Problem 24-10
Source:
Problem 8.14 - page 329 - NILSSON, James W. & RIEDEL, Susan A. -
Book: Electric Circuits - Publisher Pearson Education do Brasil - 10th edition - 2016.
The switch in the circuit shown in Figure 24-10.1 has been in position a for a long time. At t = 0 it instantly moves to position b.
Determine Vo (t) for t ≥ 0.
Solution of the Problem 24-10
Initially, we will find the equivalent resistance value in the circuit highlighted by the dashed rectangle represented in Figure 24-10.1. To do this we must calculate
the Thévenin resistance, as learned in chapter 2. Therefore, we will apply a voltage source of 20 V to this circuit. Follow the circuit on
Figure 24-10.2. Note that if through the resistor
of 60 Ω circulates i1 then through the 150 Ω resistor circulates 60 i1/ 150 = 0.4 i1. Therefore, the total current
IT circulating in the circuit is IT = - 1.4 i1. And we easily determine the value of i1 by looping the dependent source and the 60 Ω resistor.
- 20 - 10 i1 - 60 i1 = 0
Solving this equation, we find:
i1 = - 2/7 A
Knowing the value of i1 we can calculate the value of IT, or
IT = - 1.4 i1 = (- 1.4) (- 2/7) = 0.4 A
And now we can calculate the value of the Thévenin resistance, because:
Rth = V/IT = 20/ 0.4 = 50 Ω
Now let's calculate the value of α of a parallel RLC circuit. Using eq. 24-05, we have:
Note that in this case, α < ωo, confirming an underdamped response. Therefore, the two roots of the characteristic equation are complex and the solution equation will be in the form of eq. 24-14 shown below for clarity.
eq. 24-14
We need to find the value of ωd. Using eq. 24-13, we have:
ωd = √ (ωo2 - α2 ) = 937.5 rad/s
Then the system's response to Vo (t) will be given by:
Vo (t ) = Vo (∞) + e- 1250 t ( B1 cos ωd t + B2 sen ωd t )
Observing the circuit shown in Figure 24-10.1, we easily determine that Vo (∞) = 0, as in this time the inductor will behave like a short circuit. Now it is necessary to find the values of B1 and B2. Let's determine what the voltage on the capacitor is when the switch is in position a,
i.e. VC(0+). Using a resistive voltage divider we find
VC (0+) = 50 (4/(4 + 6)) = 20 V
And as we studied in the theoretical part, we know that:
VC (0+) = B1 = 20
And to find the value of B2 it is necessary to know the value of the current in the capacitor at the instant (0+). We know that when we apply a voltage like
jump in an inductor, the current flowing through it is zero. Therefore, the current that will flow through the capacitor is the current that will flow through the Thévenin resistance.
Então:
iC (0+) = - VC (0+)/ Rth = - 20/50 = - 0.4 A
Note the negative sign in iC (0+), as the current is leaving the positive terminal of the capacitor . And to determine the value of
B2 we use the relationship studied in the theoretical part and reproduced below.
dVC / dt = iC (0+)/ C = - α B1 + ωd B2
Performing numerical substitution and carrying out the calculation we find the value of B2, or:
B2 = - 26.67
Then the complete answer can be written as:
Vo (t ) = e- 1250 t ( 20 cos 937.5 t - 26.67 sen 937.5 t )