Problem 23-3 Source:
Problem 7.39 - page 302 Book: Análise de Circuitos em Engenharia - J. David Irwin - 4ª edição - Ed. Pearson - 2013.
In the circuit show in the Figure 23-03.1, the S switch has been closed for a long time. At t = 0 the switch has been opened. Find Vo (t) to t > 0.
Figure 23-03.1
Solution of the Problem 23-3
As the S switch remained on for a long time the inductor behaves like a short circuit.
This way we can calculate the current i that circulates through the inductor in t = O-. Note that in this case, the 12 volts source in series with the 2 ohms resistor does not enter the calculations because we know the voltage at the extremes of this series which is 12 volts. So:
i(0-) = 12 / 2 = 6 A
When the S switch is opened, the 12 volts source (from the left of the circuit) is removed from the circuit.
Figure 23-3.2
When the S switch is opened, the 12 volts source
(from the left of the circuit) is removed from the circuit. In the circuit shown
in the Figure 23-3.2 this situation appears, where we replace the inductor
with an impulsive current source with the previously calculated value,
that is,
6 µo (t). Note that there is a current source in parallel
with a resistor.
Then it is possible to make a new transformation of sources, getting a voltage
source impulsive in series with the resistor of 4 ohms.
Figure 23-3.3
In the Figure 23-3.3 we can see how the circuit looked after the transformation of sources. Note that the polarity of voltage sources is opposite. Then we must subtract its values to find the voltage value acting on the circuit. Thus, a value of 12 volts results with the positive pole pointing to the right, agreeing with the direction of current indicated by the blue arrow.
It should be noted that the last two circuits shown (circuits B and C) only valid for time t = (0+). For any time other than 0+, they have no validity.
So based on the C circuit, let's calculate the current value i (0+).
i (0+) = 12 / (2 + 4 + 2) = 1.5 A
Then, V (0+) will be:
Vi = V(0+) = 2 i(0+) = 3 V
This calculated value is the initial value of V(t). Let's calculate the value when
t → ∞.
For this case let's remember that the inductor is a short circuit. Looking at the A circuit,
we easily realize that the current is given by:
if = - 12 / (2 + 2) = -3 A
Then, Vf will be:
Vf = 2 x (-3) = -6 volts
It remains to calculate the value of the time constant of the circuit. Therefore, we must find what is the value of Req of the circuit. Now, looking at the circuit A, we notice that the 4 ohms resistor, which is parallel to the inductor, is also parallel to the series of two 2 ohms resistors. This results in a Req = 2 Ω. So the time constant is:
τ = L / Req = 1/6 s
Thus, with all claculated values it is possible to write the solution equation of the circuit, remembering
that the equation that will be used is eq. 23-03, reproduced below.
V (t) = Vf + (Vi - Vf ) e -t/τ
Substituting for the numerical values found, the solution equation is:
