Problem 23-4 Source:
Exercise 7-15 - page 329 - THOMAS, Roland E. ,
ROSA, Albert J. , THOUSSAINT, Gregory J. - Book: The Analysis & Design of Linear Circuits
- 6ª Edição - Ed. John Willey & Sons, Inc. - 2009.
In the circuit show in the Figure 23-04.1, the S switch has been closed for a long time
in position a. In t = 0 the switch has moved to position b.
Calculate the voltage v(t) to t > 0.
Figure 23-04.1
Solution of the Problem 23-4
As the S switch remained on for a long time the inductor behaves like a short circuit
and the current i circulating through the resistor R in t = 0- is i(0-) = 0.
Then, through the inductor circulates a current of:
iL (0-) = 10 / 50 = 0.2 A
This is the situation in t = 0-, when the S switch is in position a.
Passing the key to position b at t = 0 has the condition shown in the circuit
below. Note that we made a source transformation on the circuit consisting of the 10 volts source voltage and
the resistor of 50 Ω. We found a current source of 0.2 u-1 (t) with the arrow pointing down. And the resistor of 50 Ω paralleled the inductor.
On the other hand, we introduced an impulsive source of 0.2 A in place of the
current circulating through the inductor in t = 0-. See the
Figure 23-04.2.
Figure 23-04.2
Thus, when at t = 0 , acting on the impulsive source together with the step source, the inductor will behave as an
open circuit. In this situation, the current flowing through the resistor
R can be calculated, knowing that the two sources acting together act as a single source of
0.4 A.
Then, using a current divider, we get:
i(0) = - 0.4 (50 / 125) = - 0.16 A
Note the negative signal due to the reversal of current sources. Now you can calculate the voltage on
the resistor R, that is, v(0), using the law of Ohm.
v(0) = - 0.16 x 25 = - 4 volts
It is necessary to calculate the current over the resistor R when t → ∞. But, when t → ∞ the inductor will behave like a short circuit and the impulsive source will not act. Thus, we conclude that i(∞) = 0, because the current from the step current source will circulate completely through the inductor.
To find the solution to the problem we need to calculate the time constant of the circuit. Of
above circuit, it is noted that the resistor of 50 Ω is in parallel with the other two that
are in series. Thus, making the calculation, we find the equivalent resistance, equal to 30 Ω. So:
τ = L / Req = 0.15 / 30 = 0.005 s
To find the solution of the problem just use the eq. 23-03. Then:
