General Solution for Natural and Step Responses
Introduction: This article summarizes the previous RL and RC step and natural responses by generalizing the approach in solving current and voltage for their respective elements.
The Essentials
With the previous topics of the step and natural response of both RL and RC circuits, we can generalize the method in analyzing these circuits.
Whether it comes to the current through an inductor or the voltage across the capacitor, they are both described by a differential equation. Describing any of these kinds of circuits take this general form:
\[ \frac{dx}{dt} + \frac{x}{\tau} = K\]
X can either represent voltage or current. When x reaches its final value, the change of x will be 0, so we can rewrite the equation, where \( x_f \) is the final value of our unit x.
\[ x_f = K\tau\]
We then need to solve for the first derivative:
\[ \frac{dx}{dt} = \frac{-x}{\tau}+K = \frac{-(x-K\tau)}{\tau} = \frac{-(x-x_f)}{\tau}\]
When we integrate this we get the following general equation:
where
\( x_f = \text{final x value} \)
\( x(t_0) = \text{initial x value before change in circuit} \)
\( t_0 = \text{the time when sudden change happens} \)
\( \tau = \text{time constant} \)
To summarize, to solve for the amplitude of x (\( x_f \) and \( x(t_0) \)), we are interested in the initial and final values of the amplitude after the sudden change. To know about the rate at which the amplitude changes, we are interested in the time constant, which is dependent on the circuit elements we use.
Example
Let's do a generalized example of both an RL and RC circuit.
Firstly, an RL circuit. Figure 1. contains an example of a natural response RL circuit.
Figure 1: Natural Response RL Example
We are interested in the voltage across and the current through the inductor. Let's solve for current first.
Using the general method, we need to find the current through the inductor just before and after the switch moves. Just before the switch moves and after a long time, the inductor acts like a wire and all current goes through it. So, we divide the voltage by the resistance to get the current through it:
\[ i_L = \frac{120}{4300} = 27.91 \, \text{mA}\]
Now to solve after the switch moves. Since the switch doesn't connect to any new power sources after time t = 0, then the final state is 0 amps.
\( \tau \) can be solved by using the definition of the time constant for RL circuits, which is:
\[ \tau = \frac{L}{R} = \frac{5}{6075.61} = 823 \, \mu s\]
We get the 6075.61 term from combining resistors from the perspective of the inductor. So, the parallel combination of 2.6k and 5.6k resistor are in series with the 4.3k resistor, so the combination is 6075.61 \( \Omega \).
So the final expression for current would be the following:
\[ i(t) = 27.91e^{-1215.066t} \ \text{mA}\]
Now if we express the voltage across the inductor, we use the equation from when we first learned about inductors:
\[ v_L(t) = L\frac{di}{dt} = 5 \frac{d}{dt}(0.02791e^{-1215.066t}) = -169.56e^{-1215.066t} \, \text{V}\]
Now let's do an example of a step response of an RC circuit. Figure 2. contains an example of one such circuit.
Figure 2: Natural Response RL Example
We are interested in the voltage and current through the capacitor. In Figure 2., the switch moves into its position shown at time t = 0. Again, let's analyze the circuit just before and after the switch moves.
Firstly, just before the switch moves. After a long time, the capacitor acts like an open circuit. With the current source, now we just multiply the 2.3k resistor with the current source to get a \( v(t_0) = 6.9 \, \text{V} \).
Now let's find the voltage after the switch moves. After a long time, the voltage acts like an open circuit still and we can use voltage division to find the final voltage across the capacitor.
\[ v_f = 60\cdot\frac{7800}{7800+8100} = 29.43 \ \text{V}\]
Now for the \( \tau \) term. \( \tau = RC \) in an RC circuit, so we get the following:
\[ \tau = RC = (9800 + (8100\parallel7800))\cdot50\cdot10^{-6} = 0.6887 \, \text{s}\]
So we get the following for the voltage across the capacitor when the switch moves at time t = 0.
\[ v(t) = 29.43-22.53e^{-1.45t} \ \text{V}\]
Now to solve for current, we take the equation for solving current through a capacitor:
\[ i_C = C\frac{dv}{dt} = 50\cdot10^{-6}\frac{d}{dt}(29.43-22.53e^{-1.45t}) = 1.633e^{-1.45t} \ \text{mA}\]
Practice
- Using Figure 3., solve for the current/voltage through/across the inductor just after the switch is moved at time t = 0 using the general method described above.
- Using Figure 4., solve for the voltage/current across/through the capacitor just after the switch is moved at time t = 0 using the general method described above.
Figure 3: Problem 1
Figure 4: Problem 2
Solutions:
Problem 1 Solution.) \( i(t) = -5 + 14.73e^{-345.429t} \ \text{mA} \) and \( v(t) = -50.88e^{-345.429t} \ \text{V} \). Let's use the general solution for solving for this inductor circuit.
\( v_f \): After the switch moves and stays there for a long time, the inductor acts like a wire and will take all current that is flowing into it, so we have our 5 mA current source flowing up through the inductor. We will assign \( v_f = -5 \ \text{mA} \) since it is flowing upward against our assumed downward direction.
\( v(t_0) \): Just before the switch moves, the inductor is part of the amalgamation of resistors with the voltage source. Note that all current coming from the voltage source does not go to the inductor because of the unique combination of the resistors. The resistors after the 1.1k\( \Omega \) resistor are all in parallel, so we can do some current/voltage division (whichever one you prefer) to solve for the current through the inductor to get:
\[ v(t_0) = 9.73 \ \text{mA}\]
\( \tau \): Solving for time constant with the following equation:
\[ \tau = \frac{L}{R} = \frac{10}{7800\parallel6200} = \frac{10}{3454.29} = 0.00289\]
Now we can put together our expression for \( i(t) \).
\[ i(t) = -5 + 14.73e^{-345.429t} \ \text{mA}\]
Now for \( v(t) \) we use the definition of the voltage over the inductor.
\[ v(t) = 10\cdot\frac{d}{dt}(-0.005 + 0.01473e^{-345.429t}) = -50.88e^{-345.429t} \ \text{V}\]
Problem 2 Solution.) \( v(t) = \) and \( i(t) = \). Let's go through the same process as earlier.
\( v_f \): For this problem, when the switch opens at time t = 0, there are no additional power sources connected. So after a long time, the voltage across the capacitor will approach 0 volts.
\( v(t_0) \): To find the voltage across the capacitor just before the switch moves, we can do circuit analysis to find the voltage divided to the capacitor. When the resistors are combined correctly, we get \( v(t_0) = 19.96 \ \text{V} \)
\( \tau \): Time constant for RC circuits is defined as follows:
\[ \tau = RC = 0.1396\]
So we get the following for the voltage across the capacitor.
\[ v(t) = 19.96e^{-7.16t} \ \text{V}\]
Using the definition of the current through a capacitor, we get the following:
\[ i(t) = C\frac{dv}{dt} = 70\cdot10^{-6} \frac{d}{dt}(19.96e^{-7.16t}) = -10e^{-7.16t} \ \text{mA}\]