Problem 22-1    Source:   Example 10.9 - page 288 -    BOYLESTAD, Robert L. ,   Book: Introdução à Análise de Circuitos - Ed. Pearson - 10ª edição- 2008.

The capacitor shown in the Figure 22-01.1 has an initial voltage of 4 volts

a) Determine the mathematical expression for the voltage between capacitor terminals once the key is closed.

b)Determine the mathematical expression for the current during the transitional period.

c) Sketch the waveforms of voltage and current from initial value to final.

Solution of the Problem 22-1

Attention:  The solution to this problem has been adapted from the one in page 289 of the book (source) mentioned above.

Item a

From the problem, we already know that V_i = 4 volts. It is also easy to see that after closing the S switch, the capacitor will start charging until it reaches the final value, that is the voltage source value. So, V_f = 24 volts. We must now calculate the time constant of the circuit. Note that the two resistors of the circuit are in series. So we can add their values and find the value of the equivalent resistance. So we have R_eq = 3.4 kΩ. So the time constant is given by:

Since we know the time constant of the circuit, let's use the eq. 22-03 below to find the mathematical expression that defines the voltage between the terminals of the capacitor, that is:

So, we have:

Performing the calculation gives:

Note that if we take t = 0, we will find V_c = 4 volts (initial condition), and if t → ∞ , we will find V_c = 24 volts (final condition).

With this equation, we can now calculate the capacitor voltage for any desired time. Just replace the value of t (in ms) in the equation and perform the calculations.

Item b

To calculate the current expression, we need the value of i_c to t = O⁺ and also to t → ∞. Once we find these values, we replace in the eq.23-03 shown below.

To t= O⁺ we know that the capacitor has a voltage of 4 volts and the voltage on the capacitor cannot vary abruptly, so the current that will flow through the circuit depends only on the resistors that are in series with the capacitor. Soon the initial current will be:

Note that this electric current is the maximum current that will circulate through the circuit. The moment the S switch is turned on, the capacitor begins to charge, and its voltage rises exponentially. As a result, the current in the circuit decreases exponentially until it reaches the zero value, which will be the final current value. This will happen when the capacitor acquires its maximum charge, and therefore will have its voltage equal to the power supply voltage of 24 volts.

Therefore, the starting and ending values of the electric current in the circuit are already known. Substituting their values into the equation, one can find the mathematical expression that describes the behavior of the circuit.

Item c

See the Figure 22-1.2 for the graph of the capacitor voltage increase and the decay of the current in the circuit. Note the agreement of calculated values and those shown by the graph. The voltage on the capacitor starts at 4 volts (for t = 0 ) and tends to 24 volts over time. Similarly, the maximum current is 5.88mA (for t = 0 ) and exponentially decays to zero when t → +∞.