Problem 22-4    Source:   Adapted from Exercise 7.12 - page 158 -    NILSSON, James W. & RIEDEL, Susan A. - Book: Circuitos Elétricos - Ed. LTC - 5ª edição - 1999.

In the circuit shown in the Figure 22-04.1, the switch S1 remained closed and the switch S2 open, both for a long time. In t = 0 the key S1 is open The S2 key is closed 50 ms later. Determine:

a) v_C (t) to 0 ≤ t ≤ 0.05 s

b) v_C (t) to t ≥ 50 ms

Solution of the Problem 22-4

Item a

From key conditions S1 and S2 reported in the problem it is possible to determine the capacitor voltage to t = 0^-. Using the concept of current divider we can calculate the current flowing through the 50 KΩ resistor which is in parallel with the capacitor. From this we can calculate the voltage on the capacitor, which will be the same voltage on this resistor. So:

So the voltage on the capacitor is:

When the key S1 is open, the voltage on the capacitor cannot vary abruptly, so we can say that:

In addition, the circuit is reduced to the capacitor in parallel with the 50 kΩ resistor. This way, we can calculate what is the time constant of the circuit. Like this:

With this data we can write the circuit solution equation for the first 50 ms. That is:

Item b

We must pay attention when the key S2 close in t = 50 ms. Note that in this case, the capacitor has been discharged since t = 0 ms until t = 50 ms. Thus, to find the initial value of the capacitor voltage, for this new condition, we must use the equation found in the previous item and calculate its voltage when t = 50 ms. Then:

Of course when t → ∞ the capacitor will have zero voltage, or V_C(∞) = 0 volts

We must not forget that when the S2 switch closed, it put the 200 kΩin parallel with the resistor of 50 kΩ, what results in an equivalent resistor of 40 kΩ . So we can calculate the new time constant of the new circuit, or:

With this data and using eq. 22-03 we can write the circuit solution equation for t ≥ 50 ms, or:

Notice that in this equation we use the fact that - (t - 0.05) / 0.08 = -12.5 (t - 0.05).