Problem 23-5    Source:   Exercise 7.13 - page 158 -    NILSSON, James W. & RIEDEL, Susan A. - Book: Circuitos Elétricos - Ed. LTC - 5ª edição - 1999.

In the circuit show in the Figure 23-05.1, the S₁ switch remained open for a long time, while the key S₂ remained closed for a long time. The S₁ switch is closed when t = 0 and, after remaining closed for 1 s, it is opened again. The S₂ switch is open when t = 1 s. Determine expressions for current i in the inductor that are valid in the ranges:

a) 0 ≤ t ≤ 1.

b) t ≥ 1.

Attention - In the problem statement there is a typo error in the letter b. Where it says t ≤ 1 s read on t ≥ 1 s.

Solution of the Problem 23-5

Item a

From the statement of the problem we deduce that before t = 0 There is no source acting on the circuit. So as soon as the S₁ switch close (em t = 0), There will be no current flow through the inductor. Hence, it follows that:

Knowing the initial value of the current by the inductor, we need to know its final value and the time constant of the circuit. For this, the Thévenin equivalent of the circuit must be found. Note that only the voltage source is operating. So the open circuit voltage (Thévenin voltage) will be 6 volts, which is the voltage over the 3 ohms resistor. To meet the resistance of Thévenin, we must short circuit the voltage source. This results in the parallel of the resistors of 2 and 3 ohms, resulting a resistor of 1.2 ohm, which added to the 0.8 ohm resulting Thévenin resistance equal to 2 ohms. On the other hand, the S₂ switch closed, puts the 3 and 6 ohm resistors in parallel, resulting in a 2 ohms resistor in parallel with the inductor. This situation is shown in the figure below. We opted to make the equivalent of Thévenin to the left of the S₁ switch for clarity.

From the circuit show in the Figure 23-05.2 , it is possible to calculate the equivalent resistance to the left of the inductor. There is the parallel of the resistors of 2 ohms resulting the value of 1 Ω as equivalent resistance. So the time constant will be:

Note that for t → ∞, the circuit comes down to the inductor and the resistance equivalent of Thévenin. Therefore, it is concluded that the current value i when t → ∞ is equal to:

Thus, one can write the requested equation in the problem. Using the eq. 23-03, we get:

Item b

To t ≥ 1 whole circuit to the left of the S₁ switch is deleted. Thus, the Thévenin equivalent must be found for the circuit to the right of the inductor. Eliminating the source, we can find Thévenin's resistance. Thus, the resistors of 6 and 9 ohms which are in series result in a value of 15 ohms. This, in turn, is in parallel with the resistor of 3 ohms. This results R_th = 2.5 Ω. For the calculation of the Thévenin voltage, we must transform the current source in parallel with the 9 ohms resistor into a voltage source equal to 72 volts, which will be in series with the 9 ohms resistor. Note that the positive pole of the voltage source points downwards. Therefore, the Thévenin voltage will be the voltage over the 3 ohms resistor. After performing the calculation, we are V_th = -12 V

Knowing R_th, it is possible to calculate the time constant of the circuit, or:

We must calculate the start and end current in the inductor. For the initial current, we use the equation found in item a and calculate the current for t = 1s.

And to t → ∞, the inductor behaves like a short circuit, so:

Using the eq. 23-03, we can write:

Performing the calculation, we find the final expression:

Here we use the fact that -(t-1)/0.8 = -1.25 (t-1).