Problem 22-5 Source:
Problem 7.30 - page 300 Book: Análise de Circuitos em Engenharia - J. David Irwin - 4ª edição - Ed. Pearson - 2013.
In the circuit shown in the Figure 22-05.1, the switch S remained open for a long time. In
t = 0 the S key is closed. Determine the current i(t).
Solution of the Problem 22-5
Note that we have a capacitor association in the circuit. The bottom two are parallel, totaling a capacitance of 400 µF. This in turn is in series with the other that is connected to the point
a. Calculating the series, one finds Ceq = (800 / 6 ) µF. On the other hand, note that this capacitance is parallel to the resistor of 6 kΩ . Therefore, it is easy to see that when calculating the voltage on the capacitor, VC (t), it is possible to calculate i (t) by simply dividing this voltage by
the resistor value, that is
6 kΩ.
Following this reasoning, the voltage on the capacitor with the S switch open will be the same as the voltage on
the 6 kΩ resistor. To calculate this tension, any method studied can be employed. Using the source transformation method, it is found that it is possible to reduce to a single upward-pointing 1 mA current source. The two resistors of 2 kΩ and 3 kΩ, which after the transformation are parallel, results in a resistor of 6/5 kΩ.
With this, through a current divider, it is possible to calculate i(0-) circulating through the resistor of 6 kΩ, that is:
Therefore, using the law of Ohm it is possible to calculate the voltage on the resistor of
6 kΩ, which is exactly equal to the voltage on the capacitor. Soon:
When the S switch is closed, the resistance of 2 kΩ is connected in parallel with the capacitor and resistor of 6 kΩ. Now, as is well known, a capacitor cannot abruptly alter the voltage on its terminals. So the initial condition is:
It is now necessary to calculate the voltage on the capacitor when
t → ∞ (final condition). From the circuit, it is observed that the resistor of
2 kΩ remains in parallel with the resistor of 6 kΩ. In this case, to calculate the voltage on the capacitor, simply substitute the equation ⓵ the value of 6 kΩ by the parallel of the two resistors, ie 1.5 kΩ. Note that there was no change in the current source value. So:
With this information, the value of the time constant of the circuit is missing. For this, the voltage sources are annulled and the equivalent resistance is calculated. Note that all resistors are connected in parallel. Thereby, Req = 1.059 kΩ. So the time constant τ of the circuit is:
The eq. 22-03 allows you to write the resulting voltage over the capacitor, or:
As stated at the beginning of the problem solution, to calculate i (t) just use the equation below:
Therefore, the solution equation for the problem is:
