Problem 24-7    Source:   Adapted from Question 1 of the PUCRS II Electrical Circuits Test - 2019.

a) Determine the value of i_L (t) for t > 0 knowing that the circuit is in steady state when the switch is in the position shown in Figure 24-07.1 for t < 0 .

b) Graph the system response.

Solution of the Problem 24-7

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⁺ ) = 20/50 = 0.4 A. We can also calculate the value of i_R (0⁺) = (20 + 5)/50 = 0.5 A and, applying LKC to node a, we can determine the value of i_c (0⁺) = 0.5 - 0.4 = 0.1 A. These are the initial conditions of the problem.

Note that in Figure 24-07.2, 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 i_L (t) will be in the form of eq. 24-14 plus i_f = 0.4 A.

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 i_L (0⁺) = 0.4 A and t = 0, substituting In the equation above we find that:

And to calculate the value of B₂, we will use the initial condition v_L (0⁺ ) = v_c (0⁺) = - 5 V . Since v_L is related to the first derivative of i_L, then deriving i _L 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 24-07.3, the graph shows a very small damping due to the small value of α. This shows a very unstable system. Note that by varying the value of α we can control the oscillations in the circuit's response. This small value of α is associated with the large value of R. Note that when t → ∞ the final value of the current in the inductor tends to 0.4 A, as shown in the solution to the problem. Compare this result with that shown in problem 25-5. It is worth noting that circuits similar to this were widely used in the production of sounds from musical instruments. in an analogue way.