At t = 0 +, the switch moves downward
disconnecting the source while connecting R 1 to the rest of the circuit.
- Obtain analytically the loop differential equation set and the initial conditions
- Using the MATLAB symbolic solver, create the script fi le transient_2loops that
returns the transient currents for each loop, and their respective plots, for t ≥ 0 - Also obtain simplify and pretty expressions for each of the transient loop currents of
part 2
FIGURE 2.105
Network of Example 2.31.
V 1 = 6 V
R 3 = 6 Ω
R 2 = 3 Ω
R 1 = 4 Ω
Switch moves down at t = 0
+ −
+
−
L 1 = 1 H
i 1 (t) i 2 (t)
L 2 = 2 H
+
−
+
−
ANALYTICAL Solution
The conditions at t = 0 are i 1 (0) = 3 A and i 2 (0) = 1 A. Applying KVL to the two-mesh
network of Figure 2.105, for t ≥ 0, results in the following set of simultaneous differen-
tial equations:
()()
()
RRitL ()
di t
di
121 11 Ri t 22 ^0
Ri t R R i t L
di t
21 3 2 2 (^2) di
() () ()^2 () 0
Replacing the elements by their values yields the following set of differential equations:
730 it 1 1 2
di t
di
() it
()
()