Problema 24-5    Source:   Adapted from Question 1 of the PUCRS II Electrical Circuits Test - 2019.

a) Determine the value of v_c (t) for t > 0 knowing that the circuit is in steady state when the switch is in the position shown in Figure 24-05.1 for t < 0 .

b) Graph the system response. Making R = 5 Ω, redo the graph and compare with the previous one.

Solution of the Problem 24-5

Item a

Note that for t < 0 we have two independent circuits in steady state. In this way, we can calculate v_c (0⁺) = 5 V and i_L (0⁺) = 20/10 = 2 A. We can also calculate the value of i_R (0⁺) = (20 - 5)/10 = 1.5 A and, applying LKC to node a, we can determine the value of i_c (0⁺) = 1.5 - 2 = - 0.5 A. These are the initial conditions of the problem.

Note that in Figure 24-05.2, we have a parallel RLC circuit that is perfectly possible to be solved by applying the theory already studied. Then, with the values provided by the problem, we can calculate the operating frequency of the circuit using eq. 24 - 06 and, we find:

We easily conclude that:

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 < ω_o, confirming an underdamped response. Therefore, the two roots of the characteristic equation are complex and the equation for v_c (t) will be in the form of eq. 24-14 plus v_f = 0. For clarity, below we repeat the equation.

Now we need to calculate the value of ω_d, which is given by eq. 24-13. Then:

So the system response is:

At this point it is necessary to discover the values of B₁ and B₂. To do this, we will use the initial conditions of the problem. Therefore, as we know that v_c (0⁺) = 5 V and t = 0, substituting into the equation above we found that:

And to calculate the value of B₂, we will use the initial condition i_c (0⁺ ) = 1.5 - 2 = - 0.5 A. Since i_c is related to the first derivative of v_c, then deriving v_c we find the relationship below:

Therefore, performing numerical substitution, we find:

Solving this equation, we find the value of B₂, or:

And now we can write the solution equation of the system, or:

Item b

Note that, in Figure 24-05.3, the graph in blue represents the answer to the problem solved in item a. When we make R = 5 Ω, the system response is represented by the curve in red. In this case, we have α = 2 rad/s. As α represents the damping factor of the circuit, observe that the response tends to steady state more quickly. So, when we decrease the value of R, we increase the value of α, making the circuit more stable, that is, less oscillating. There is a value of R where the circuit stops being underdamped and becomes critically damped. This fact happens when R = √10. Check!!!