Problem 24-6
Source:
Question 3 - 3rd test Electrical Circuits I - Ufrgs - 2018.
In the circuit shown in Figure 24-06.1, the switch was in the closed position for a long time.
At t = 0 the switch is opened.
Determine iL(t) for t ≥ 0.
Figure 24-06.1
Solution of the Problem 24-6
Item a
First, let's find the current that flows through the inductor when the switch is closed. Since the inductor behaves like a short circuit, it is clear that all the current from the source will flow through the inductor. So, iL(0-) = 12 A. Let's now determine the value of the current in the inductor when the switch is open for a long period, that is, we will calculate the final value of the current in the inductor, or iL(∞). As the inductor behaves like a short circuit, we can apply a current divider, as in this case the 3 Ω resistor is inserted in the circuit. Thus, we obtain:
iL(∞) = 12 x 1 / (1 + 3 ) = 3 A
These are the initial conditions of the problem. On the other hand, we are facing a parallel RLC circuit. So, let's calculate the damping factor, α, that is:
α = 1 / ( 2 R C ) = 1 rad/s
Note that the value of R is the sum of the values of the two resistors, or R = 4 Ω. We will determine the value of
ωo, or
ωo = 1 / √ (L C ) = 1 rad/s
Clearly we have α = ωo which characterizes a circuit that has a critically damped response. Then the complete response of the circuit will be in the form:
iL (t ) = A e r t + B t e r t + i( ∞ )
Note that the roots of the characteristic equation are equal to r and given by r = - α = - 1. In this way, we have the equation:
iL (t ) = A e-t + B t e-t + 3
Now let's apply the initial conditions to this equation. We know that iL(0-) = iL(0 +) = 12 A, because as we studied, the inductor does not accept sudden changes in current. And making t = 0, we obtain:
12 = A + 3 ⇒ A = 9
To calculate the value of B, we must find the expression for:
vL (t ) = L di/dt
We know that vL (0 ) = 0, as the inductor is a short circuit. So, we have to:
vL (t ) = L [-A e-t + B (-t e-t + e-t )]
Where the expression in brackets is the derivative of the current in the inductor with respect to time. This way, making t = 0, we obtain:
0 = 8 [-9 e-0 + B (-0 e-0 + e-0 )]
Solving this expression, we obtain:
B = 9
Then the general expression for the circuit is given by:
