Problem 23-7 Source:
Problem 3.9 - 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 Vo in the circuit shown in
Figure 23-07.1, when
key S is in position 2, knowing that it has been in position 1 for a long time.
Figure 23-07.1
Solution of the Problem 23-7
Note that this problem can be solved by finding the expression for iL. To find the final result, we use the fact that
V
o = 6 i
L
eq. 23-07.1
When the S switch is in the 1 position, we can easily see that iA = 3 A.
Thus, the dependent source assumes the value of 2 iA = 6 volts . As the inductor behaves
like a short circuit, we quickly find that
iL (0-) = iL (0+) = 3 A
To calculate the value of iL for time t → ∞, we should consider
the S key in position 2. In this case, we must find the equivalent Thévenin
the circuit. So, we must work with the circuit shown in Figure 23-7.2.
Figure 23-7.2
Note that we indicate the inductor as a short circuit. So, we must find the value of
iL (∞). To do this, we will calculate the values of iA and
i1.
Then, making the mesh indicated by the green arrow, we have the following relation:
- 36 + 2 i
1 + 4 i
A = 0
eq. 23-07.2
On the other hand, by making the mesh indicated by the arrow blue, we obtain the following relation:
- 4 iA + 6 (i1 - iA ) - 2 iA = 0
eq. 23-07.3
Algebraically working the equation eq. 23-07.3 and along with the equation
eq. 23-07.2 we get a system of two equations with two unknowns that are easy to solve and the values found are:
iA = 4.50 A and i1 = 9.0 A
So, we have to
iL ( ∞ ) = 4.50 A
To find the value of Rth we must put in place of the inductor a convenient
current source, such as I = 3 A, as shown in Figure 23-7.3.
Figure 23-7.3
Note that we use the fact that if the iA circulates in the 4 ohm resistor,
then in the 2 ohm resistor should circulate twice the current or, 2 iA.
So, we easily find the value of iA, because of the circuit we have to
3 iA = 3 so iA = 1 A
To find the value of Vab just make the mesh, or
Vab = 4 iA + 2 iA + 6 x 3
But as iA = 1 A , then performing the calculation we find
Vab = 24 volts
At this point, we can calculate the value of Rth, because
Rth = Vab / I = 24 / 3 = 8 ohms
And now we are going to calculate the time constant of the circuit, that is
τ = L / Rth = 3 / 8 = 0.375 s
With all the calculated data, we can write the equation of iL (t), or
iL (t) = 4.5 - 1.5 e - t / 0.375 A
Now using the equation eq. 23-07.1 developed at the beginning of the solution of the problem, we have the solution requested in the statement of the problem, or