Natural Response First Order RL Circuits
Introduction: This article focuses on the nature of a first order RL circuit and how to analyze a given circuit.
The Essentials
A first order circuit is a circuit where its characteristics are described by a first-order differential equation. We've seen scenarios where we use derivatives to describe the current through a capacitor and voltage across an inductor, and we will be applying this to our circuit analysis. For these kinds of circuits, there are lots of moving parts and it may seem like a lot, but be patient and follow the process and you will be proficient in this topic.
When it comes to analyzing these kinds of circuits, we are interested in two instances of time: 1) before a switch moves at \( t < 0 \), and 2) after the switch moves at \( t > 0 \). These two conditions are crucial in understanding the nature of each element in the circuit. This is important because the voltage across the inductor is described as the change of current. If current is a DC source, the voltage across the inductor is 0, so the only chance for voltage to be greater/less than 0 is when there is a change in the current.
So just before the switch moves, the current reaches an initial state where we analyze the circuit. This initial state would just be seen as \( I_0 \). In this initial state, there is no current through the other elements in the circuit where it makes sense. To simply state, after a long time in the initial state, the inductor acts like a wire.
Once the switch moves, the current through the inductor will change but not instantaneously. Since this change is not instantaneous, we can assume the initial value of the current after the switch moves as the following:
\[ i(0^-) = i(0^+) = I_0\]
So the initial current just before the switch moves is the same value as the current just after the switch moves.
Once again, the current does not instantly change once the switch moves. As described in the textbook, the current of a first order RL circuit after the switch moves at \( t = 0 \) can be represented as the following:
A term that is helpful to know about is the time constant \( \tau \), where
So
The time constant describes the rate at which the current approaches zero. Because of this, we can get an idea of how long a system takes in order for the circuit to approach a steady state after the switch moves. Usually for a system to approach this state takes about 5 values of \( \tau \). When \( t = 5\tau \), the output of the inductor current will be so close to its steady state, the error is less than 1\( \% \).
Now this is a lot but there are three steps to follow in order to find the equation for the current \( i(t) \). It goes as follows: 1) Find \( I_0 \), 2) find the time constant \( \tau \), and 3) use equation 1 to find \( i(t) \) from \( I_0 \) and \( \tau \).
Example
Let's analyze an example of an RL circuit. Figure 1. has this example.
Figure 1: First-Order RL Circuit
Let's say the current \( i_s = 10 \, \text{A} \). So let's go through our step-by-step process. 1) solve for \( I_0 \). We know from earlier that \( I_0 = i(0^-) = i(0^+) \), so we need to know the initial current through the inductor. We're assuming that the switch has been closed for a long time so it has reached its initial state. This means that there is no change in the current, so the voltage across the inductor is 0 V and all current goes through the inductor. So, \( I_0 = 10 \).
2) Find \( \tau \). Since \( \tau = \frac{L}{R} \), we need to solve for the resistance from the perspective of the inductor. We'll need to combine resistors to the right of the inductor since after the switch is opened the resistors on the left side will not experience the stored energy. Finding R:
\[ R = 30 \parallel 15 = 10 \, \Omega\]
So \( \tau = \frac{5}{10} = \frac{1}{2} \).
Now finally, 3) solve for \( i(t) \). We just need to use equation 1 and substitute values.
\[ i(t) = I_0e^{-t/\tau} = 10e^{-t/0.5} = 10e^{-2t} \, \text{A}\]
A graph shown in Figure 2. shows the equation we just solved for over time.
Figure 2: First-Order RL Circuit Graph
Note: since we solved for \( \tau \) to be 1/2, we can solve for the time when the current through the inductor to reach its final state. Since we know that it takes \( 5\tau \) to get to a very close approximation, we simply do \( 5\cdot\frac{1}{2} = 2.5 \, \text{seconds} \). So at 2.5 seconds, the current output is 0.0674 A, which is 0.674\( \% \) of the change to the final state current.
Practice
- Given that the circuit has been closed for a long time, solve for the current \( i(t) \) through the inductor when \( t = 0 \), when the switch opens.
Figure 3: Problem 1
- Given that the circuit has been closed for a long time, solve for the current \( i(t) \) through the inductor when \( t = 0 \), when the switch opens. Also solve for the voltage across the 60 \( \Omega \) resistor when the switch is opened.
Figure 4: Problem 2
- Given that the circuit has been closed for a long time, solve for three characteristics; 1) the current \( i(t) \) through the inductor when the switch is opened and 2) the voltage across the 60 \( \Omega \) resistor, and 3) solve for current through inductor after 5\( \tau \).
Figure 5: Problem 3
Solutions:
Problem 1 Solution.) \( i(t) = 6e^{-3.63t} \, \text{A} \). Let's go through the steps. 1) Solve for \( I_0 \). We know that after a long time, the inductor acts like a wire and all current passes through it, so \( I_0 = 6 \, \text{A} \). 2) solve for tau. Since \( \tau = \frac{L}{R} \), so \( \tau = \frac{4}{14.52} = \frac{100}{363} \). Finally, 3) write the current equation based on equation 3. If we put this together, we should get the solution shown above.
Problem 2 Solution.) \( i(t) = 1.18e^{-3.59t} \, \text{A} \) and \( v(t) = 33.86e^{-3.59t} \, text{V} \). Now we have a voltage source connected to this circuit, but this should be fairly simple to solve for our first step: finding \( I_0 \). We need to find the current through the 35 \( \Omega \) resistor and from there all of the current through that resistor will be the current through the inductor.
We can simply combine the 35 \( \Omega \) and 15 \( \Omega \) resistor in parallel and divide the voltage across the outcome resistor and the 20 \( \Omega \) resistor. Once we do this, we get a voltage value 41.3 volts across the 35 \( \Omega \) resistor. Then use Ohm's law to solve for the current, which is 1.18 amps. Since after a long time the inductor acts like a wire, all current goes through that, so that is our \( I_0 \).
Now, solve for \( \tau \).
\[ \tau = \frac{8}{55 \parallel 60} = 0.279\]
Now, let's finally solve for \( i(t) \) using the values we solved for.
\[ i(t) = 1.18e^{-3.59t} \, \text{A}\]
Now let's solve for the voltage across the 60 \( \Omega \) resistor. Now, the voltage across that resistor is the same as the voltage across the 55 \( \Omega \) resistor as well. Because of this, we can combine the resistors in parallel and solve for the \( V_0 \) of \( v(t) \).
\[ V_0 = I_0 \cdot (55\parallel60) = 33.86 \, \text{V}\]
Now let's put this as an expression for \( v(t) \). We don't need to solve for a new exponential term since current and voltage are related linearly.
\[ v(t) = 33.86e^{-3.59t} \, \text{V}\]
Problem 3 Solution.) \( i(t) = 2e^{-5.8t} \, \text{A} \), \( v(t) = 26.06e^{-5.8t} \), and \( i(5\tau) = 13.4 \, \text{mA} \). With these two current sources in parallel, they can be added up and since they are in opposite directions, they will sum to 2 amps up, so \( I_0 = 2 \, \text{A} \). Next let's solve for \( \tau \). The total resistance seen from the inductor is
\[ R = (90 \parallel 25) + 50 = 69.56 \, \Omega\]
So \( \tau = \frac{12}{69.56} = \frac{300}{1739} \).
Now we can write our expression for \( i(t) \), so
\[ i(t) = 2e^{-5.8t} \, \text{A}\]
Now let's solve for the voltage across the 60 \( \Omega \). We take \( I_0 \) from earlier and solve for \( V_0 \). We can do some voltage division a couple of times to solve for \( V_0 \) across the 60 \( \Omega \) resistor to be 26.09 volts. So,
\[ v(t) = 26.08e^{-5.8t} \, \text{V}\]
Finally, finding the current after \( 5\tau \). The significance of \( 5\tau \) is that that length of time is considered "long enough" for the circuit elements to reach their final state, whether the switch is opened or closed. To do this, we simple multiply the tau value we found earlier by 5 and substitute it into our \( i(t) \) equation.
\[ 5\tau = 5\cdot\frac{12}{69.56} = \frac{1500}{1739}\]
\[ i(5\tau) = 2e^{-5.8\cdot\frac{1500}{1739}} = 13.4 \, \text{mA}\]