For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

### RC Circuits: Transient Analysis

- The time constant in an RC circuit, tau, is equal to the circuit’s resistance multiplied by its capacitance.
- The time constant indicates the time at which the quantity under observation has achieved 63% of its final value.
- By 5 time constants, the quantity under observation is within one percent of its final value.

### RC Circuits: Transient Analysis

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 0:00
- Objectives 0:13
- Charging an RC Circuit 1:11
- Basic RC Circuit
- Graph of Current Circuit
- Graph of Charge
- Graph of Voltage
- Mathematically Describe the Charts
- Discharging an RC Circuit 13:29
- Graph of Current
- Graph of Charge
- Graph of Voltage
- Mathematically Describe the Charts
- The Time Constant 20:03
- Time Constant
- By 5 Time Constant
- Example 1 20:39
- Example 2 28:53
- Example 3 27:02
- Example 4 44:29
- Example 5 55:24

### AP Physics C: Electricity and Magnetism Online Course

I. Electricity | ||
---|---|---|

Electric Charge & Coulomb's Law | 30:48 | |

Electric Fields | 1:19:22 | |

Gauss's Law | 52:53 | |

Electric Potential & Electric Potential Energy | 1:14:03 | |

Electric Potential Due to Continuous Charge Distributions | 1:01:28 | |

Conductors | 20:35 | |

Capacitors | 41:23 | |

II. Current Electricity | ||

Current & Resistance | 17:59 | |

Circuits I: Series Circuits | 29:08 | |

Circuits II: Parallel Circuits | 39:09 | |

RC Circuits: Steady State | 34:03 | |

RC Circuits: Transient Analysis | 1:01:07 | |

III. Magnetism | ||

Magnets | 8:38 | |

Moving Charges In Magnetic Fields | 29:07 | |

Forces on Current-Carrying Wires | 17:52 | |

Magnetic Fields Due to Current-Carrying Wires | 24:43 | |

The Biot-Savart Law | 21:50 | |

Ampere's Law | 26:31 | |

Magnetic Flux | 7:24 | |

Faraday's Law & Lenz's Law | 1:04:33 | |

IV. Inductance, RL Circuits, and LC Circuits | ||

Inductance | 6:41 | |

RL Circuits | 42:17 | |

LC Circuits | 9:47 | |

V. Maxwell's Equations | ||

Maxwell's Equations | 3:38 | |

VI. Sample AP Exams | ||

1998 AP Practice Exam: Multiple Choice Questions | 32:33 | |

1998 AP Practice Exam: Free Response Questions | 29:55 |

### Transcription: RC Circuits: Transient Analysis

*Hello, everyone, and welcome back to www.educator.com.*0000

*In this lesson we are going to talk about RC circuits, specifically about the transient analysis.*0003

*What happens as a function of time in these circuits?*0010

*Our objectives include calculating and interpreting the time constant τ of an RC circuit.*0013

*Sketch or identify graphs of stored charge voltage for the capacitor or resistor.*0020

*Write expressions that describe the time dependence of the stored charge voltage or current for elements in RC circuit.*0025

*Analyze the behavior of circuits containing several capacitors and resistors.*0031

*As we get into this lesson, please understand this one is going to be pretty math heavy.*0036

*The transient analysis of RC circuits is one of the more challenging portions of the E and M course.*0040

*This may get in depth a little bit, probably a great time to pause, to go back, to take really good notes.*0047

*You may have to come back to this one a couple of times.*0053

*It is not easy stuff and the first time you see it, it looks like there is a lot a math involved.*0055

*What you will find as we go through is you will see a bunch of the same patterns repeating and repeating.*0060

*But the first time you see it, it can look a little bit scary.*0064

*Do not be daunted, you can get through it.*0067

*Let us take a look at what happens when we charge an RC circuit.*0070

*Here I have a basic RC circuit, we have our source of potential difference resistor.*0075

*We have defined our current direction, our capacitor with the voltage across our capacitor, and the time T = 0 we are going to close the switch.*0082

*It also had graphs down here of current in the circuit, charge on our capacitor,*0090

*and the voltage across the capacitor that we are going to be filling out as we go through here.*0096

*We have done these before, but let us take a minute before we get heavy into the math.*0099

*Just to think about what these are going to look like.*0103

*We know the current in the circuit initially is going to be high because the capacitor acts like a wire.*0106

*So we are going to have all the current I = VT/ R.*0112

*We are going to start here at this high level of current and by the time we get roughly to 5 τ,*0117

*the current is going to dwindle down to less than 1% of its initial value because our capacitor starts to act like an open.*0123

*Our graph is going to have an exponential decay, something like that.*0133

*The charge on our capacitor is going to start at 0, it is uncharged.*0138

*After a long time that being again 5 τ, it is going to be CVT.*0143

*We will have an exponential rise toward that asymptote.*0150

*The voltage across our capacitor starts at 0 acting like a wire and over time approaches VT.*0154

*Our graph should look something like that.*0166

*Our goal here is going to be to mathematically describe those by actually figuring out what happens rather than just quick estimations.*0168

*To do that, let us start by looking at Kirchhoff’s voltage law.*0176

*We are going to do that, I'm going to start down here and go clockwise around our circle.*0179

*As I look at our potential drops, I see - VT first + IR + the voltage across our capacitor, brings us back to our starting point.*0185

*All of that must equal 0.*0197

*We also know here that C = Q/ V, therefore, V = Q/ C.*0201

*I can rewrite my equation now as, let me arrange this a little bit to VT is equal to IR + I’m going to replace VC with Q/ C.*0209

*This implies then though, Q and I are related because I = DQ DT.*0227

*We can write this as VT is equal to R × DQ DT, replacing I with DQ DT + we still have our Q/ C.*0237

*We have a differential equation, we have Q and the derivative of Q in the same equation.*0258

*How do we deal with something like that?*0264

*The first thing I’m going to do is called separation of variables.*0267

*I'm going to try and get all the variables of the same type on the same side.*0270

*I need to get Q and DQ together.*0273

*This is going to take a little bit of algebraic manipulation.*0275

*The reason I know how to do this is I have done it quite a few times.*0279

*You really just have to practice it and dive in, and do it again and again and again.*0282

*Let us rearrange this, I'm going to get DQ DT all by itself by dividing both sides by R.*0288

*I will have VT/ R is equal to, we have our DQ DT + Q/ RC.*0293

*Rearranging this again, we are trying to get RQ together .*0308

*We are going to write this as DQ DT is equal to VT/ R -Q/ RC, just a quick algebraic manipulation,*0312

*which implies then that DQ DT =, I can multiply C down here to combine these on the right hand side.*0325

*They give us a common denominator of RC so I would get C VT -Q/ RC.*0336

*Getting those variables separated, I have DQ/ C VT -Q must equal DT/ RC.*0346

*Just a little bit more algebraic manipulation.*0361

*I have my Q on one side, I got my variable T on the other, that means I can go when I can try and integrate both sides.*0364

*Not try, we are going to.*0372

*Integrate both sides, we will take the left hand side first.*0374

*The integral and our variable of integration is Q, which is going to go from some value 0 initial charge on our capacitor is 0,*0377

*to some final charge Q of DQ/ CVT – Q, must equal the integral of the right hand side DT / RC.*0383

*Our variable of integration on the right hand side T goes from some initial time T = 0 to some final time which we are going to call T.*0401

*As we integrate this, the left hand side we have got a problem of the form DU/ U.*0411

*The integral of DU/ U is nat log of U.*0418

*We have got DU / -U so we are going to get - the log of, we have CVT - Q evaluated from 0 to Q.*0422

*The right hand side of, 1/ RC is a constant, we are just going to get T/ RC evaluated from 0 to T.*0440

*A little bit more algebra, this implies then that, we will substitute in our values, our limits here.*0452

*- the log of CVT -Q - the opposite of the log which is going to be + the log of CVT -0, which is just CVT = T / RC.*0458

*If we got a difference in the logs, log of that + log of that,*0477

*we can say that we have the log of CVT - Q/ CVT equal to, we got that negative sign over here to the right, - T / RC.*0481

*We can simplify this left hand side a little bit too.*0504

*That means that the log of, that could be 1 -Q/CVT = - T/ RC.*0508

*I’m starting to run out of room, let us carry this over onto our next screen here.*0524

*We have the log of 1 -Q/ CVT = - T/ RC.*0531

*This log is troubling, how do we get rid of a log?*0542

*We take E and raise it to that power.*0545

*E ⁺log of 1 -Q/ CVT is what we are going to do.*0547

*Raise both sides to that, so E ⁺log of 1 -Q/ CVT must be equal to E ⁻T/ RC,*0551

*doing the same thing on both sides to maintain that equality.*0562

*E ⁺log of something is just that something.*0567

*Our left hand side gets a little simpler.*0569

*We have 1 - Q/ CVT =, right hand side E ⁻T/ RC.*0571

*Let us see if we can get to that one on that side and do a little bit of rearrangement to say that Q/ CVT = 1- E ⁻T / RC,*0583

*which implies then that getting just Q by itself, Q is a function of time is equal to CVT × (1 - E ⁻T/ RC).*0598

*We were able to solve for the charge on the capacitor as a function of time.*0612

*When you do these types of problems, you are going to see very similar forms come up again and again.*0619

*Some constant multiplied by either 1 – E ⁻T/ RC or just that constant × E ⁻T / RC.*0624

*You are going to see this come up again and again and again,*0633

*to the point where you can almost predict the answer before you go and actually solve it.*0637

*If that is our charge, let us see if we can find the voltage across C.*0642

*VC = Q/ C, we just found Q was CVT × 1 - E ⁻T/ RC ÷ C.*0648

*Our C make a ratio of 1, we get that the voltage across our capacitor or potential difference is just VT × 1 - E ⁻T/ RC.*0668

*How about current flow?*0685

*The trick to getting current flow then is realizing that I is the current is equal to the derivative of the charge.*0688

*Current I is DQ DT which is the derivative with respect to T of our Q which was CVT × 1 - E ⁻T/ RC*0696

*which is going to be, we can pull our constant out, - CVT × E ⁻T/ RC × -1 / RC E ⁺U DU or DU is –T/ RC.*0715

*The E ⁺U DU, the derivative is -1/ RC, excuse me.*0736

*That is going to be, CVT / RC × E ^-¬T/ RC, which implies then,*0741

*our capacitance is C is going to make a ratio of 1, that our current then = VT/ R E^ -¬T/ RC.*0755

*By the way, VT/ R that was our initial current.*0769

*I is actually equal to its initial value × E ¬⁺T/ RC.*0772

*We were able to find the values for the current, the voltage, and the charge on the capacitor,*0783

*all as functions of time, much more exactly.*0790

*You see that the exponential relationship over and over again.*0792

*Or again, RC is your time constant τ.*0796

*Sometimes you will actually even see this written as E ⁻T / RC is written as E ^-¬T/ τ.*0799

*Let us take a look at the opposite side of the storage, discharging an RC circuit.*0810

*Just like we did in our steady state analysis, we have pulled our source of potential difference out of the circuit.*0815

*We have our charged up capacitor and at the time T = 0, we are going to close the switch.*0821

*What happens?*0827

*We know that our current starts out as V/ R because initially that capacitor*0829

*is going to act like a source of potential difference, it discharges.*0835

*We are going to start over here with our current with VC / R and that is going to decay down to 0.*0839

*Our charge starts at Q0 and also decays down.*0848

*Our voltage DC starts at its initial value, let us call that V0 and is also going to decay down.*0856

*We have done that before, now let us fill in the detail.*0865

*What happens in between?*0867

*We use KVL again, Kirchhoff’s voltage law.*0870

*This time I'm going to start here and go this way, counterclockwise.*0873

*As I write that, that will be -VC + IR = 0, which implies then because we know capacitance = Q/ V, therefore V = Q/C.*0878

*That Q/ C is equal to IR, just moving it to the other side as well.*0895

*In this case, we know that I is the rate of change of charge with respect to time*0905

*but now we are discharging, we are going the opposite way.*0911

*Do not forget we got to have a negative there, I = – DQ DT.*0914

*We have Q/ C equal, we still have our R, we will put -R in there because our next piece is that DQ DT.*0920

*The - coming from our I = –DQ DT.*0933

*This implies then, let us do our separation of variables again.*0940

*DQ ÷ Q is going to be equal to - DT/ RC, that piece was a little simpler.*0944

*But we can integrate both sides now, we integrate the left hand side.*0956

*The integral of DQ/ Q starting from some Q = 0 all the way to Q =, - the integral from T = 0 to T of DT/ RC.*0959

*We call that Q₀ to some final value Q.*0978

*Q = Q₀, whatever the Q initial is to its final value, that is a little bit better.*0983

*Which implies then that the log of Q, the integral of DU/ U is the natural log,*0988

*so the log of Q going from some its initial value Q0 to some final value Q.*0994

*What we are going to have over here is - T/ RC.*0998

*Taking this left hand side and putting in our limits, we have the log of Q/ Q0 is going to be equal to - T/ RC,*1009

*which implies that if we raise both of these to the E, that the left hand side becomes Q/ Q₀.*1022

*The right hand side becomes E ¬⁻T/ RC, and solving for just Q, Q = its initial value Q₀ E ^-¬T/ RC.*1029

*There we see that exact function, that exponential decay.*1043

*Once again, same sort of form, we are looking at some constant × E ⁻T/ RC or constant ×(1- E ⁻T/ RC).*1051

*Just like we did before, let us take a look at voltage.*1063

*Voltage is Q/C which is going to be in this case, Q₀ E ^-¬T/ RC ÷ C.*1066

*But we know that our initial value of voltage was Q₀/ C.*1081

* Really this is just saying that VC = its initial value V₀ × E ^¬T/ RC.*1086

*Here we see our function that shows the same thing, the same shape that exponential decay.*1099

*Let us take a look and see if we can do current just to maintain some symmetry here.*1105

*As I look at current, we have I = – DQ DT, which is - the derivative with respect to time of*1113

*we had previously Q₀ E ^-¬T/ RC, which is -Q₀ E ^-¬T/ RC × -1/ RC, which will be Q₀/ RC E ⁻T/ RC.*1125

*We can go a little bit further with that because we know that I initial, I0 was V₀/ R which is Q₀ / RC.*1153

*This is just saying then that current is a function of time is its initial value I₀ E^ -¬T/ RC.*1166

*Once again, there is our exponential decay.*1177

*A pretty heavy stuff in there, and not easy to do the first time you go through it.*1182

*It take some time but keep working on that until you can do it repeatedly.*1186

*It is an expectation for the AP Physics C E and M exam but you can do that.*1190

*Not easy, give it some time, give it some practice.*1194

*Let us talk for just a minute more about that time constant again.*1199

*The time constant in RC circuit τ = RC is the time when the quantity has reached 1 – E⁻¹ or 63% of its final value, just like we said.*1203

*By τ time constant it is more than 99% of the way to its final value.*1214

*If you want to be sure that your circuit is in pretty close to a steady state condition, wait at least 5 time constants then you will be there.*1218

*Let us finish this up by, let us do a bunch of old AP problems that involve this process because really it is all about practice.*1230

*We will start with the 2013 exam free response number 2.*1237

*You can go download it, take a minute, print it out.*1243

*Read it over, give it a shot, then come back here when you are done and hit play again, then we will keep on.*1246

*This exam question starts off with a circuit and says indicate the position to which the switch should be moved to charge the capacitor.*1256

*I'm going to draw where it should be, the answer is going to be B,*1265

*The charge that the capacitor so that you keep the source of potential difference in the loop there.*1269

*But I'm going to draw that as something like this.*1275

*We have got our source of potential difference, our resistor R.*1280

*We come from here over to our capacitor and there we go.*1288

*There is our basic circuit.*1297

*It asks on the diagram, draw a voltmeter that is properly connected to the circuit, the line to measure the voltage across the capacitor.*1299

*Here we need to remember that voltmeters are connected in parallel.*1307

*I'm just going to put a nice happy little voltmeter right there in parallel with my capacitor and I should be all set for part A.*1311

*Here it says, it gives us some information of graph.*1322

*Your partner has does some stuff with the stop watch, it asks you to determine the time constant of the circuit from the slope of the linear graph.*1328

*How are we going to do that?*1335

*Let us take a look, as we start to examine what data we have, we have got time and voltage.*1338

*I'm going to look at the relationship for this as we are discharging the capacitor,*1343

*realizing that V is equal to its initial value × that exponential decay portion E ⁻T/ RC.*1349

*I want to see how I can get this into a linear graph based on time and the potential data that we have.*1357

*If I take the log of both sides, I can say that the log of potential is equal to, the log of that is going to be log of V₀ + - T/ RC.*1366

*Or with a little bit more manipulation, the log of V will equal -1 / RC × T + log of V₀.*1381

*Why did I put it that way?*1396

*I’m trying to make it match the form for a line, the equation of a line that says Y = MX + B.*1397

*Really, I'm saying that we are going to have log V on the Y.*1409

*Our X is going to be T, our Y intercept is going to be log of V₀ which means that the slope of that line that we end up with is going to be -1/ RC.*1414

*We made it match that form.*1424

*It says use the rows in the table of the data to record any quantities that you indicated that are not given,*1427

*label each row used and include units.*1433

*We have got time which is going to go on our X, we do not have is the log of V.*1436

*I would add a row in there for the log of the potential and put that data in there right under the voltage.*1441

*The units of that are going to be log volts.*1449

*Take the value that is in there for B, take the nat log of it and make one more row and you should be all set for part B2.*1453

*Next we go on to part C and we are actually going to be graphing this.*1462

*Let us give ourselves a bunch of room here.*1469

*For part C, let us see if we can do a quick approximation of the graph.*1472

*Of course, you guys will do a much better job plotting points carefully.*1477

*We have got time in seconds over here and we have got the log V and log volts over there.*1489

*On our X, we go 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, something like that.*1501

*Heading up, we need to make it to about 6V, it looks like something like that 1234567.*1513

*When we go and actually plot our points, I end up with something that looks roughly,*1524

*I'm going to draw my rough line first and then fill in the points so I'm not plotting them.*1530

*But you get the general idea of what the shape of that is going to look like.*1537

*And I have got a couple of points scattered above and below the line.*1540

*I’m drawing the best fit line something kind of like this.*1544

*That would be the data, draw a straight line that best fits your data points.*1552

*For part D, from your line in part C, obtain the value of the time constant τ from the circuit.*1556

*And we are going to use what we already talked about.*1562

*We are going to use the fact for part D.*1564

*Get the slope which is Y2 –Y1/ X2 – X1.*1568

*As we pick a couple points on our line, that data points, points on our line is, in my case I ended up with something right around -0.079.*1577

*That slope is also has a meaning, it is -1/ RC, which implies that is -1/ τ = -0.079.*1589

*Therefore, τ = 1/ 0.079 or about 12.7 s in my calculation.*1603

*Let us come check out part E, E1 in the experiment that the capacitor had a capacitance of 1.5 µf,*1616

*calculate an experimental value for the resistance R.*1625

*That is easy, if τ is 12.7 s and that is RC, it means RC = 12.7 s.*1628

*Therefore, R = 12.7 s/ C, which it gave us as 1.5 µf, 1.5 × 10⁻⁶ F which is about 8.47 mega ohms with my values from the slope.*1635

*Part E2, let us go to black here.*1657

*On the axis that we made already, use a dashed line to sketch a possible graph*1665

*if the capacitance was greater than 1.5 µf but the resistance R was the same.*1669

*Justify your answer.*1675

*Let us see, the capacitance was greater then RC would go up if capacitance was greater since C goes up.*1678

*Our τ, our time constant is also going to go up.*1688

*The slope is the inverse of τ.*1693

*Our slope would have to decrease.*1701

*I would draw a line that has a shallower slope, something like that perhaps.*1706

*That is part E2 right there and that is how I would answer that one.*1719

*I think that would leave you in pretty good stead on a problem like that.*1724

*Let us move on and take a look at the 2012 exam free response number 2.*1732

*In this one, we have an experimental setup where a student is measuring resistivity of slightly conductive paper.*1740

*It gives us the thickness and we have got some data there.*1747

*The first thing we are asked to do is use the grid to plot a linear graph of the data points from which the resistivity can be determined.*1750

*Include labels and scales for both axis and draw the straight line that best represents the data.*1758

*The way I would start with this, you have got data for resistance and length.*1763

*Let us see what we need to graph there.*1768

*I'm thinking R = ρ L / A, there is a relationship that includes the variables that we have,*1770

*which by the way is ρ L/ the area, is just the width × the thickness.*1778

*Then, resistance = ρ/ WT × L.*1785

*If we want this to fit the equation of a line that is similar to Y = MX.*1793

*If we plot L on our X axis or on our Y axis, the slope should be ρ/ WT.*1799

*Let us take a look here and see if we can sketch in that graph.*1808

*Over here, we are going to have our resistance in ohms.*1827

*Here we will have our length in meters.*1831

*We will have you guys pick appropriate, I have 200 J, 400 J, appropriate intervals there, labeled well 600 K.*1834

*For length in meters, let us say that is 0.1, that is 0.2, and fill in some values in between.*1845

*When I go and do this and actually plot the data, I get something that, with a couple of data points above, below.*1855

*You are drawing your best fit line there.*1872

*That looks remotely like what you should be finding there.*1876

*In part B, says using the graph, calculate the resistivity of the paper.*1879

*We have done the hard work there because we just said the slope is ρ/ ω T.*1885

*If slope is ρ/ WT, that implies then that the resistivity ρ = WT × the slope which is 0.02 × our thickness 10⁻⁴.*1889

*Our slope Y2 – Y 1/ X2 – X1, picking some points from your line not data points,*1911

*wherever they happen to be, whatever your slope is, figure that out.*1918

*I ended up with a value that for the whole thing of the resistivity of about 8.75 ohm meters.*1924

*It should be somewhere in that ballpark.*1937

*We are changing things up a little bit in this problem.*1945

*A student uses those resistors R4 and R5 to build a circuit using wire, a 1.5V battery,*1948

*an uncharged capacitor, and an open switch as shown.*1953

*Calculate the time constant of the circuit.*1957

*I really do not like the way that is drawn, circuit schematics tend make me happy, warm and fuzzy inside.*1961

*I'm going to draw that in a way that is a little more comfortable for me.*1967

*We will do that for part C.*1970

*With a source of potential difference 1.5V, there is the positive and negative side.*1973

*We then have a couple resistors in parallel, R4 and R5.*1979

*We have our capacitor C and we have a switch.*1991

*Something like that.*2001

*Looking at that as an equivalent, we have two resistors in parallel.*2004

*If we wanted to do that to find the equivalent resistance, R4 and R5,*2009

*the equivalent resistance for R4 and R5 is just going to be R4 × R5/ R4 + R5.*2013

*I should say that that is going to be our equivalent for all of those, we have 370,000 ohms or 440,000 ohms R5/ the sum of those two.*2025

*What is it going to be, 781000 ohms.*2041

*Which gives you about 200 kl ohms or 200,000 ohms.*2048

*If we want to find the time constant of the circuit τ, our time constant is RC, is going to be our 200,000 ohms,*2054

*the equivalent resistance of our circuit × our capacitance 10 µf or 10 ×⁻⁶ F, which is about 2 s.*2067

*Part D, at time T = 0, the student closes the switch.*2083

*On the axis, sketch the magnitude of the voltage across the capacitor*2087

*and the voltages VR4 and VR5 across each resistors functions of time.*2092

*The nice thing there is these are going to have the same voltages across them.*2097

*That is going to be just one plot because they are connected together on the ends.*2101

*Label each curve according to the circuit element and on the axis explicitly label any intercepts, things like that.*2105

*Let us make ourselves another nice, happy, little graph.*2112

*Part D, something like that, where we have our potential and time.*2116

*If we start uncharged and then we are going to close that circuit,*2138

*we know that the voltage across the capacitor is going to have an exponential increase to some final value,*2141

*that is going to be the value of the voltage of our circuit.*2147

*We can draw an asymptote here with about 1.5V where our voltage across our capacitor is going to do something like that.*2151

*Where right around here, let us call that around approximately 5 τ or about 10s.*2164

*That is easy.*2172

*VR4 and VR5, are all of a sudden going to drop down.*2176

*They no longer have that voltage across them so they start at 1.5V*2181

*and they are going to have the corresponding exponential decay.*2184

*I think I can draw it a little prettier, that was not the best job ever.*2188

*Something perhaps kind of like that, where this is the voltage across R4 which is equal to the voltage across R5.*2194

*That should cover you for that one.*2209

*The general graph above the voltage across our resistors and across your capacitor as a function of time.*2211

*Let us take a look now at the 2007 exam.*2223

*Take a minute, printout, look it over.*2225

*Hit the pause if you need to, come back.*2228

*E and M number 1 from 2007.*2231

*Here we have another RC circuit, it says a student sets up that circuit in the lab.*2237

*The values of the resistance and capacitance are as shown but the constant voltage delivered by the ideal battery is not known.*2242

*At time T = 0, the capacitor is uncharged and the student closes the switch.*2250

*It gives us a graph of the current as a function of time measured using a Computer System*2254

*and you get that nice, happy, little graph that they show you.*2258

*For part A, using the data, calculate the battery voltage E.*2262

*The current at time T = 0, we can see right from the graph is about 2.25 milliamps or 0.00225 amps.*2268

*To find the battery voltage, that is just going to be equal to IR in our circuit by Ohm’s law*2282

*which will be our current 0.00225 amps × our resistance, we have got 550 ohms there, for a voltage of about 1.24.*2290

*Part B, calculate the voltage across the capacitor at time T = 4s.*2306

*Let us take a second and draw our circuit.*2313

*Here we go, there is E.*2316

*We have got our switch, we have got our resistor that is 550 ohms, we have our capacitor which is 4000 µf.*2319

*What I'm going to do is, I'm going to go through and do a KVL around the loop, starting right there going clockwise.*2337

*I would write that - E +, we will define to the right is the direction of our current - E + IR + the voltage across our capacitor = 0.*2344

*Therefore, DC = E – IR.*2363

*But we know that the current at 4s I at T = 4s is equal to 2.35 milliamps.*2369

*Roughly right from the graph.*2382

*That then means that VC is going to be equal to, we have got our 1.24V – our I 0.00035 amps × our resistance 550 ohms.*2385

*So I would get a VC of around 1.05V when I put that into my calculator.*2402

*That should cover part B.*2411

*Moving on to part C, calculate the charge on the capacitor at T = 4s.*2413

*Q on the capacitor is charge × voltage, which is going to be our 4000 µf, 4000 × 10⁻⁶ F × our potential 1.05V or about 0.0042 C.*2423

*It looks like we are making some headway here.*2448

*Let us take a look at part D, give ourselves more room here in the next page.*2450

*It asks us to sketch a graph of the charge on the capacitor as a function of time.*2455

*Let us draw our axis in here.*2461

*Our Y axis, our T axis.*2470

*There we have T, there is our Q.*2479

*What is that going to look like, the charge on the capacitor as a function of time?*2483

*We can do this over and over by now.*2487

*We are going to have that exponential rise as it charges up.*2490

*There is part D, let us take a look at E.*2494

*Calculate the power being dissipated as heat on the resistor at 4s.*2500

*Power = I² R which is going to be our current 4s at 0.00035 amps² × our resistance 550 ohms or about 6.74 × 10 ⁻5W.*2504

*Onto F, the capacitor is not discharged but it is dielectric of constant K κ = 1 is replaced by dielectric of constant K = 3.*2532

*It triple the dielectric constant.*2543

*The procedure is repeated, is the amount of charge on one plate of the capacitor at 4s greater than,*2545

*less than, or the same as before, and we have to justify our answer.*2551

*I would start by looking at the formula for capacitance K ϵ₀ A/ D.*2555

*But in this case, we know that K triples.*2563

*All those other things are not going to change.*2567

*It is the same capacitor geometry.*2570

*Therefore, we can say that our capacitance is going to triple which implies then that we are going to have a greater amount of charge.*2572

*We can look at that as, with 3 κ, Q3 is going to be E × 3, the initial capacitance, × 1 – E ⁻T/ RC.*2587

*RC is our initial capacitance still.*2602

*That would be 3 EC × 1 - E⁻⁴/ 3 × our resistance 550 × our capacitance 4000 × 10⁻⁶.*2604

*Just to give you a feel for the numbers, complies then that that is going to be,*2619

*Q3 will be, 3EC 5454 µf × about 1.24V or 6.76 × 10⁻³ C which is 0.00676 C.*2624

*Which by the way, as we suggested already when we said it was greater,*2641

*it is greater than what we initially had the initial 0.0042 C.*2648

*Plug in right through these, let us move on and take a look at the 2006 exam free response number 2.*2664

*Take a minute, find the problem, download it, print it out.*2673

*Give it a try, hit pause, comeback.*2676

*If you have had a few minutes to look it over and we will go through this one.*2678

*This one is a bit of a doozy.*2681

*The circuit that they show you has a capacitor of capacitance C, a power supply of EMF given E,*2687

*two resistors R1 and R2, and a couple of switches.*2692

*Initially the capacitor is uncharged when both switches are open.*2696

*Switch S1 then gets close at time T = 0.*2700

*Part A says, write a differential equation that can be sought to obtain the charge on the capacitor as a function of time T.*2705

*We started out the lesson with something like this but let us do it.*2711

*A, let us take a look.*2714

*We have our source of EMF, our potential difference.*2716

*I’m just going to redraw this a little bit so it is a little easier to see.*2723

*R1, we have to capacitor and it should look like that initially based on where the switches are.*2727

*That both are switches are open and S1 gets close to T = 0.*2740

*Right at T = 0, this is basically our functioning circuit.*2743

*Makes it look a little simpler.*2746

*We also know of course, τ was going to be RC in the circuit.*2748

*Let us do KVL around the loop again, - E + IR1 + VC = 0.*2753

*Which implies then, since VC = Q/ C, that we have - E + IR1 + Q/ C = 0.*2763

*We also know that I is DQ DT, that is charging up.*2774

*Then, we get - E + R1 DQ DT + Q/ C = 0.*2779

*There is our differential equation.*2792

*We have Q and the derivative of Q in the same equation.*2794

*That should cover us for part of A.*2798

*Typically, that is as far as they ask you to go on most questions.*2801

*In this one, they ask you actually go and solve it.*2804

*In part B, solve the differential equation to determine the charge on the capacitor as a function of time.*2806

*Exact same thing we have been doing.*2811

*Chances are you could probably even, based on this tell me what the answer is without even going through the derivation.*2814

*However, to get full credit you probably got to go through and do all the work.*2821

*Let us do that.*2825

*Let us pull the R1, divide everything by r1.*2828

*We have - E / R1 + DQ DT + Q/ R1 C = 0.*2830

*A little bit of rearrangement which implies then that DQ DT = E/ R1 -Q/ R1 C, which implies that DQ DT =,*2843

*we will get a common denominator here, multiply this one by C.*2857

*We have our 1C in the denominator.*2861

*That will give us EC -Q/ R1C.*2864

*Then separating our variables, this implies that DQ/ EC -Q = DT/ R1C.*2873

*Which implies then that DQ and is going to switch the order here with a negative sign,*2886

*multiplying through Q - EC = - DT/ R1C.*2891

*Now we can integrate both sides.*2900

*We will integrate the left hand side from some Q = 0 to final value Q.*2903

*The right hand side will be from some T = 0 to final value T*2908

*which is going to give us the left hand side integral of the DU/ U will be the nat log of U.*2914

*We will get something that looks like the log of Q - EC evaluated from 0 to Q = - T / R1C.*2921

*Which implies then that the log of Q - EC - the log of - EC = - T/ R1C.*2936

*The left hand side, the log of the difference is equal to the log of the quotient.*2953

*The difference of logs is the log of the quotient.*2958

*That will give us the log of Q - EC/ -EC = Q – T/ R1C, raising both of these to the E.*2961

*E ⁺log of that just gives us that piece.*2980

*We will have Q - EC/ -EC = E ^¬T/ R1C.*2983

*Multiply it through by that EC so we get Q - EC is equal to – EC E ^¬T/ R1C.*2993

*Or getting Q by itself, Q = EC × 1 - E ^-¬T/ R1C.*3008

*You earn your points on that one.*3024

*Or we could also take a look at that Q = EC 1 - E -T / R1C.*3029

*I think that works.*3035

*Let us give ourselves more room before we move on to part C.*3040

*For part C, it says determine the time at which the capacitor has a voltage of 4V across it.*3047

*To do that, we know Q, let us see if we can solve for V and then we can back out the time from the voltage.*3055

*If Q = CV which is equal to EC × 1- E ^-¬T/ R1C, that implies then that V is just equal to E × 1 – E ^-¬T/ R1C.*3069

*Or V/ E we will pull out that -1, = –E ^-¬T/ R1C.*3090

*Which implies then that, let us switch our negative signs around.*3102

*1 - V / E = E ^-¬T/ R1C, which implies then that the log of 1 - V/ E = – T/ RC.*3106

*Or getting T all by itself, T = – RC × the log of 1 - V/ E.*3127

*We can substitute in our values to find that time.*3138

*T is equal to -4700 × 0.06 × our log of 1 - 4/ 12.*3142

*Put that all on my calculator and I come up with a time of about 114.3s.*3157

*After switch S1 has been closed for a long time, switch S2 now gets close in new time T = 0.*3170

*For part D, we have got a new T = 0 configuration which looks kind of like this.*3177

*We have got our source of potential difference, over here we have got our resistor R1.*3185

*We come down here, we have our capacitor C.*3194

*We come down here, we have got R2.*3201

*There we go, those two are in parallel.*3208

*Sketch graphs of the current I1 and R1 vs. Time and of the current I2 and R2 vs. Time.*3212

*Clearly label which is I1 and I2.*3220

*I guess that is not so bad.*3223

*Let us draw our axis here.*3224

*As I look at this, it looks like I1, the current through R1 is going to start at 0,*3243

*it is going to go to some center point to some final value toward an asymptote.*3248

*It looks like I2 is going to start at a maximum current when we first do that and approach the same point,*3254

*so from slightly different directions here.*3260

*As I take a look at that because those resistors are equal, they are going to split the voltage across them and end up with the same current.*3264

*I would probably draw this something kind of like this.*3271

*I would expect that I1 to go like this.*3274

*I will label my axis here, there is our current, here is our time.*3280

*I want to do something like that and I2 to come in from about the same point and do something like that,*3285

*where they are getting closer and closer to each other but not quite meeting.*3295

*Those should be symmetric but we have got to think you have got the right idea of what that graph should look like.*3302

*You can perhaps a little closer to that.*3309

*And that should cover part D.*3311

*Alright, that finishes up that problem and let us see if we can do one more.*3315

*Let us look at the 2003 AP physics C E and exam free response number 2.*3324

*As always, we will have you take a minute, look it up, print it out if you can.*3330

*Give it a shot, come back here, and we will see how this one goes.*3334

*In a lab, you can connect a resistor and a capacitor with unknown values in series with a battery of EMF 12V.*3342

*You include a switch in a circuit, and when the switch is closed the circuit is completed.*3348

*You measure the current through the resistor as a function of time, as they would show you in a plot below.*3351

*Using common symbols, draw the circuit that you have constructed that does this.*3358

*Show the circuit before the switch is closed and include whatever other devices*3363

*you need to measure the current through the resistor to obtain that plot.*3366

*Label the components in the diagram.*3370

*That is not so bad, we need to set up an RC circuit where we can measure the current through the capacitor and the switch S1.*3373

*We are showing it before the switch has been closed.*3381

*We have got our source of potential difference E and we will label that.*3384

*We will label that, we would make it nice and clear.*3390

*We have some switch S1, we have a resistor R, we have a capacitor C.*3398

*Somewhere in this series configuration, I will put E in here.*3412

*Just to be safe, I would probably go through and actually write capacitor, resistor, switch, ammeter.*3421

*I will let you guys do that.*3426

*Having obtained the curve shown, determine the value of the resistor at two placed in the circuit.*3428

*The way I would do that one is I would first look at the point where you have got T = 0 because you can use a ohms law R = VI.*3434

*At T=0, the voltage across the capacitor is 0.*3443

*Let me write that at T = 0, VC = 0, which implies that the voltage across R is equal to EMF,*3447

*which implies that the resistance is just going to be the EMF ÷ the current which is 12V/ 0.01 amps or about 1200 ohms.*3456

*C, what capacitance you need to insert in the circuit to give that result?*3474

*I would take a look here and say, our time constant RC must be 4s from the formula*3479

*that they gave us for the current as a function of time.*3490

*Therefore, the capacitance must be 4s/ R which is 4s/ 1200 ohms or about 3.3 × 10⁻³ F.*3494

*Let us check part D, give ourselves some more room here*3515

*because it looks like we got a significantly different portion of the question.*3518

*For part D, you are now asked to reconnect the circuit with a new switch so as to charge*3524

*and discharge the capacitor and be able to get a graph like that when you are switching between positions A and B.*3529

*Draw that schematic and label anything you might need.*3537

*It looks like in part A, it is charging in the switches at A and at part B it is discharging.*3541

*There are bunch of ways you can draw this.*3547

*Electrically they should all be equivalent though.*3549

*I would start off with something like our battery.*3552

*Be good and label this, our battery with an EMF.*3558

*Let us put this switch over here and we will call that position A.*3562

*We will put the switch maybe right here, we will call that S1.*3566

*We will have another position for it down here called B.*3574

*As we go through here, we need our resistor in the circuit.*3580

*We need our capacitor, they want to measure the current flow so we will put an ammeter in here in line with that in series.*3585

*We will connect there so different position B.*3599

*All we have is this loop to discharge.*3601

*I will just continue that up here.*3604

*In position A, this does not make any difference, it is an open circuit.*3605

*In position B, the whole left hand side is missing.*3609

*We have got our battery and we have got our ammeter here.*3612

*We have got our capacitor.*3617

*If we want to measure the voltage across all of that, let us put our voltmeter in parallel with our capacitor.*3622

*We have got our resistor up there and there is our switch.*3636

*I think that should cover all the requirements.*3645

*Hopefully, you got pretty good feeling for transient analysis of RC circuits.*3648

*It takes some time, you might want to go back and do some of these analyses a couple of times*3654

*until it starts to make sense to you, until you start doing the patterns.*3659

*Thank you very much for watching www.educator.com.*3662

*We will see again real soon, make it a great day everybody.*3665

1 answer

Last reply by: Professor Dan Fullerton

Tue Jan 26, 2016 7:06 AM

Post by Shehryar Khursheed on January 25, 2016

What would happen if you had an RC circuit with the capacitor and resistor in parallel, not series as all the examples indicated?

1 answer

Last reply by: Professor Dan Fullerton

Tue Jan 26, 2016 7:03 AM

Post by Shehryar Khursheed on January 25, 2016

When you discussed discharging a curcuit, I have a question regarding the negative sign on the rate of change of the charge on the capacitor. I understand why you used the negative sign- because the charge is decreasing, therefore the current will also be decreasing. However, doesn't the derivative of the charge already imply that it is decreasing? Because we see that charge is leaving the capacitor, so they derivative itself must already be negative without the negative sign in front of it.

2 answers

Last reply by: Professor Dan Fullerton

Thu Apr 9, 2015 1:50 PM

Post by Thadeus McNamara on April 9, 2015

@54:00, i dont understand how you got the shape of those graphs. could you explain again? thanks

1 answer

Last reply by: Professor Dan Fullerton

Thu Apr 9, 2015 12:55 PM

Post by Thadeus McNamara on April 9, 2015

@46:01 can you explain how you got the + or - for E, IR1, and Vc

1 answer

Last reply by: Professor Dan Fullerton

Thu Apr 9, 2015 12:47 PM

Post by Thadeus McNamara on April 9, 2015

@1:12, shouldn't the C have the negative charge on top and the positive charge on the bottom? because i thought the positive terminals of the battery and capacitor cant face each other and the same thing with the negative terminals of the battery and capacitor

1 answer

Last reply by: Professor Dan Fullerton

Tue Apr 14, 2015 6:25 PM

Post by Thadeus McNamara on April 7, 2015

are there other axi that you could use for the Example 1? because i checked the grading rubric and it seemed to do the problem differently

2 answers

Last reply by: Thadeus McNamara

Tue Apr 7, 2015 11:01 PM

Post by Thadeus McNamara on April 7, 2015

can you explain how you would go about graphing the y = mx + b equation you came up with?

1 answer

Last reply by: Thadeus McNamara

Tue Apr 7, 2015 10:35 PM

Post by Thadeus McNamara on April 7, 2015

@ 19:20, how do you know I = Q / RC ?

2 answers

Last reply by: Thadeus McNamara

Tue Apr 7, 2015 10:14 PM

Post by Thadeus McNamara on April 7, 2015

@around 16:10, the bounds are from q=0 to Q. Why wouldn't it be from Q to q=0 (since it is discharging)? is it because the I = -dq/dt already accounts for the discharge?