Natural Response First Order RC Circuits
Introduction: This article covers the topic and analysis of a natural response first order RC circuit
The Essentials
Similar to the natural response of an RL circuit, we are curious to know the natural response of an RC circuit. However, we do not treat this circuit in the same way that we did for the RL circuit since we are dealing with a capacitor.
To illustrate this, we will use Figure 1. to make explaining the circuit analysis easier.
Figure 1: First-Order RC Circuit
For RC circuit analysis we first assume that the switch has been on position a for a long time. Because a capacitor acts like an open circuit after a long time, the initial voltage \( V_0 \) of the capacitor is the same as the voltage source. And since there is no instantaneous change in voltage after the switch goes from position a to position b, we can assume the following:
\[ v(0^-) = v(0^+) = V_g = V_0\]
So the voltage just before the switch moves is the same just after the switch moves.
The textbook delves more into analyzing the natural response of an RC circuit, but this equation is derived using node voltage analysis in the circuit:
Like the RL circuits, we can also solve for the time constant \( \tau \), which is defined as the following:
So, after rewriting the natural response equation above, we get the following:
v(t) = V_0e^{-\frac{t}{\tau}}
\]This is a similar process to the RL circuit analysis, and we even have very similar steps in solving for the response of an RC circuit. It can be broken down into 3 steps. 1) solve for initial voltage \( V_0 \), 2) solve for the time constant \( \tau \), and 3) using equation 3 described above, use \( V_0 \) and \( \tau \) to generate a \( v(t) \) to describe the circuit behavior.
Example
Let's do an example of a problem that you may encounter. We will go through step by step how to get a natural response expression for an RC circuit. Figure 2 has an example that you may encounter.
Figure 2: First-Order RC Circuit Example
The schematic may seem confusing, but to clarify the switch is moved from position a to position b at time t = 0. Now let's analyze step by step.
1) Find \( V_0 \). We need to know the voltage across the capacitor when the switch is at position a after a long time initially. Since we know that after a long time a capacitor is seen as an open circuit, we can safely assume the voltage across the capacitor before t = 0 is 120 volts. So, \( V_0 = 120 \, \text{V} \).
2) Find \( \tau \). Using equation 2, we can solve for \( \tau \) very easily. We just need to combine the resistors in parallel to find our R after the switch is moved to the b position. Use the equation above to get the following:
\[ \tau = RC = (60\parallel75)\cdot470\cdot10^{-6} = \frac{37}{3000}\]
3) Use \( V_0 \) and \( \tau \) to write an expression for \( v(t) \). We substitute values to get the following:
\[ v(t) = V_0e^{-\frac{t}{\tau}} = 120e^{-63.83t} \, \text{V}\]
To give us an idea of what the voltage across the capacitor looks over time, Figure 3. shows this relationship.
Figure 3: RC Circuit Example Graph
As we can see, the dissipation of the voltage is very quick, even less than a quarter of a second. We can see how long it would take for the voltage to reach a stable point by using the same \( 5\tau \) method as we did in the RL circuits. Calculating for \( 5\tau \), we get 78.3 milliseconds.
Note: there are scenarios where you may need to analyze the natural response of an RC circuit with multiple capacitors. This is possible and there will be multiple exponential terms to take into account. There is a practice problem with this very scenario, give it a shot and see if you can get it correct.
Practice
- Given that the switch as been in the 'a' position for a long time, find the voltage of the capacitor when the switch moves to position 'b'.
- Given that the switch has been closed for a long time, find an expression for \( v(t) \) that describes both capacitor voltages. (Hint: treat each pair of capacitors with resistors as their own expressions, then add them together to find the final \( v(t) \))
- Given that the switch has been closed for a long time, solve for the voltage across the capacitor when the switch opens. Also solve for the current through the resistors.
Figure 4: Problem 1
Figure 5: Problem 2
Figure 6: Problem 3
Solutions:
Problem 1 Solution.) \( v(t) = 60e^{-t/4.41} \, \text{V} \). Let's go through step by step.
1) Solve for \( V_0 \). This is done by noticing that the capacitor, after a long time, acts like an open circuit applied to a 60 V power source. So, \( V_0 = 60 \, \text{V} \).
2) Solve for \( \tau \). Using equation 2, we get the following:
\[ \tau = ((102k\parallel47k)+56k) \cdot 50 \cdot10^{-6} = 4.41\]
3) Put together our \( v(t) \) using the values we just calculated. Using \( V_0 \) and \( \tau \), we get the following:
\[ v(t) = 60e^{-t/4.41}\]
Note: for this capacitor, the time it takes for the circuit to reach its steady state, whether the switch is in position a or b, take a much longer time than usual. Our example above 78.3 milliseconds, while this one will take about 22 seconds. This is in part due to both the larger value of the equivalent resistance we're seeing. Figure 7. contains a graph to illustrate the dissipation of the capacitor voltage over time.
Figure 7: Problem 1 Voltage graph
Problem 2 Solution.) \( v(t) = 21.3e^{-1.042t} + 14.2e^{-1.786t} \, \text{V} \). For this, it may seem intimidating to solve for the voltage across the capacitors but let's look at our hint given to us. When the switch is closed for a long time, the capacitors act like an open circuit. While that is happening, the flow of charges go through each resistor. Because of this, each capacitor will have an initial voltage directly linked to the voltage across each resistor. So we do voltage division to find the voltage across each capacitor like so;
\[ V_{80\mu} = \frac{12}{12+8+25}\cdot80 = 21.3 \, \text{V}\]
\[ V_{70\mu} = \frac{8}{8+12+25}\cdot80 = 14.2 \, \text{V}\]
Now with our respective \( V_0 \) values, we can now find their respective \( \tau \) values.
\[ \tau_{80\mu} = 12000 \cdot 80 \cdot 10^{-6} = 0.96\]
\[ \tau_{70\mu} = 8000 \cdot 70 \cdot 10^{-6} = 0.56\]
Now with both values \( V_0 \) and \( \tau \) solved for each capacitor, now we can solve for \( v(t) \). Remember, as we solve for \( v(t) \) we sum both equations into one to describe the equation.
\[ v_1(t) = 21.3e^{-1.042t} \, \text{V}\]
\[ v_2(t) = 14.2e^{-1.786t} \, \text{V}\]
\[ v(t) = v_1(t) + v_2(t) = 21.3e^{-1.042t} + 14.2e^{-1.786t} \, \text{V}\]
Problem 3 Solution.) \( v(t) = 2.61e^{-59.52t} \, \text{V} \) and \( i(t) = 1.24e^{-59.52t} \, \text{mA} \). Let's go through the steps.
1) Solve for \( V_0 \). After a long time, let's find the voltage applied across the capacitor. We can do current division between the two different branches, the 9.5k\( \Omega \) resistor and the sum of 7.5k, 1.2k, and 900 \( \Omega \) resistor.
\[ i_{right} = 2.5\cdot10^{-3}\cdot\frac{9500}{9500+7500+1200+900} = 1.24 \, \text{mA}\]
Then to find the voltage across the capacitor, we need to multiply that current by the sum of 1.2k and 900 \( \Omega \) resistor.
\[ V_0 = 1.24\cdot10^{-3}\cdot(1200+900) = 2.61 \, \text{V}\]
2) Solve for \( \tau \). Using equation 2,
\[ \tau = 8 \cdot 10^{-6}\cdot2100 = 0.0168\]
3) Use \( V_0 \) and \( \tau \) to solve for v(t). We get the following:
\[ v(t) = V_0e^{-t/\tau} = 2.61e^{-59.52t}\, \text{V}\]
Now let's solve for the current through the resistors. This can be done by dividing v(t) by the resistance. So,
\[ i(t) = \frac{v(t)}{R} = 1.24e^{-59.52t} \, \text{mA}\]