Problem-Solving Strategy for solving the First-Order Circuit(Using the Step-By-Step Approach)
STEP 1. We assume a solution for the variable $x(t)$ of the form $x(t) = K_1 + K_2e^{-\frac{t}{τ}}$.
STEP 2. Assuming that the original circuit has reached steady state before a switch was thrown, draw this previous circuit with the capacitor replaced by an open circuit or the inductor replaced by a short circuit. Solve for the voltage across the capacitor, $v_C(0−)$, or the current through the inductor, $i_L(0−)$, prior to switch action.
STEP 3. Voltage across a capacitor and the current flowing through an inductor cannot change in zero time. Draw the circuit valid for $t = 0+$ with the switches in their new positions. Replace a capacitor with a voltage source $v_C (0+) = v_C (0−)$ or an inductor with a current source of value $i_L (0+) = i_L(0−)$. Solve for the initial value of the variable $x(0+)$.
STEP 4. Assuming that steady state has been reached after the switches are thrown, draw the equivalent circuit, valid for $t > 5τ$, by replacing the capacitor by an open circuit or the inductor by a short circuit. Solve for the steady-state value of the variable
$x(t)|_{t > 5τ} ≐ x(∞)$
STEP 5. Since the time constant for all voltages and currents in the circuit will be the same, it can be obtained by reducing the entire circuit to a simple series circuit containing a voltage source, resistor, and a storage element by forming a simple Thévenin equivalent circuit at the terminals of the storage element. This Thévenin equivalent circuit is obtained by looking into the circuit from the terminals of the storage element. The time constant for a circuit containing a capacitor is $τ = R_{Th}C$, and for a circuit containing an inductor it is $ τ = \frac{L}{R_{Th}}$.
STEP 6. Using the results of steps 3, 4, and 5, we can evaluate the constants in step 1 as
$x(0+) = K_1 + K_2$, $x(∞) = K_1$
Therefore, $K_1 = x(∞)$, $K_2 = x(0+) − x(∞)$, and hence the solution is
$x(t) = x(∞) + [x(0+) − x(∞)]e^{−\frac{t}{τ}}$
Keep in mind that this solution form applies only to a first-order circuit having dc sources. If the sources are not dc, the forced response will be different. Generally, the forced response is of the same form as the forcing functions and their derivatives.
PROBLEM:
Note: This post is based on Irwin's Basic Engineering Circuit Analysis 11th ed.
If there are any errors in solution, please let me know in the comments
SOLUTION:
We can find that
$i_L(0-)=\frac{24}{6+6}=2A$
Since $i_L(0+)=i_L(0-)=2A$,
$v_o(0+)=2\times 6=12V$
The KCL equation is
$-2+\frac{v_o( \infty )}{6}+\frac{v_o( \infty )}{6}+\frac{v_o( \infty )-24}{6}=0$
$v_o( \infty )=12V$
The thevenin equivalent resistance is
$R_{Th}=\frac{6\times 6}{6+6}+6=9\Omega$
Therefore, time constant is
$\tau=\frac{L}{R_{Th}}=\frac{4}{9}$
$v_o(t)$ is of the form $K_1+K_2e^{-\frac{t}{\tau}}$. From the previous analysis we find that
$K_1=v_o( \infty )=12$
$K_2=v_o(0+)-v_o( \infty )=0$
Hence, $v_o(t)=12V$
Related posts: First-Order Circuit(Using the Differential Equations)
Comments
Post a Comment