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CLAIMED BY MARK RUSSELL SPRING 2016
Claimed by Adam Teske Fall 2025


An RC circuit (resistor-capacitor circuit) is a type of [[Non Steady State|non steady state]] circuit.
[[File:Rc_circuit.JPG|400px|right|]]
==The Main Idea==
==The Main Idea==


The figure below shows a capacitor, ( C ) in series with a resistor, ( R ) forming a RC Charging Circuit connected across a DC battery supply ( Vs ) via a mechanical switch. When the switch is closed, the capacitor will gradually charge up through the resistor until the voltage across it reaches the supply voltage of the battery. The manner in which the capacitor charges up is also shown below.
The number of independent energy storage elements in the RC circuit determine the circuit's order, for this course first order RC circuits are the most prevalent. First order RC circuits can have series and parallel elements. The figure in the top right of the page depicts an RC circuit.  
[[File:Rc_circuit.JPG|300px|thumb|left|]]


Let us assume left, that the capacitor, C is fully “discharged” and the switch (S) is fully open. These are the initial conditions of the circuit, then t = 0, i = 0 and q = 0. When the switch is closed the time begins at t = 0 and current begins to flow into the capacitor via the resistor.
RC Circuits contain charging and discharging phases. For charging phases, a voltage source is connected to the capacitor, with capacitance C and max charge of Q. When the switch is disengaged, the capacitor discharges its stored energy to the resistors with resistance R.
 
The time it takes for the capacitor to charge and discharge can be represented by the circuit's time constant, τ.
 
[[File:Rc_switch.JPG|400px|center|]]
 
Remember that potential difference across the capacitor is delta <math>{V} = {Q}/{C}</math>, where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at <math> {time} = {0}</math>. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of <math>{I}_{0} = {emf}/{R}</math> to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time <math>{t} = {0}</math>. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.
 
Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is <math>{V} = {emf(1-e^{\frac{-t}{RC}})}</math>. <math>{V}</math> is defined as the voltage across the capacitor. <math>{emf}</math> is equal to the emf of the DC voltage source. The units of <math>{RC}</math> are in seconds. <math>{τ} = {RC}</math>. <math>{τ}</math> is the constant of time in the RC circuit.
 
The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance (<math>{I} = {V/R}</math>). Also the smaller the capacitor (C), the less time it will need to charge. <math>{τ} = {RC}</math> explains both of these.
 
[[File:images.png|100px|left|]]
Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.
 
 
[[File:images2.jpeg|100px|right|]]
Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.  


Since the initial voltage across the capacitor is zero, ( Vc = 0 ) the capacitor appears to be a short circuit to the external circuit and the maximum current flows through the circuit restricted only by the resistor R.


As the capacitor charges up, the potential difference across its plates slowly increases with the actual time taken for the charge on the capacitor to reach its maximum possible voltage. The maximum voltage the capacitor reaches is equal to the voltage supply of the battery. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.
===A Mathematical Model===
===A Mathematical Model===
These three equations are helpful in solving and understanding RC circuit problems
[[File:Rc123.jpg]]
To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:
[[File:Rc234.jpg]]
The graph of Voltage vs. Time in a RC Circuit is given by:
[[File:Outputrc.jpg]]
===A Computational Model===
Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.
For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)
[[File:rcvoltage.gif|300px|center|]]
For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)


ΔV = I * R
[[File:rccurrent.gif|300px|center|]]


Q=CV
== Analyzing a Switched Circuit with One Capacitor ==


[[File:images.png|100px|thumb|left|]]
The basic series RC circuit is a useful starting point, but many circuit problems have several resistors and a switch. A reliable way to analyze any linear circuit with one capacitor is to replace the rest of the circuit, as seen from the capacitor’s terminals, with its Thevenin equivalent. The resulting circuit has one voltage source, one resistance, and the capacitor.
Node Rule-Current is assigned (positive or negative) quantity reflecting direction towards or away from a node.  
This principle can be solved by adding all the currents up to equal 0. The Node Rule is important to RC Circuits
because finding the current flow in the different states of an RC Circuit often relies on nodes when the  
capacitor is in parallel.
[[File:images2.jpeg|100px|thumb|right|]]


This method works when the circuit contains one capacitor and resistors with constant values. It also works if several capacitors can first be combined into one equivalent capacitor. If there are multiple independent capacitors, the circuit may need more than one differential equation.


Loop Rule-Similar to the node rule; however, the voltage jumps and drops are added up around the circuit to equal to 0.
=== The General Voltage Equation ===


Suppose a switch changes position at time <math>t_0</math>. Define:


* <math>V_i</math> as the voltage across the capacitor immediately before the switch changes.
* <math>V_f</math> as the final, steady-state capacitor voltage after the switch changes.
* <math>R_{\text{th}}</math> as the Thevenin resistance seen from the capacitor’s terminals after the switch changes.
* <math>C</math> as the capacitance.


For times after the switch changes, the capacitor voltage is
[math]\displaystyle{ V_C(t)=V_f+(V_i-V_f)e^{-(t-t_0)/(R_{\text{th}}C)} }[/math]
The time constant is
[math]\displaystyle{ \tau=R_{\text{th}}C }[/math]
This equation describes both charging and discharging. It also handles a capacitor that starts with a nonzero voltage and moves toward a different final voltage.
The capacitor’s voltage cannot change instantly. Therefore,
[math]\displaystyle{ V_C(t_0^+)=V_C(t_0^-)=V_i }[/math]
The current can change instantly when the switch changes. If positive current is defined as entering the capacitor terminal labeled positive, then
[math]\displaystyle{ I_C(t)=C\frac{dV_C}{dt}=\frac{V_f-V_i}{R_{\text{th}}}e^{-(t-t_0)/\tau} }[/math]
A negative value means the actual current flows opposite to the chosen positive direction.
=== How to Find the Thevenin Equivalent ===
1. Draw the circuit as it is connected after the switch changes. The initial voltage <math>V_i</math> comes from the circuit before the switch changes; <math>R_{\text{th}}</math> and <math>V_f</math> come from the circuit afterward.
2. Remove the capacitor and find the open-circuit voltage across its terminals. This is <math>V_f</math>, the voltage the capacitor approaches after a long time.
3. Find <math>R_{\text{th}}</math> looking into the capacitor’s terminals. Turn off independent ideal voltage sources by replacing them with wires, and turn off independent ideal current sources by opening them. Leave dependent sources active.
4. Calculate <math>\tau=R_{\text{th}}C</math>.
5. Substitute <math>V_i</math>, <math>V_f</math>, and <math>\tau</math> into the general voltage equation.
6. Use <math>Q=CV_C</math> or <math>I_C=C\,dV_C/dt</math> if the problem asks for charge or current.
At one time constant, the voltage has moved about 63.2% of the way from its initial value toward its final value. At three time constants, it has moved about 95.0% of the way. At five time constants, it is more than 99% of the way there.
=== Worked Example: A Capacitor Connected to a Voltage Divider ===
A <math>12\ \text{V}</math> ideal battery is connected through a switch and a <math>2.0\ \text{k}\Omega</math> resistor, <math>R_1</math>, to a node. A <math>4.0\ \text{k}\Omega</math> resistor, <math>R_2</math>, and a <math>1000\ \mu\text{F}</math> capacitor are both connected from that node to ground. Before the switch closes, the capacitor is disconnected from the battery and has voltage <math>0\ \text{V}</math>. At <math>t=0</math>, the switch closes.
Find the capacitor’s final voltage, the time constant, the voltage after one time constant, and its initial charging current.
;Step 1: Find the initial voltage.
The capacitor begins uncharged, so <math>V_i=0\ \text{V}</math>.
;Step 2: Find the final voltage.
After a long time, the capacitor current is zero. The resistors form a voltage divider, so
[math]\displaystyle{ V_f=12\ \text{V}\left(\frac{4.0\ \text{k}\Omega}{2.0\ \text{k}\Omega+4.0\ \text{k}\Omega}\right)=8.0\ \text{V} }[/math]
;Step 3: Find the Thevenin resistance.
Turn off the ideal battery by replacing it with a wire. From the capacitor’s terminals, <math>R_1</math> and <math>R_2</math> are in parallel:
[math]\displaystyle{ R_{\text{th}}=R_1\parallel R_2=\frac{R_1R_2}{R_1+R_2}=\frac{(2.0)(4.0)}{2.0+4.0}\ \text{k}\Omega=1.33\ \text{k}\Omega }[/math]
;Step 4: Find the time constant.
Since <math>1000\ \mu\text{F}=1.00\times10^{-3}\ \text{F}</math>,
[math]\displaystyle{ \tau=R_{\text{th}}C=(1.33\times10^3\ \Omega)(1.00\times10^{-3}\ \text{F})=1.33\ \text{s} }[/math]
;Step 5: Write the voltage as a function of time.
For <math>t\geq0</math>,
[math]\displaystyle{ V_C(t)=8.0\left(1-e^{-t/(1.33\ \text{s})}\right)\ \text{V} }[/math]
At one time constant,
[math]\displaystyle{ V_C(\tau)=8.0(1-e^{-1})\ \text{V}\approx5.06\ \text{V} }[/math]
;Step 6: Find the initial current into the capacitor.
Using the current direction into the capacitor’s positive terminal,
[math]\displaystyle{ I_C(0^+)=\frac{V_f-V_i}{R_{\text{th}}}=\frac{8.0\ \text{V}}{1.33\times10^3\ \Omega}\approx6.0\ \text{mA} }[/math]
The current decreases as the capacitor charges. After a long time, the capacitor current approaches zero, even though current still flows through the two resistors in the voltage divider.
=== What to Check in a Circuit Sketch ===
* Mark the capacitor’s positive terminal and use the same voltage direction throughout the calculation.
* Use the circuit before switching to determine <math>V_i</math>. Capacitor voltage is continuous at the instant the switch changes.
* Use the circuit after switching to find <math>V_f</math> and <math>R_{\text{th}}</math>.
* At steady state in a DC circuit, an ideal capacitor carries no current. It acts like an open circuit after it has finished charging.
* The resistance in <math>\tau=R_{\text{th}}C</math> is the resistance seen from the capacitor’s terminals. It may combine several resistors and is not always the resistor drawn next to the capacitor.
* Use seconds for time, ohms for resistance, and farads for capacitance. Since <math>\Omega\cdot\text{F}=\text{s}</math>, the time constant has units of seconds.
=== Practice Check ===
A capacitor initially has voltage <math>6.0\ \text{V}</math>. After a switch changes position, the Thevenin equivalent seen by the capacitor is <math>10\ \text{V}</math> in series with <math>3.0\ \text{k}\Omega</math>. The capacitance is <math>200\ \mu\text{F}</math>.
# Find the time constant.
# Find the capacitor voltage after one time constant.
# Find the initial current into the positive terminal.
;Answers:
[math]\displaystyle{ \tau=(3.0\times10^3\ \Omega)(200\times10^{-6}\ \text{F})=0.60\ \text{s} }[/math]
[math]\displaystyle{ V_C(\tau)=10+(6.0-10)e^{-1}\approx8.53\ \text{V} }[/math]
[math]\displaystyle{ I_C(0^+)=\frac{10-6.0}{3.0\times10^3}\approx1.33\ \text{mA} }[/math]
=== Further Reading ===
OpenStax, ''University Physics Volume 2'', Section 10.5, “RC Circuits”: https://openstax.org/books/university-physics-volume-2/pages/10-5-rc-circuits


==Examples==
==Examples==


===Simple===
===Simple===
[[File:RCcircuitSimple.gif|100px|thumb|left|]]
In a circuit where time t = RC, the factor <math>{e^{\frac{-t}{RC}}}</math> has fallen from value <math>{e^0} = {1}</math> to the value:  
In the steady state what is the potential difference across the plates of the capacitor? The circuit has a resistor, R, capacitor, C, and battery, Vs.
 
e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37
 
Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.  


Solution:


<math> {RC}= {12*1} = {12 seconds}</math>


===Middling===
Using the information from the problem above:


Show that the power dissipated in the bulb at t=RC is only 14% of  the original power?


Solution:


Using <math>{IΔV} = {RI^2}</math>, reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369


Solution: The key concept in this problem is that in the steady state time is approaching infinity. In an RC circuit as time approaches infinity, the current is 0 because the potential difference across the capacitor equals that of the battery. Therefore, the potential difference across the plates of the capacitor equals Vs.
===Difficult===
===Difficult===
The circuit below has been in position a for a long time. At time t = 0 the switch is thrown to position b. DATA: Vb = 12 V, C = 10 mF, R = 20 W
The best example of difficult RC circuit problems come from past exams. Take a look at this RC circuit.
[[File:rccircuit.gif|200px|thumb|left|]]
[[File:Primary1.jpg]]




a.) What is the current through the resistor just BEFORE the switch is thrown?
The first step in these problems is to identify all the parts on the circuit and draw in nodes at necessary locations. You want to choose spots for the nodes where the wire splits so you can apply Kirchoff's law and relate the current at these locations.
[[File:Primary2.jpg]]


The circuit is in a steady state so I = 0


b.) What is the current through the resistor just AFTER the switch is thrown?
Once you correctly label the diagram you can begin the problem.


Solution: I = V/R I = 0.6 amps since the capacitor has a potential difference and is creating the current.
1. Write down the three energy conservation equations for the circuit. Also write down the charge conservation equation for the circuit you would need to determine the current through the bulbs and capacitor. You do not need to solve these equations.


c.) What is the charge across the capacitor just BEFORE the switch is thrown?
2.Determine the initial current passing through each bulb.


Solution: Q = CV
3. Determine the current passing through each bulb for a long time after the switch has been closed.
Q = 120 mC


d.) What is the charge on the capacitor just AFTER the switch is thrown?
[[File:Workwork.jpg]]


Solution: Charge does not change instantaneously
Q = 120mC
==Connectedness==
==Connectedness==
#How is this topic connected to something that you are interested in?
1. How is this topic connected to something that you are interested in?
:: RC Circuits can be used in musical instruments such as guitars and amplifiers. I play the guitar so it is cool to see how the guitar actually works.
:: I enjoy listening to music and RC circuits are used in speakers and headphones. They allow the current to be regulated and for the music to to played at the right pitch and volume.
#How is it connected to your major?
[[File:Speakercirc.jpg]]
:: Mechanical Engineering requires the use of many electronics and learning how various circuits work is necessary for creating new parts and machines.
 
#Is there an interesting industrial application?
2. How is it connected to your major?
RC circuits can be used in many industrial areas to allow certain frequencies to pass through a circuit.
:: I'm a Mechanical Engineering major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled. I would be designing the machines not wiring the curcuits, but it is important to understand how the RC circuits are used in order to properly design and implement the machines.
3. Is there an interesting industrial application?
:: RC circuits are used as filters for almost all types of electronics. The initial input is taken across the RC circuit and the output is taken across the capacitor. This allows the capacitor in the circuit to be charged to different values and used when the desired one is obtained. It is an easy way to allow multiple energy outputs across a single circuit rather than using a bunch of different circuits.
[[File:Fil42.gif]]
 
==History==
 
The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists. Once his work was recognized and accepted by the scientific community people began to realize the many applications of RC circuits as filters and began to further research and implement them in their own work.


== See also ==
== See also ==
Line 91: Line 241:


http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/
http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/
http://buphy.bu.edu/~duffy/semester2/c11_RC.html
http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf


==References==
==References==


http://www.electronics-tutorials.ws/rc/rc_1.html
http://www.compadre.org/portal/items/detail.cfm?ID=9986


https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html
https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html
https://openstaxcollege.org/textbooks/college-physics
http://www.animations.physics.unsw.edu.au/jw/RCfilters.html


[[Category:Simple Circuits]]
[[Category:Simple Circuits]]

Latest revision as of 08:57, 26 October 2026

Claimed by Adam Teske Fall 2025

An RC circuit (resistor-capacitor circuit) is a type of non steady state circuit.

The Main Idea

The number of independent energy storage elements in the RC circuit determine the circuit's order, for this course first order RC circuits are the most prevalent. First order RC circuits can have series and parallel elements. The figure in the top right of the page depicts an RC circuit.

RC Circuits contain charging and discharging phases. For charging phases, a voltage source is connected to the capacitor, with capacitance C and max charge of Q. When the switch is disengaged, the capacitor discharges its stored energy to the resistors with resistance R.

The time it takes for the capacitor to charge and discharge can be represented by the circuit's time constant, τ.

Remember that potential difference across the capacitor is delta [math]\displaystyle{ {V} = {Q}/{C} }[/math], where Q is charge on the plate and C is the capacitance. When the switch is closed, voltage on the capacitor rises rapidly at first, due to the high current at [math]\displaystyle{ {time} = {0} }[/math]. The voltage opposes the battery, increasing from zero to the max emf when the capacitor is fully charged. The current decreases from its initial value of [math]\displaystyle{ {I}_{0} = {emf}/{R} }[/math] to zero as the voltage on the capacitor reaches the same value as the emf. The current is initially at its max at time [math]\displaystyle{ {t} = {0} }[/math]. Once the potential difference across the plates of the capacitor equals the battery's voltage supply, current will stop flowing through the circuit. This is known as the steady state of an RC circuit; it is reached when time goes to infinity.

Using derived calculus, the equation for voltage versus time when the capacitor is charged through resistor R is [math]\displaystyle{ {V} = {emf(1-e^{\frac{-t}{RC}})} }[/math]. [math]\displaystyle{ {V} }[/math] is defined as the voltage across the capacitor. [math]\displaystyle{ {emf} }[/math] is equal to the emf of the DC voltage source. The units of [math]\displaystyle{ {RC} }[/math] are in seconds. [math]\displaystyle{ {τ} = {RC} }[/math]. [math]\displaystyle{ {τ} }[/math] is the constant of time in the RC circuit.

The smaller the resistance, the faster a capacitor will be charged. It takes longer to charge than to discharge. This is because a larger current flows through a smaller resistance ([math]\displaystyle{ {I} = {V/R} }[/math]). Also the smaller the capacitor (C), the less time it will need to charge. [math]\displaystyle{ {τ} = {RC} }[/math] explains both of these.

Kirchoff's Node Rule is important to RC Circuits because finding the current flow in the different states of an RC Circuit often relies on nodes when the capacitor is in a parallel circuit.


Kirchhoff’s loop rule explains that the sum of changes in potential around any closed loop must be zero.


A Mathematical Model

These three equations are helpful in solving and understanding RC circuit problems


To find the equation for voltage versus time when charging a capacitor through a resistor, you start with rearranging the energy equation and solving for I:


The graph of Voltage vs. Time in a RC Circuit is given by:

A Computational Model

Using a simulation where R and C were equal, the capacitors were charged and then discharged and the two images show the voltage and current of this versus time.

For the gif below, the capacitor is charged and then discharged. This shows the voltage over time. The red line is the voltage and the gray line is the emf. (May have to click image to see gif)


For the gif below, the capacitor is charged and then discharged. This shows the current over time. The blue line is the current. (May have to click image to see gif)

Analyzing a Switched Circuit with One Capacitor

The basic series RC circuit is a useful starting point, but many circuit problems have several resistors and a switch. A reliable way to analyze any linear circuit with one capacitor is to replace the rest of the circuit, as seen from the capacitor’s terminals, with its Thevenin equivalent. The resulting circuit has one voltage source, one resistance, and the capacitor.

This method works when the circuit contains one capacitor and resistors with constant values. It also works if several capacitors can first be combined into one equivalent capacitor. If there are multiple independent capacitors, the circuit may need more than one differential equation.

The General Voltage Equation

Suppose a switch changes position at time [math]\displaystyle{ t_0 }[/math]. Define:

  • [math]\displaystyle{ V_i }[/math] as the voltage across the capacitor immediately before the switch changes.
  • [math]\displaystyle{ V_f }[/math] as the final, steady-state capacitor voltage after the switch changes.
  • [math]\displaystyle{ R_{\text{th}} }[/math] as the Thevenin resistance seen from the capacitor’s terminals after the switch changes.
  • [math]\displaystyle{ C }[/math] as the capacitance.

For times after the switch changes, the capacitor voltage is

[math]\displaystyle{ V_C(t)=V_f+(V_i-V_f)e^{-(t-t_0)/(R_{\text{th}}C)} }[/math]

The time constant is

[math]\displaystyle{ \tau=R_{\text{th}}C }[/math]

This equation describes both charging and discharging. It also handles a capacitor that starts with a nonzero voltage and moves toward a different final voltage.

The capacitor’s voltage cannot change instantly. Therefore,

[math]\displaystyle{ V_C(t_0^+)=V_C(t_0^-)=V_i }[/math]

The current can change instantly when the switch changes. If positive current is defined as entering the capacitor terminal labeled positive, then

[math]\displaystyle{ I_C(t)=C\frac{dV_C}{dt}=\frac{V_f-V_i}{R_{\text{th}}}e^{-(t-t_0)/\tau} }[/math]

A negative value means the actual current flows opposite to the chosen positive direction.

How to Find the Thevenin Equivalent

1. Draw the circuit as it is connected after the switch changes. The initial voltage [math]\displaystyle{ V_i }[/math] comes from the circuit before the switch changes; [math]\displaystyle{ R_{\text{th}} }[/math] and [math]\displaystyle{ V_f }[/math] come from the circuit afterward. 2. Remove the capacitor and find the open-circuit voltage across its terminals. This is [math]\displaystyle{ V_f }[/math], the voltage the capacitor approaches after a long time. 3. Find [math]\displaystyle{ R_{\text{th}} }[/math] looking into the capacitor’s terminals. Turn off independent ideal voltage sources by replacing them with wires, and turn off independent ideal current sources by opening them. Leave dependent sources active. 4. Calculate [math]\displaystyle{ \tau=R_{\text{th}}C }[/math]. 5. Substitute [math]\displaystyle{ V_i }[/math], [math]\displaystyle{ V_f }[/math], and [math]\displaystyle{ \tau }[/math] into the general voltage equation. 6. Use [math]\displaystyle{ Q=CV_C }[/math] or [math]\displaystyle{ I_C=C\,dV_C/dt }[/math] if the problem asks for charge or current.

At one time constant, the voltage has moved about 63.2% of the way from its initial value toward its final value. At three time constants, it has moved about 95.0% of the way. At five time constants, it is more than 99% of the way there.

Worked Example: A Capacitor Connected to a Voltage Divider

A [math]\displaystyle{ 12\ \text{V} }[/math] ideal battery is connected through a switch and a [math]\displaystyle{ 2.0\ \text{k}\Omega }[/math] resistor, [math]\displaystyle{ R_1 }[/math], to a node. A [math]\displaystyle{ 4.0\ \text{k}\Omega }[/math] resistor, [math]\displaystyle{ R_2 }[/math], and a [math]\displaystyle{ 1000\ \mu\text{F} }[/math] capacitor are both connected from that node to ground. Before the switch closes, the capacitor is disconnected from the battery and has voltage [math]\displaystyle{ 0\ \text{V} }[/math]. At [math]\displaystyle{ t=0 }[/math], the switch closes.

Find the capacitor’s final voltage, the time constant, the voltage after one time constant, and its initial charging current.

Step 1
Find the initial voltage.

The capacitor begins uncharged, so [math]\displaystyle{ V_i=0\ \text{V} }[/math].

Step 2
Find the final voltage.

After a long time, the capacitor current is zero. The resistors form a voltage divider, so

[math]\displaystyle{ V_f=12\ \text{V}\left(\frac{4.0\ \text{k}\Omega}{2.0\ \text{k}\Omega+4.0\ \text{k}\Omega}\right)=8.0\ \text{V} }[/math]

Step 3
Find the Thevenin resistance.

Turn off the ideal battery by replacing it with a wire. From the capacitor’s terminals, [math]\displaystyle{ R_1 }[/math] and [math]\displaystyle{ R_2 }[/math] are in parallel:

[math]\displaystyle{ R_{\text{th}}=R_1\parallel R_2=\frac{R_1R_2}{R_1+R_2}=\frac{(2.0)(4.0)}{2.0+4.0}\ \text{k}\Omega=1.33\ \text{k}\Omega }[/math]

Step 4
Find the time constant.

Since [math]\displaystyle{ 1000\ \mu\text{F}=1.00\times10^{-3}\ \text{F} }[/math],

[math]\displaystyle{ \tau=R_{\text{th}}C=(1.33\times10^3\ \Omega)(1.00\times10^{-3}\ \text{F})=1.33\ \text{s} }[/math]

Step 5
Write the voltage as a function of time.

For [math]\displaystyle{ t\geq0 }[/math],

[math]\displaystyle{ V_C(t)=8.0\left(1-e^{-t/(1.33\ \text{s})}\right)\ \text{V} }[/math]

At one time constant,

[math]\displaystyle{ V_C(\tau)=8.0(1-e^{-1})\ \text{V}\approx5.06\ \text{V} }[/math]

Step 6
Find the initial current into the capacitor.

Using the current direction into the capacitor’s positive terminal,

[math]\displaystyle{ I_C(0^+)=\frac{V_f-V_i}{R_{\text{th}}}=\frac{8.0\ \text{V}}{1.33\times10^3\ \Omega}\approx6.0\ \text{mA} }[/math]

The current decreases as the capacitor charges. After a long time, the capacitor current approaches zero, even though current still flows through the two resistors in the voltage divider.

What to Check in a Circuit Sketch

  • Mark the capacitor’s positive terminal and use the same voltage direction throughout the calculation.
  • Use the circuit before switching to determine [math]\displaystyle{ V_i }[/math]. Capacitor voltage is continuous at the instant the switch changes.
  • Use the circuit after switching to find [math]\displaystyle{ V_f }[/math] and [math]\displaystyle{ R_{\text{th}} }[/math].
  • At steady state in a DC circuit, an ideal capacitor carries no current. It acts like an open circuit after it has finished charging.
  • The resistance in [math]\displaystyle{ \tau=R_{\text{th}}C }[/math] is the resistance seen from the capacitor’s terminals. It may combine several resistors and is not always the resistor drawn next to the capacitor.
  • Use seconds for time, ohms for resistance, and farads for capacitance. Since [math]\displaystyle{ \Omega\cdot\text{F}=\text{s} }[/math], the time constant has units of seconds.

Practice Check

A capacitor initially has voltage [math]\displaystyle{ 6.0\ \text{V} }[/math]. After a switch changes position, the Thevenin equivalent seen by the capacitor is [math]\displaystyle{ 10\ \text{V} }[/math] in series with [math]\displaystyle{ 3.0\ \text{k}\Omega }[/math]. The capacitance is [math]\displaystyle{ 200\ \mu\text{F} }[/math].

  1. Find the time constant.
  2. Find the capacitor voltage after one time constant.
  3. Find the initial current into the positive terminal.
Answers

[math]\displaystyle{ \tau=(3.0\times10^3\ \Omega)(200\times10^{-6}\ \text{F})=0.60\ \text{s} }[/math]

[math]\displaystyle{ V_C(\tau)=10+(6.0-10)e^{-1}\approx8.53\ \text{V} }[/math]

[math]\displaystyle{ I_C(0^+)=\frac{10-6.0}{3.0\times10^3}\approx1.33\ \text{mA} }[/math]

Further Reading

OpenStax, University Physics Volume 2, Section 10.5, “RC Circuits”: https://openstax.org/books/university-physics-volume-2/pages/10-5-rc-circuits

Examples

Simple

In a circuit where time t = RC, the factor [math]\displaystyle{ {e^{\frac{-t}{RC}}} }[/math] has fallen from value [math]\displaystyle{ {e^0} = {1} }[/math] to the value:

e^(-t/RC) = e^1 = 1/e = 1/2.718 = 0.37

Calculate the time constant for the RC circuit with R = 12 ohms and C = 1 farad.

Solution:

[math]\displaystyle{ {RC}= {12*1} = {12 seconds} }[/math]

Middling

Using the information from the problem above:

Show that the power dissipated in the bulb at t=RC is only 14% of the original power?

Solution:

Using [math]\displaystyle{ {IΔV} = {RI^2} }[/math], reduction in current by factor of 0.37 gives a reduction in power by a factor of (0.37)^2 = 0.1369

Difficult

The best example of difficult RC circuit problems come from past exams. Take a look at this RC circuit.


The first step in these problems is to identify all the parts on the circuit and draw in nodes at necessary locations. You want to choose spots for the nodes where the wire splits so you can apply Kirchoff's law and relate the current at these locations.


Once you correctly label the diagram you can begin the problem.

1. Write down the three energy conservation equations for the circuit. Also write down the charge conservation equation for the circuit you would need to determine the current through the bulbs and capacitor. You do not need to solve these equations.

2.Determine the initial current passing through each bulb.

3. Determine the current passing through each bulb for a long time after the switch has been closed.

Connectedness

1. How is this topic connected to something that you are interested in?

I enjoy listening to music and RC circuits are used in speakers and headphones. They allow the current to be regulated and for the music to to played at the right pitch and volume.

2. How is it connected to your major?

I'm a Mechanical Engineering major and RC currents are directly related to every job, but in warehouses where heavy machinery and appliances are involved, RC currents are used in parts either at the warehouse or ones that are being assembled. I would be designing the machines not wiring the curcuits, but it is important to understand how the RC circuits are used in order to properly design and implement the machines.

3. Is there an interesting industrial application?

RC circuits are used as filters for almost all types of electronics. The initial input is taken across the RC circuit and the output is taken across the capacitor. This allows the capacitor in the circuit to be charged to different values and used when the desired one is obtained. It is an easy way to allow multiple energy outputs across a single circuit rather than using a bunch of different circuits.

History

The RC circuits have been in use for a long time. Georg Simon Ohm, was someone who spent alot of time researching RC Circuits and Ohm's Law was founded by him. He did his work from 1830s-1840s and he received recognition, after a long time of being mocked by other scientists. Once his work was recognized and accepted by the scientific community people began to realize the many applications of RC circuits as filters and began to further research and implement them in their own work.

See also

Further reading

Charge in a RC Circuit

Current in a RC circuit

Loop Rule

Node Rule

Current

External links

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/rcimp.html

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-4/series-resistor-capacitor-circuits/

http://buphy.bu.edu/~duffy/semester2/c11_RC.html

http://ocw.mit.edu/courses/physics/8-022-physics-ii-electricity-and-magnetism-fall-2006/lecture-notes/lecture32.pdf

References

http://www.compadre.org/portal/items/detail.cfm?ID=9986

https://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html

https://openstaxcollege.org/textbooks/college-physics

http://www.animations.physics.unsw.edu.au/jw/RCfilters.html

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