Problem 25-3 Source:
Problem 8.47 -
page 332 - NILSSON, James W. & RIEDEL, Susan A. -
Book: Circuitos Elétricos - Editora LTC - 10ª edition - 2016.
In the circuit shown in Figure 25-03.1 below, the switch was in the closed position for a long time.
At t = 0 the switch is opened.
Determine Vo (t) for t ≥ 0.
Solution of the Problem 25-3
Note that according to the circuit shown in Figure 25-03.1, we notice that the circuit highlighted by the orange rectangle,
containing resistors R1, R2 and R3, form a delta circuit.
This way, it is possible to find the equivalent circuit star according to the circuit
shown in Figure 25-03.2.
To transform a delta circuit into a star circuit, studied in
Chapter 5, we use the equations eq 05-01,
eq 05-02 and eq 05-03. For greater clarity, we repeat eq 05-01 below.
eq. 05-01
See Figure 25-03.3 how the circuit looked for t < 0 after the transformation. With this we can
establish the initial conditions of the problem.
As a capacitor behaves like an open circuit when fully charged and an inductor like a short circuit,
we easily determine VC (0+) and iL (0+ sup>). To calculate
VC (0+) let's use a resistive voltage divider, or
VC (0+) = 100 (60/(20 + 20 + 60)) = 60 V
And to calculate iL (0+) we will directly use Ohm's law, or
iL (0+) = 100 / (R5 + RC + Rb) = 1 A
Now let's analyze the circuit for t > 0. In this case, the voltage source and resistors R5
and RC are disconnected from the circuit. Then, we obtain a circuit RLC Series, as shown in
Figure 25-03.4. Based on the figure we can calculate the circuit parameters.
α = R /2 L = 500 rad/s
ωo = 1 / √ ( L C ) = 400 rad/s
In this case, we see that α > ωo. So the circuit has overdamped behavior. Then,
The response of the circuit is given by
eq. 25-04, repeated below. Since the circuit, for t > 0, has no independent sources, then, If = 0.
eq. 25-04
The roots r1 and r2 are given by eq. 25-07 and eq. 25-08 and are worth:
r1 = - 200 rad/s
r2 = - 800 rad/s
So, we can write the answer as:
iL (t) = A1 e- 200 t + A2 e- 800 t
To determine the values of A1 and A2 let's apply the initial conditions of the problem.
iL (0) = A1 + A2 = 1
We also know that:
d iL (0) = VL(0) / L
Therefore, we must calculate the value of VL(0). To do this, we will mesh the circuit in Figure 25-03.4.
- VC (0) + 200 iL(0) + VL(0) = 0
By numerically substituting the values and carrying out the calculation, we find:
VL(0) = - 140 V
Thus, we have:
d iL (0) = VL(0) / L = - 700 = -200 A1 - 800 A2
Now we have two equations with two unknowns that are easy to solve. The values are:
A1 = 166.67 mA
A2 = 833.33 mA
Thus, we have:
iL (t) = 166.67 e- 200 t + 833.33 e- 800 t mA
And as the problem statement asks to determine the value of the voltage across the inductor, we can
derive the equation for the current through the inductor and obtain the voltage across it, as
the eq. 23 - 04. This way, we obtain: