For more information, please see full course syllabus of AP Physics 1 & 2

For more information, please see full course syllabus of AP Physics 1 & 2

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### RC Circuits

- Capacitors store charge on their plates.
- The voltage across capacitors in parallel is the same. The charge stored on capacitors in series is the same.
- Parallel: Ceq=C1+C2+C3+...
- Series: 1/Ceq=1/C1+1/C2+1/C3+…
- Uncharged capacitors in a circuit act like wires. Charged capacitors in a circuit act like open circuits.

### RC Circuits

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Objectives
- Capacitors in Parallel
- Capacitors Store Charge on Their Plates
- Capacitors In Parallel Can Be Replaced with an Equivalent Capacitor
- Capacitors in Series
- Charge on Capacitors Must Be the Same
- Capacitor In Series Can Be Replaced With an Equivalent Capacitor
- RC Circuits
- Comprised of a Source of Potential Difference, a Resistor Network, and One or More Capacitors
- Uncharged Capacitors Act Like Wires
- Charged Capacitors Act Like Opens
- Charging an RC Circuit
- Discharging an RC Circuit
- Example 1: RC Analysis
- Example 2: More RC Analysis
- Example 3: Equivalent Capacitance
- Example 4: More Equivalent Capacitance

- Intro 0:00
- Objectives 0:08
- Capacitors in Parallel 0:34
- Capacitors Store Charge on Their Plates
- Capacitors In Parallel Can Be Replaced with an Equivalent Capacitor
- Capacitors in Series 2:42
- Charge on Capacitors Must Be the Same
- Capacitor In Series Can Be Replaced With an Equivalent Capacitor
- RC Circuits 5:40
- Comprised of a Source of Potential Difference, a Resistor Network, and One or More Capacitors
- Uncharged Capacitors Act Like Wires
- Charged Capacitors Act Like Opens
- Charging an RC Circuit 6:23
- Discharging an RC Circuit 11:36
- Example 1: RC Analysis 14:50
- Example 2: More RC Analysis 18:26
- Example 3: Equivalent Capacitance 21:19
- Example 4: More Equivalent Capacitance 22:48

### AP Physics 1 & 2 Exam Online Course

### Transcription: RC Circuits

*Hi everyone and I would like to welcome you back to Educator.com.*0000

*Our next lesson is going to be on RC circuits or circuits that have resistors and capacitors in them.*0003

*Our objectives are going to be to calculate equivalent capacitances for capacitors in both series and parallel configurations, describing how stored charges divided between capacitors in parallel, determining the ratio of voltages for capacitors in series...*0009

*...and calculating the voltage or stored charge under steady state conditions for a capacitor connected to a circuit consisting of a battery and resistor network, an RC circuit.*0023

*To start with, capacitors in parallel -- remember that capacitors store charge on their plates and therefore, they are storing electrical energy in their electrical fields.*0035

*Capacitors in parallel can be replaced with an equivalent capacitor, just like we can do that with resistors.*0046

*If we take a look -- capacitor is a symbol like that, two even lines and we will give it a positive and negative side and it has a potential difference across it with Q1 on one side and we would have -Q1 on the other side.*0051

*If we were to see this in a parallel configuration, we could draw one capacitor here and one capacitor over here where we have +/- voltage right here and we will call this C1 and that will be C2 and the voltage drop across both of them must be the same.*0071

*Since we know that C = Q/V, then Q = CV, so Q1 = C1V and Q2 = C2V.*0092

*If we want the equivalent capacitance, the equivalent capacitance is going to be the total charge (Q) divided by the total voltage (V) and our total charge is going to be Q1 + Q2, so that is going to be Q1 + Q2/V...*0108

*but Q1 = C1V, so that is going to be C1V and Q2 = C2V, so C2V/V and I can factor a V out, so that is V, C1 + C2/V or just C1 + C2.*0125

*In short, for capacitors in parallel to replace them with an equivalent capacitance, all you do is add the individual capacitances, so the equivalent capacitance formula is C1 + C2 and you just keep on adding them for however many capacitors you happen to have in that parallel configuration.*0141

*Capacitors in series -- The charge on the capacitors must be the same, so we can replace capacitors in series also with an equivalent capacitance.*0162

*But now our diagram is going to look a little bit different.*0170

*Here is our first capacitor, our second capacitor in series and we will call this the V1 across C1 and V2 across C2 and we will label this C1, C2...*0175

*...and over here we must have charge Q1+, which must mean we have Q1- over here and we have Q2+ and Q2-.*0191

*What is important to note is if we pay attention here, that part of the capacitor, those plates are isolated.*0202

*A charge cannot get on or off the plate, so if we have Q1- here, that must be equal to Q2+ over here.*0211

*The total charge is 0 in that region, therefore Q1 must equal Q2.*0218

*So to start our derivation, define how we get the formula for capacitors in series, the equivalent capacitance, C1 = Q/V1, therefore, V1 = Q/C1.*0226

*We can do the same thing for C2; C2 = Q2/V2, therefore, V2 = Q/C2.*0243

*Now the equivalent capacitance must be the charge divided by the total voltage, therefore the total voltage must be equal to (Q) over the equivalent capacitance.*0253

*The total voltage must be equal to V1 + V2, so if we write V-total = V1 + V2, I can replace V-total with Q/C-equivalent.*0269

*V1, we just found out is Q/C1 and V2 is Q/C2, so if I divide both sides by Q, what I then get is 1/C-equivalent, which is 1/C1 + 1/C2 and that will go on and work for as many capacitors as you happen to have in series as well.*0285

*Capacitors in series combine like resistors in parallel and capacitors in parallel, combine like resistors in series.*0310

*Just like we did for resistors -- for two capacitors in series, if you have just two, the equivalent capacitance is going to be C1 × C2/C1 + C2, but that only works for two capacitors.*0321

*If you have more than that, you have to go back to the general formula.*0335

*RC circuits are circuits comprised of a source of potential difference, a resistor network, and one or more capacitors.*0342

*We are going to look at RC circuits from the steady state perspective.*0348

*We are going to look at what happens when we first turn them on and we are going to look at them again after they have been on for a long, long time and what we call a long time is a special definition when we are talking about circuits.*0352

*The key to understanding RC circuit performance is that uncharged capacitors act like wires, while charged capacitors act like open circuits.*0364

*When it is fully charged, you can pretend it is open; when it is completely uncharged, it acts like a wire.*0374

*This is the key to our entire analysis coming up.*0379

*Charging an RC circuit -- let us first draw and RC circuit where we will start out with a battery with some terminal voltage, VT, we will put a resistor here (R)...*0384

*...we will add a capacitor to our circuit (capacitor C +/- for potential difference across it VC), and we will add a switch here.*0397

*Here is our switch and at time (t = 0), we are going to close the switch.*0412

*Now initially that is uncharged so (C) initially acts like a wire.*0419

*As it acts like a wire -- if we take a look at Kirchhoff's Voltage Law around the circuit going this way, when that is first closed and the capacitor is acting like a wire...*0423

*...KVL is going to tell us that we are going to see -VT first and if we define our current direction as that way, we will have -VT + IR + the voltage across the capacitor and all of that must equal 0 as we come back to our starting point.*0433

*But we also know that C = Q/VC, so VC, therefore must equal Q/C.*0455

*If we write this a little bit differently, we have -VT + IR + VC, which I am going to write as Q/C must equal 0.*0469

*But at time (t = 0), the charge on our capacitor is 0, so at t = 0 and Q = 0 -- if that is the case, that term is going to go to 0 and we get that -VT = IR or VT = IR.*0484

*What does that mean? That is the same as if there were no capacitor in the circuit.*0505

*Initially, you could pretend the capacitor is not there, it is just a wire.*0508

*After a long, long time however, now that is going to act like an open and in that case as we do KVL around our circuit, we see -VT + IR + VC = 0.*0513

*But if that is an open, then no current can flow, so that means I = 0, therefore, -VT + VC = 0 or VT = VC.*0534

*The voltage across the capacitor is the same as what you have across your battery; there is no current flowing.*0549

*Let us take a look at a couple of graphs of what is going on.*0555

*Let us start by looking at current flow.*0558

*I will make a graph of current versus time and initially we have a lot of current the moment that switch is closed.*0563

*We are going to start off here, but after a long, long time, current goes to 0, so we are going to have this approach 0.*0571

*We are going to say that at about 5 time constants (5 τ), that is about 99% of the way to its final value.*0579

*It is almost gotten to 0, but it has not quite reached it yet and what is that time constant τ?*0587

*τ is equal to the resistance of your circuit times the capacitance of your circuit.*0592

*So that will give you a time -- multiply that by 5 and by 5 times that, that is what we are talking about when we say a long, long, long time.*0599

*Usually it is really not that long in terms of time we are used to.*0605

*How about charge on our capacitor?*0611

*If we look at charge versus time, initially we have no charge on our capacitor and after a long, long time it is completely charged up, so we are going to have that go something like that and after a long, long time, what is the charge on our capacitor?*0615

*Well, if C = Q/V, then Q = CV, so this is going to be C × V after a long, long time, but after a long, long time, the voltage across the capacitor is V-terminal, so it is going to look like that.*0634

*Again, you are 99% of the way there to your final value right about the time you get to 5 τ or 5RC, whatever R and C happen to be for your circuit.*0647

*Let us also look at the voltage across our capacitor, VC versus time.*0659

*Initially, it acts like a wire, so the voltage across it must be 0 and after a long, long time, it works its way up until VC = VT, so we will have an asymptote here as well at VT and again this happens right around 5 τ or 5RC, which is where you get to 99% of that final value.*0670

*How about when you want to discharge an RC circuit?*0696

*Now we are going to pull that battery out of our circuit and we are going to have our resistor (R), we will start off with a charge capacitor, and we will have our switch.*0700

*So over here we have our capacitor (C), voltage across it (+/- VC) and we will define our current as going this way and again at t = 0 is when we are going to close our switch.*0718

*If we close that switch at t = 0, we know right away as we go around our circuit -- Kirchhoff's Voltage Law -- as we go around our circuit in this direction, a clockwise manner, I see +VC - IR and I get back to where I start, must equal 0.*0732

*Therefore, the voltage across our capacitor must equal IR.*0753

*That looks like what we get if this was a battery there instead, so initially our capacitor acts like a voltage source.*0759

*As time increases, as (t) gets bigger and bigger, as (t) approaches infinity, the charge on our capacitor is going to bleed off.*0773

*The charge on our capacitor is going to start to approach 0, therefore, the voltage across our capacitor is going to approach 0 and therefore the current in our circuit is going to approach 0.*0781

*If we made some graphs of what we have going on now, let us start with current (I) versus (t).*0794

*We start off with the full current and over time we are going to bleed off that charge, that current and approach 0 and guess how long that takes again -- about 5 τ until you get to 99% of the way to your final destination or here in this case, 0.*0803

*If we took a look at the charge on our capacitor versus time -- well initially it starts fully charged over here and over time that is going to bleed off again, so it is going to have the same basic shape as your current and it basically has 1% of the charge left after 5 τ (99%) of the way to its final destination.*0821

*Finally, let us look at the voltage across our capacitor, VC versus time, and you can probably already guess the shape of the graph.*0845

*We are going to start at the full voltage, which was VT when we were charging it and over time we are going to get to about 5 τ again as we go close to 0 -- 5RC and we are 99% of our way to that 0 level, so charging and discharging capacitors.*0857

*This is really straightforward when we are looking at them at the steady state when the circuits are first turned on, when they are uncharged, or when they are discharged, or fully charged.*0872

*The time varying element, what happens between 5RC, we are going to save that until we have a little bit of Calculus under our belt.*0881

*Let us take a look at our first example here.*0890

*What is the current through R2 here when the circuit is first connected and what is the current through R2 a long time after it has been connected?*0893

*Well, when it is first connected -- in that case, C1 acts like a wire, so I am going to draw the circuit as if C1 were a wire at that point to make it a little easier to analyze.*0901

*In that case, we are going to have a 20-volt battery, we will have up here R1 (200 ohms), then we are going to go to...*0918

*...we have R2 here, which is 400 ohms and we also have in parallel with it, R3 (300 ohms) and that will complete our circuit.*0932

*We will draw it as if the capacitor was not even there because it acts like it is not there when it is first turned on.*0944

*The way I could figure this out is, if we want to know the current through R2 -- that is going to be the current through that branch of the circuit right there.*0950

*If that is I2, well I = V/R and that is going to be 20 V over -- well what is our equivalent resistance going to be?*0961

*The equivalent resistance of these two in parallel -- 300 × 400/700 -- well that is just going to be an equivalent resistance of about 171 ohms.*0971

*My total resistance, if I treat this as an equivalent resistor in parallel, is 200 + 171 or 371 for a total current of 0.0539 A.*0984

*I2 then, is going to be V2/R2 and our V2 -- well if we have 0.0539 A going through 200 volts, what we are going to have over here is...*0998

*We have 0.0539 A through 200 volts and that is going to give us a voltage here of about 9.22 volts on this side.*1020

*We have dropped 10.78, so that is 9.22 volts/R2 (400 ohms) to give us a current of about 0.0231 A through that 400-ohm resistor, so there is the current through R2.*1030

*How about after the circuit has been connected a long, long, long time?*1054

*At that point C1 is no longer acting like a wire, it is acting like an open.*1058

*Let us go analyze that down in this bottom left region.*1063

*When that acts like an open, that means this whole branch has no current flowing through it.*1067

*We can ignore all of that.*1072

*That becomes a nice, simple series circuit, so the current flowing through R2 at that point is going to be the total current in the circuit, which is V/R, 20 volts over our equivalent resistance, for which we have two in series now, 200 + 400 or 600 ohms for a current of about 0.0333 A.*1074

*Simplify the circuit and figure out what you need to using what we have already learned about circuit analysis.*1099

*Let us do some more RC analysis with a little more practice.*1107

*What is the current through R3 when the circuit is first connected and what is the current through R2 a long time after it has been connected?*1110

*When it is first connected again, let us draw it that way.*1117

*When it is first connected, we have 10 volts, we have 100 ohms R1 up here and when it is first connected, that is going to act like a wire, so we have 200 ohms here, the capacitor acts as if it does not exist, so it is a wire.*1120

*And we have in parallel with that another 200 ohms and we want to know the current through R3 -- we will call that I3.*1140

*What I would do right there is replace these parallel resistors with one equivalent resistance.*1154

*That is going to be 100 ohms, so my total current is going to be -- let us redraw that again.*1161

*We have 10 volts then, from our power supply and we have 100 ohms and 100 ohms and a nice simple series circuit.*1171

*Well, it should be easy to see if that is 10 volts and we have two resistors right between them here, we must have a voltage of 5 volts and if we go back to this version, that means that we have 5 volts here and 5 volts here because anywhere on the wire must be the same.*1186

*So the current flow through I3, which is V/R3 is going to be the drop across I3, 5 volts/200 ohms, which implies then that I3 = 0.025 A.*1201

*What is the current through R2 a long time after the circuit has been connected?*1220

*Let us take a look there.*1228

*After it has been connected a long time, now this acts like an open and we can pretend that all of that does not exist.*1230

*We have a very simple circuit now of 10 volts, 100 ohms, 200 ohms, and that is all there is to it there.*1237

*If this is R3 and we want the current through R3, that is just going to be the total current because we have a series circuit.*1251

*So I = V/R, which is going to be 10 volts over our total resistance, 200 + 100, since they are in series, or 300 ohms for a total of 0.0333 A.*1257

*Let us try two more.*1275

*What is the equivalent capacitance of the capacitor network shown right here?*1279

*Well we have one in series, two in parallel, and one in series, so I am going to redraw this with our 5 microfarad capacitor.*1284

*Our two capacitors in parallel are very easy to calculate their equivalent capacitance because we just add them.*1296

*So 10 + 10 microfarads will be 20 microfarads and then we go to our third capacitor in series, 5 microfarads.*1302

*That is an equivalent network there, so let us find out the equivalent capacitance of this network.*1312

*One over the equivalent capacitance is going to be 1/5 × 10 ^{-6} (5 microfarads) + 1/20 × 10^{-6} (20 microfarads) + 1/5 × 10^{-6} again.*1318

*Or 1/C-equivalent = 450,000, therefore C-equivalent = 1/450,000 or C-equivalent = 2.22 × 10 ^{-6} farads, which is 2.22 microfarads.*1335

*One last problem to try here.*1366

*What is the equivalent capacitance of the capacitor network here shown below?*1370

*We have one lead here and one lead here.*1374

*This looks kind of complicated, so let us redraw it and see if we cannot simplify it.*1377

*These capacitors at 45 degree angles kind of trouble me, so if we redraw it, we could have -- let us call this C1, C2, C3, and C4.*1381

*We could redraw this as C1, and from C1 we go straight to C2 and then straight to C3 and then we come to the other end.*1392

*Now we also have C4, which if I drew it this way, probably looks a lot simpler.*1405

*What we really have here is we have three capacitors in series and all of those are in parallel with one, so let us replace these in series by one equivalent capacitance, an equivalent capacitance for C1, 2, and 3.*1418

*To do that I am going to say 1/C-equivalent is 1/C + 1/C + 1/C or 1/C-equivalent and that is going to be 3/C, therefore, C-equivalent must be C/3.*1431

*Now I could redraw this as one capacitor (C/3) in parallel with our other capacitor, which we still have C4 or just C.*1446

*Capacitors in parallel -- well all you have to do is just add them to get their equivalent capacitance, so our total equivalent capacitance is going to be C/3 + C or 4/3 C.*1463

*Hopefully that gets you a good start on RC circuits.*1479

*Thanks so much for your time. Make it a great day everyone!*1483

1 answer

Last reply by: Professor Dan Fullerton

Wed Jun 21, 2017 7:36 AM

Post by Tope Adedolapo on June 21, 2017

Wouldn't it be easier to do the VIRP table during these examples?

1 answer

Last reply by: Professor Dan Fullerton

Sun Mar 6, 2016 5:02 PM

Post by Sarmad Khokhar on March 6, 2016

In example 4 how did you figured out that C4 is in parallel with other capacitors while other three capacitors are in series.

2 answers

Last reply by: Professor Dan Fullerton

Thu Aug 20, 2015 8:07 AM

Post by Anh Dang on August 19, 2015

In example 1, could you explain in detail how you got to 9.22.

1 answer

Last reply by: Professor Dan Fullerton

Thu Mar 27, 2014 3:48 PM

Post by Milan Ray on March 27, 2014

Yes.. where did the 5v come out from on example 2, you said it is easy to know, but I don't know...so could you please explain... :)

0 answers

Post by ibrahim shawi on March 4, 2014

for example two where did the 5v come from?

3 answers

Last reply by: Professor Dan Fullerton

Wed Mar 5, 2014 5:41 AM

Post by ibrahim shawi on March 4, 2014

im confused on how we came about to the 9.22 please help,, thank you.

1 answer

Last reply by: Professor Dan Fullerton

Wed May 8, 2013 6:14 AM

Post by Nawaphan Jedjomnongkit on May 8, 2013

in example 2 the question ask about current through R2 after connect long time so that should be 0 right?

1 answer

Last reply by: Professor Dan Fullerton

Mon Apr 15, 2013 2:05 PM

Post by Walid Shreim on April 15, 2013

For problem solving in the AP, is it acceptable to use the VIRP table? More precisely, is the VIRP table enough work (show work) for a student to grant a full credit on the free response questions?

Many thanks for the great explanation.