Step Response of RL and RC Circuits
Introduction: This article focuses on the step response of RL and RC circuits
The Essentials
Up to this point, we have analyzed the RL and RC circuits in cases where we suddenly remove the power source from them. Now we are curious to know what happens when we suddenly apply a power source, also known as a step response.
Firstly the RL circuit. Figure 1. illustrates the step response of an RL circuit.
Figure 1: Step Response RL Circuit
Using similar methods in the natural response, we are given the following expression for the step response.
Now for the step response of an RC circuit. Figure 2. shows the general schematic for an RC circuit. In the case for an RC circuit, using similar analysis methods, we get the following equation.
Figure 2: Step Response RC Circuit
For both of these equation, these describe the current or voltage the moment the circuit experiences immediately after a power source is applied.
Example
We'll do two examples, one for RL circuits and one for RC circuits.
First, RL circuits. Figure 3. contains an example of the step response of an RL circuit.
Figure 3: Step Response RC Circuit
To clarify for the circuit diagram, the switch shown in its position just before it switches at time t = 0.
Let's use equation 1, but let's go through how we'd fill it in. For reference,
\[ \frac{V_s}{R} + (I_0-\frac{V_s}{R})e^{-(R/L)t} \]
For \( V_s \), we use the voltage applied to the inductor after the switch has moved from right to left. So, \( V_s = 20 \, \text{V} \).
For R, we use the total resistance of the circuit with relation to the source we just switched to. So, in the scenario of this circuit, the only resistor affected is the 5.6 k\( \Omega \). The other resistors we can ignore. So, \( R = 5600 \)
Finally, for \( I_0 \). This essentially means the initial current through the inductor before the switch it moved. This shouldn't be too hard, especially since we had practice with the natural response circuits. For this example, after a long time, the inductor acts like a wire just before switching, so the initial current is the current source applied. So, \( I_0 = 5 \, \text{mA} \).
For the variable \( R/L \) in the exponent is also the same as the \( \tau \) covered previously for RL first order circuits. When it comes to calculating \( \tau \), which is represented by:
\[ \tau = \frac{L}{R}\]
We use the same R as described earlier and use the inductor value, so \( \tau = \frac{300\cdot10^{-3}}{5600} = \frac{3}{56000} \)
Now just take this inverse to get \( R/L \), so \( \frac{1}{\tau} = \frac{56000}{3} \).
Now we can fill in values for the equation.
\[ i(t) = \frac{20}{5600}+(5\cdot10^{-3}- \frac{20}{5600})e^{-(56000/3)t} = 3.57 + 1.43e^{-(18666.67)t} \, \text{mA}\]
To get an idea of the current, Figure 4. contains a graph of this current with respect to time.
Figure 4: Step Response RC Circuit
As we can see, the current initially is set to 5 mA and after a long time (meaning 5\( \tau \), since a long time is less than a quarter of a millisecond). Then after the switch is moved the current settles at 3.57 mA.
Now let's do an example with a capacitor. Figure 5. has an example.
Figure 5: Step Response RC Circuit
Let's analyze the circuit and fill in equation 2 described above. For reference, the equation is the following:
\[v_C = I_sR + (V_0 - I_sR)e^{-t/RC}\]
Let's look through this in a logical way. The \( I_sR \) is the open circuit voltage just after the switch is moved. In this scenario, just before the switch moves, all of the 5 mA current source is going through the 3 k\( \Omega \) resistor, so the open circuit voltage is 15 volts.
Then we need to find the initial voltage \( V_0 \). This is the voltage across the capacitor just before the switch moves. In this case, the voltage due to the 60 volt power source. Using voltage division, we get \( V_0 = 32 \, \text{V} \).
Finally, our \( \tau \) term. \( \tau \) is the same as the natural response of an RC circuit, so \( \tau = RC \). Solving for \( \tau \) we get:
\[ \tau = RC = ((2700\parallel3000) + 2000) \cdot 470 \cdot10^{-6} = 1.61\]
Now putting together our equation, we get the following:
\[ v_C(t) = 15 + 17e^{-t/1.61}\]
To give us an idea of what \( v_C(t) \) looks like, Figure 6. has a graph showing the voltage across the capacitor when the switch is moved.
Figure 6: Step Response RC Graph
After a long time (5\( \tau \)), we stabilize our new voltage as 15 volts.
Practice
- Using Figure 7., solve for the current through the inductor given that after a long time the switch was moved at time t = 0.
- Using Figure 8., solve for the current through the inductor given that after a long time the switch was moved at time t = 0.
- Using Figure 9., solve for the voltage across the capacitor given that after a long time the switch was moved at time t = 0.
- Using Figure 10., solve for the voltage across the capacitor given that after a long time the switch was moved at time t = 0.
Figure 7: Problem 1
Figure 8: Problem 2
Figure 9: Problem 3
Figure 10: Problem 4
Solutions:
Problem 1 Solution.) \( i(t) = -2,18 + 45.18e^{-1022.39t} \, \text{mA} \). Let's find the values for the equation and then substitute into equation 1.
\( V_s \): To solve for \( V_s \), we determine the voltage across the inductor. To do this, we divide the current being applied to the inductor just after the switch is moved. To do this, divide the 5 mA current source amongst the 2.7k and 3.5k \( \Omega \) resistor. We get a current value of 2.18 mA, leading to a \( V_s = 7.62 \, \text{V} \). However, since \( V_s/R \) is simply the current through the inductor after the switch is moved, that value is -2.18 mA. This value is negative because we assume the current through the inductor just before the switch moves is downward.
\( I_0 \): Solving for \( I_0 \) involves the current just before the switch moves. Since the inductor acts like a wire after a long time, we add the 3 mA current source and the current resulting from the voltage source and 2 k\( \Omega \). Calculating the current, we get the following:
\[ I_0 = (3+40)\cdot10^{-3} = 43\cdot10^{-3} \, \text{A}\]
\( R \): Solving for R, we find the equivalent resistance from the perspective of the inductor. We get the following:
\[ R = 3500 + (2700\parallel4000) = 5.112 \, \text{k}\Omega\]
Now we can substitute our values to get our expression for \( i(t) \).
\[ i(t) = -2.18 + (43 - (-2.18))e^{-\frac{5112}{5}t} \, \text{mA}\]
Problem 2 Solution.) \( i(t) = 21.43 + 13.19e^{-252.55t} \, \text{mA} \). Again, let's use equation 1 and find values.
\( V_s \): This is the current through the inductor after the switch moves. After a long time, the inductor acts like a wire and takes all current from the 120 volts source.
\[ \frac{V_s}{R} = \frac{120}{5600} = 21.43 \, \text{mA}\]
\( I_0 \): This is the current through the inductor just before the switch moves. After a long time, the inductor acts like a wire and takes all current from the 90 volts source.
\[ I_0 = \frac{90}{2600} = 34.62 \, \text{mA}\]
\( R \): For R, we need to find the equivalent resistance after the switch moves.
\[ R = 4600\parallel5600 = 2525.5 \, \Omega\]
Now put together the equation with our solved values.
\[ i(t) = 21.43 + 13.19e^{-252.55t} \, \text{mA}\]
Problem 3 Solution.) \( v(t) = -19.5 + 44.5e^{-5.118t} \, \text{V} \). Again, use equation 2 to build the expression for the voltage across the capacitor.
\( I_sR \): For this expression, this is the voltage just after the switch is moved at time t = 0. So looking at the circuit and knowing that after a long time a capacitor acts like an open circuit, no current is passed through and goes all through the 6.5k \( \Omega \) resistor. So, the voltage just after the switch moves is the current multiplied to the 6.5 k\( \Omega \) resistor, resulting in -19.5 volts. This is the case because we assume positive terminal for the capacitor is in the top node.
\( V_0 \): This is the voltage across the capacitor just before the switch moves. For this, multiplying the current by the 5k\( \Omega \) resistor will suffice for finding \( V_0 \).
\[ V_0 = 5\cdot10^{-3} \cdot5000 = 25 \, \text{V}\]
\( R \): For this, solve for the equivalent resistance from the perspective of the capacitor. For this, it is simply a parallel combination of the two resistors.
\[ R = 6500\parallel9800 = 3907.97\, \Omega\]
Then we can substitute into equation 2 to find \( v(t) \).
\[ v(t) = -19.5 + (25 - (-19.5))e^{-t/3907.97\cdot50\cdot10^{-6}}\]
Problem 4 Solution.) \( v(t) = 42.4 + 36.2e^{-1.062t} \, \text{V} \). Use equation 2 to derive an expression for voltage across the capacitor.
\( I_sR \): This is the voltage across the capacitor just after the switch is moved at time t = 0. For this, we can use voltage division over the 5.6k and 6.3k resistor since the capacitor acts like on open circuit after a long time. After doing voltage division, the voltage across the 6.3k, and in turn the capacitor, is 42.4 volts.
\( V_0 \): This is the voltage across the capacitor just before the switch is moved at time t = 0. Since the capacitor acts like an open circuit after a long time, all current goes through the 3.1k and 2.3k \( \Omega \) resistors. Knowing the voltage is set in the node between the two resistor, we just need to find the resistor across the 3.1k\( \Omega \) resistor. All of the current from th current source travels through the 3.1k\( \Omega \) resistor and the voltage across that is 18.6 volts. After adding 60 volts, we get a total \( V_0 = 78.6 \, \text{V} \).
\( R \): Now let's solve for R after the switch is moved from the perspective of the capacitor.
\[ R = (5600\parallel6300)+7500 = 10464.71 \Omega\]
Now we can make our \( v(t) \) expression:
\[ v(t) = 42.4 + (78.6 - 42.4)e^{-t/10464.71\cdot 90 \cdot10^{-6}} \, \text{V}\]