Problem 25-1    Source:   Exercise 8.7 - page 190 - NILSSON, James W. & RIEDEL, Susan A. - Book: Circuitos Elétricos - Editora LTC - 5ª edição - 1999.

In the circuit shown in the Figura 25-01.1, the switch S has been in position 1 for a long time. At t = 0 the switch is shifted to the position 2. Determine:

a) i (0⁺)

b) v_C(0⁺)

c) d i(0⁺) / d t

d) the roots of the characteristic equation that describes the transient behavior of the circuit.

Solution of the Problem 25-1

Item a

Note that while the S switch was in position 1 the current i was null, because the right side of the circuit was open. So when we pass the switch to position 2, The inductor cannot abruptly change its value. Therefore, we conclude that:

Item b

While the S switch was set to 1, the capacitor was carried by the voltage across the 12.8 kΩ and is behaving like an open circuit. Then we can apply a resistive voltage divider and calculate this voltage. Soon:

So when we turn the S switch to position 2, the capacitor is charged with this voltage and since a capacitor cannot change its voltage abruptly, we conclude that:

Item c

As we studied in the theoretical part, the derivative of the current in the inductor is given by:

Remember that V_o is the voltage on the inductor in t = 0⁺. At this moment the current is null , so V_o is the difference between the source voltage and the voltage on the capacitor. Now, performing the calculation we have V_o = 100 - 20 = 80 volts. Substituting this value in the equation above we have:

Item d

Using the eq. 25-05 it is possible to calculate the values of α e ω_o. Then:

In this circuit we have α < ω_o, because ω_o = 7 071 rad/s, that is, an under-damped response. Thus, the roots are complex and conjugated. On the other hand, we can calculate the value of ω_d using the eq. 25-12, or:

Using the eq. 25-11 As a reference, we can write the roots of the characteristic equation like: