Problem 24-8    Source:   Adapted from Question 1 of the PUCRS II Electrical Circuits Test - 2019.

Determine the value of i_L (t) to t > 0 knowing that the circuit is in steady state when the switch is in the position shown in Figure 24-08.1 for t < 0 .

Solution of the Problem 24-8

Item a

Note that for t < 0 we have the circuit shown in Figure 24-08.2, where we replace the inductor with a short circuit. With the current source and resistor R₁ we did a source transformation obtaining a voltage source of 11 volts in series with R₁. Thus, we can add the values of R₁ and R₂, obtaining the value of R = 3 kΩ. Because the inductor behaves like a short circuit, we easily obtain the value of the voltage across the capacitor, that is, V_c (0⁺) = 4 V. Following the indication of the currents according to the circuit, we obtain i (0⁺) = - (11 - 4) / 3 k = - 7/3 mA. And i_L (0⁺) = (11 - 4) / 3 k = 7/3 mA, since no electric current flows through the capacitor . These are the initial conditions of the problem.

Para t > 0 a chave está na condição fechada. Como resultado, a fonte de corrente e o resistor R₁ são eliminados do circuito e o resistor R₂ é colocado em paralelo com o capacitor, conforme mostra a Figura 24-08.3.

Thus, 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, 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 solution equation will be in the form of eq. 24-14 and eq. 24-15.

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

Then the system's response to i_L (t) will be given by:

From the circuits shown in the figures above, we have that i_L (0⁺ ) = 7/3 mA and i_L (∞ ) = - 2 mA. Therefore, for t = 0⁺ let's find the equation below.

As we know, sin 0 = 0, cos 0 = 1 and e⁰ = 1. So, the equation reduces to:

Carrying out the calculation we find the value of B₁, or:

And to calculate the value of B₂, we will use the initial condition v_L (0⁺ ) = 0. 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 for the current in the inductor, or:

Remember that the value of this current is in milliampere.