Problem 24-3 Source:
Problem 8-49 - page 324 - Book: Fundamentos de Circuitos Elétricos - ALEXANDER, Charles K. & SADIKU, Matthew N. - Ed. McGraw Hill - 5ª edição - 2013.
In the circuit shown in the Figure 24-03.1, determine i(t) to t > 0 knowing that the switch S has been closed for a long time.
Figure 24-03.1
Solution of the Problem 24-3
To solve this problem we will use eq. 24-14. See below:
eq. 24-14
Note that the first part of the equation is the current flowing through the inductor after the S
switch has been open for a long time. It is easy to see that all current from the 3 A source will
circulate by the inductor. Thus, we conclude that If = 3 A.
Note that with the opening of the S switch, we have a parallel RLC circuit. So,
you can determine the value of α, ie:
α = 1 / (2 R C ) = 1 / [2 x 5 x (1/20)] = 2
The value of ωo2 will be:
ωo2 = 1 / L C = 1/ [5 (1/20)] = 4
Extracting the square root:
ωo = 2
Note that in this case, α = ωo, that is, the two roots
of the characteristic equation are equal and therefore the circuit has a response
of the critically damped type. Thus, the equation for i(t) will be of the form of
eq. 24-10 plus If, or:
i (t) = If + B1 e-2 t + B2 t e-2 t
eq. 24-10
As already known the value of If, we must find the value of
i (0-) in the inductor.
This can be done by transforming the 12 V source in series with the resistor.
4 Ω on a current source of 3 A (= 12/4) in parallel with the resistor of
4 Ω . So there will be two current sources of 3 A
each in parallel. This results in a single current source of 6 A .
As before t = 0 the circuit was on, it is known that the inductor behaves like a short circuit.
Therefore, we conclude that i (0 - ) = 6 A . As the electric current in the inductor cannot vary
abruptly, then i (0 + ) = 6 A. Now replacing t with zero
in the above equation and using the calculated values we get:
6 = 3 + B1
Hence, one easily calculates the value of B1, or:
B1 = 3
To find the solution equation, it is necessary to calculate the value of B2.
For this, the second boundary condition given by the derivative of the function i (t) must be used,
this when t = 0 . To find the derivative of the current in the inductor with respect to time, we must
follow the following line of reasoning:
"We know the values of i(0-) and i(0+), points
these that differ from Δt from each other. By the concept of derivative, we must make
Δt → 0.
Then, the tangent of the angle of the line with the t axis that connects these two points represents the derivative.
Now how i(0-) = i(0+), so the connecting line
these two points are parallel to the t axis. Therefore, the angle value is zero. And the tangent of zero
is zero. It is concluded that the derivative of the function i(t) when t = 0 is zero."
Now that we know the numerical value of the derivative of i(t), it is necessary to find the
analytical derivative of the function i(t). For simplicity of notation we will represent the derivative
in i(t) by i'(t). Then:
i'(t) = -2 B1 e-2 t + B2 e-2 t (1- 2 t)
In the above equation, replacing t with zero and knowing that i'(0) = 0 and
B1 = 3 , we find the following relation:
0 = -2 B1 + B2 ⇒
B2 = 6
We now have all the values needed to write the system solution equation. Based on eq. 24-3a (above)
and after some algebraic arrangement, we come to:
i(t) = 3 + (3 + 6 t) e-2 t A
Note that when t = 0, we get i(0) = 6 A and when
t → ∞, we get
i(∞) = 3 A, exactly the values calculated at the beginning
of the problem. The graphic in the Figure 24-03.3
illustrates the system response to the current in the inductor.
