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Lecture Comments (21)

1 answer

Last reply by: Professor Dan Fullerton
Tue Jan 26, 2016 7:06 AM

Post by Shehryar Khursheed on January 25 at 10:21:40 PM

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 at 06:31:27 PM

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?

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

Transcription: RC Circuits: Transient Analysis

Hello, everyone, and welcome back to

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 of1113

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

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