Problem 23-6 Source:
Problem 3.5 - page 11 - Prof. Dr. João Costa Freire - Instituto Superior Técnico - Portugal Book: Analysis of Circuits - 2006.
Available at: http://web.ist.utl.pt/pedro.m.s.oliveira/ProbAC.pdf
Determine the expression for iL in the circuit shown in
the Figure 23-06.1.
Solution of the Problem 23-6
Note that this problem uses an independent source through step function.
In this way, we know that it will act only after t = 0. Before t = 0, its value is
null. Then, the dependent font 1.5 iA will also be null . So, we already know that
iL (0-) = iL (0+) = 0 A
To find the current in the inductor for time t → ∞ , we must calculate the
Thévenin equivalent to the left of the inductor. Removing the inductor from
the circuit, we are left with the circuit shown in the Figure 23-6.2.
We need to find a relation between iA and i1.
Then, using the law of nodes for node a, we get:
- i1 + iA - 1.5 iA = 0
Resolving this relation, we have
i1 = - 0.5 iA
eq. 23-06.1
On the other hand, applying KVL to the circuit formed by the two resistors and the independent source, we have:
- 100 + 10 i1 + 20 iA = 0
eq. 23-06.2
Replacing eq. 23-06.1 in eq. 23-06.2 and performing the calculation, we get
iA = 20/3 A
With the value of iA we can calculate the value of Vab (open circuit voltage), which is the value of Vth. So
Vth = Vab = 20 iA = 400/3 V
To find the value of Rth, we must short-circuit the ab terminals on the circuit in Figure 23-6.2 and calculate the short-circuit current
Isc. We noticed that when making the short-circuit, the dependent source and the resistance of 20 ohms come out of the circuit. In this way, we have
Isc = 100/10 = 10 A
With these data and using the eq. 15-01 we can calculate the value of Rth, or
Rth = Vth / Isc = (400/3)/ 10 = 40/3 Ω
In the Figure 23-6.3 we can see what the Thévenin equivalent circuit
looked like.
At this point, we can calculate the current in the inductor for time t → ∞ , remembering that for this time the inductor behaves like a short circuit. Soon:
iL (∞) = Vth / Rth = 10 A
And now, based on the circuit of Figure 23-6.3 , we are going to calculate the time constant of the circuit, that is
τ = L / Rth = 0.05 / (20/3)
= 3.75 x 10-3 s
With all the calculated data in hand, we can write the equation that expresses the circuit
behavior shown in Figure 23-6.1. So