Problem 24-11
Source:
Problem 8.35 - page 331 - 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 24-11.1 has been in position a for a long time. At t = 0 it instantly moves to position b.
Determine VC (t) and iL (t) for t ≥ 0.
Figure 24-11.1
Solution of the Problem 24-11
When the switch is in position a, we can get two data i.e. VC (0+) and iL (0+). As the inductor behaves like a short circuit, we conclude that:
iL (0+) = 0.1 A
And by making a resistive voltage divider we will find the value of VC (0+), or
VC (0+) = (100 . 3)/( 1 + 3 ) = 75 V
When the switch moves to position b, we have a parallel RLC circuit with an independent current source. Based on this information we can now find the values of VC (∞) and iL (∞) . Knowing that for t = ∞ the inductor behaves like a short circuit, then it is easy to conclude that:
iL (∞) = 0.1 A
VC (∞) = 0 V
Now let's calculate the value of α of a parallel RLC circuit. Using eq. 24-05, we have:
α = 1/ (2 R C ) = 1 / (2 . 40 . 25 . 10-6) = 500 rad/s
And the value of ωo, using eq. 24-06 is:
ωo = √(1/ (L C )) = √(1 / ( 0.25 . 25 .10-6))
Carrying out the calculation we have:
ωo = 400 rad/s
Note that in this case, α > ωo, confirming an overdamped response. Therefore, the two roots of the characteristic equation are real and the solution equation will be in the form of eq. 24-04.
The values of r1 and r2 are given by the equations eq. 24-07 and eq. 24-08.
r1 = - α + √ (α2 - ωo2) = - 200 rad/s
r2 = - α - √ (α2 - ωo2) = - 800 rad/s
Then the system's response to iL (t) will be given by:
iL (t ) = iL (∞) + A1 e- 200 t + A2 e- 800 t )
We have already calculated that iL (∞) = 0.1, because during this time the inductor will behave like a short circuit. Now it is necessary to find the values of A1 and A2. We know that iL (0+) = 0.1. Therefore, for t = 0:
iL (0+) = 0.1 + A1 + A2 = 0,1 ⇒ A1 = - A2
The other information we have is:
diL (0+)/dt = VL (0+)/L = r1 A1 + r2 A2 = 75/0.25
Doing algebraic work and substituting numerical values we find
A1 = 0.5 e A2 = - 0.5
So, we can write:
iL (t ) = 0.1 + 0.5 e- 200 t - 0.5 e- 800 t )
To find VC (t) keep in mind that we have a parallel circuit. Therefore, all components are under the same voltage. Therefore, by calculating the voltage across the inductor we can say that this voltage is the same across the capacitor and the resistor. So, we can use eq. 23-04, transcribed below.
eq. 23-04
VC (t ) = L diL (t)/dt = 0.25 ( - 100 e- 200 t + 400 e- 800 t)
Therefore, the general solution for the circuit is given by:
