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Lecture Comments (13)

1 answer

Last reply by: Professor Dan Fullerton
Thu Jul 19, 2018 5:22 PM

Post by Laura Darrow on July 19 at 04:50:33 PM

Hello Mr. Fullerton,

Can you elaborate further on your solution to part (f) in Example 6? (Sorry, I don't mean to overcomplicate this).  I appreciate your expertise.

1 answer

Last reply by: Professor Dan Fullerton
Thu Jul 19, 2018 7:51 AM

Post by Laura Darrow on July 18 at 07:12:31 PM

Hello Mr. Fullerton,

In Example 6 (f), if you simplify the circuit by ignoring the capacitor part, then you are left with two resistors in series.  Decreasing the value of the resistor will decrease the voltage drop across it based on Ohm's Law.  This will lead to an increase in Voltage when it reaches the Capacitor.  This would imply that there would be more energy on the Capacitor due to the smaller voltage drop from the resistor. In practical application, the total resistance drops but the current increases to balance it.  The flow of charge to the Capacitor is therefore increased, leading to an increase in potential.  Thus the energy that the Capacitor gains will be greater than the initial energy by Resistor 2. Am I anywhere close to being right?  I feel like I'm missing something here.  Thank you.

1 answer

Last reply by: Professor Dan Fullerton
Mon Mar 26, 2018 9:08 AM

Post by Kevin Fleming on March 26 at 08:08:52 AM

Prof. Fullerton,
For Example 5, part e, can you also determine the voltage on the 5uF capacitor to be 20V by realizing that since a voltage drop of 10 V already occurred at the 20 ohm resistor, a voltage drop of 20v must occur at the 5uF capacitor in order to ensure that KVL is not violated. I did not make a new diagram like you did, but I considered the wire containing both the 20 ohm resistor and the 5uF capacitor to be in one loop, so I just applied KVL. Thanks

2 answers

Last reply by: Professor Dan Fullerton
Tue Jan 10, 2017 10:35 AM

Post by Alen Gacic on January 10, 2017

i cannot access the exam, navigating it on the college board is kind of difficult because they require me to log in. and i do think i am registered with them. unless i have to be?

1 answer

Last reply by: Professor Dan Fullerton
Sun Jan 24, 2016 3:23 PM

Post by Shehryar Khursheed on January 24, 2016

On the 2010 FRQ, for part d, how are you able to just switch the resistors and capacitors so that they are in series?

0 answers

Post by Professor Dan Fullerton on December 17, 2015

Hi Parth.  The convention I follow in solving these problems is to use the sign that I see first across the element.  So if I'm going clockwise starting at the bottom left of the circuit (roughly 2:44 into the video), the first side of vT that I see is the negative side, so I write negative vT.  I see the more positive side of R first, so the voltage written there is + IR

0 answers

Post by Parth Shorey on December 16, 2015

Why would it be -Vt when you're going from high terminal to low terminal ?

RC Circuits: Steady State

  • 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.
  • The time constant in an RC circuit describes how long it takes for the circuit to reach 63% of its final steady state value. The circuit reaches 99% of its final steady state condition by 5 time constants.

RC Circuits: Steady State

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 0:17
  • Capacitors in Parallel 0:51
    • Store Charge on Plates
    • Can Be Replaced with an Equivalent Capacitor
  • Capacitors in Series 1:12
    • Must Be the Same
    • Can Be Replaced with an Equivalent Capacitor
  • RC Circuits 1:30
    • Comprised of a Source of Potential Difference, a Resistor Network, and Capacitor
    • RC Circuits from the Steady-State Perspective
    • Key to Understanding RC Circuit Performance
  • Charging an RC Circuit 2:08
  • Discharging an RC Circuit 6:18
  • The Time Constant 8:49
    • Time Constant
    • By 5 Time Constant
  • Example 1 9:45
  • Example 2 13:27
  • Example 3 16:35
  • Example 4 18:03
  • Example 5 19:39
  • Example 6 26:14

Transcription: RC Circuits: Steady State

Hello, everyone, and welcome back to

I'm Dan Fullerton and in this lesson, we are going to talk about RC circuits, circuits that have resistors and capacitors.0003

We are going to focus on them in the steady state as opposed to the transient analysis.0010

What happens as a function of time, we will come up in our next lesson.0013

Our objectives include calculating equivalent capacitance for capacitors in series and parallel configurations.0017

We have done that before, we are just going to review it a little bit.0024

Describe how stored charges divided between capacitors in parallel.0027

Determine the ratio of voltages for capacitors in series.0031

Calculate the voltage or stored charge under steady state conditions for capacitor connected to a circuit consisting of a battery and resistor network.0034

Finally, sketch graphs of current stored charge and voltage for a capacitor or resistor in one of these RC circuits.0043

Very quickly, let us review capacitors in parallel.0050

We know capacitor store charge on plates and capacitors in parallel can be replaced with an equivalent capacitor.0054

We talked about doing that, the equivalent capacitance for capacitors in parallel is just the sum of the individual capacitances.0060

For capacitors in series, the charge on them must be the same and they can be replaced with an equivalent a capacitor, following 1/ the equivalent capacitance is 1/ C1 + 1/ C2, and so on.0071

As we get into RC circuit specifically then, RC circuits are comprised of a source of potential difference to make the current flow, a resistor network and 1 or more capacitors.0087

We are going to look at RC circuits from the perspective now of what happens when their first turned on and what happens after a long time.0097

What is in the middle, we are going to leave in the next lesson.0104

At least we are starting out.0107

The key to understanding RC circuit performance, uncharged capacitors act like wires.0109

Charged capacitors act like opens.0116

You got to know these 2 facts, it is going to help you immensely with your analysis of capacitor circuits.0119

Let us take a look at charging an RC circuit.0127

Here we have a source of potential difference some VT, we have got a resistor R, we have a capacitor C,0130

and we need to find the positive and negative side of our capacitor,0136

and at time T = 0 we are going to close the switch and see what happens.0139

Initially, a capacitor acts like a wire when it is uncharged.0146

If we were to take a look at this and we are going to follow a Kirchhoff’s voltage law path around that way,0150

I could write that -VT because starting here we see the negative side first.0157

+ IR + the voltage across our capacitor is going to equal 0.0163

We also know that our capacitor is charge ÷ voltage across our capacitor.0173

Therefore, the voltage across our capacitor is Q ÷ C.0183

I can rewrite my equation as - VT + IR + Q/ C must equal 0.0189

But we also said that time T = 0.0203

When we first closed that switch, the charge on our capacitor is going to be equal to 0, it is uncharged.0205

That simplifies our analysis a little bit.0213

And we have - VT + IR equal 0 which implies that VT = IR.0217

Probably not a surprise.0227

VC therefore, must equal 0.0229

If we were to do our plot of the different things we have down here, initially the current flowing through our circuit is going to be VT/ R.0231

We can make 1 point over here for the current through our circuit and call that VT/ R time equal 0.0241

We also know the charge on our capacitor at time equal 0 is equal to 0.0248

We also know the voltage across our capacitor because Q is 0, we will start at 0.0254

There is our initial analysis.0260

Let us take a look and assume that the thing is charge up, it has been a long time.0264

The long time is something we are going to define here a couple of minutes.0268

After a long time, the switch has been closed for a while, the capacitor now acts like an open.0273

It does not allow current to flow .0281

As T approaches infinity, as it gets very big and we apply Kirchhoff’s voltage law to the same loop,0285

go around the same way, I have -VT + IR + VC = 0.0292

Since I equal 0, we find that VT = VC.0303

No current flowing so we have the same voltage across the source of potential difference that we do across the capacitor.0311

Our current went down to 0, it is going to follow a path something like this and exponentially up to this point that we are going to call 5 τ 5 time constant.0318

We will talk about exactly what that time is here in a moment.0330

The capacitor must be charging up during this time.0334

When it is fully charged, we know that it must have a charge of CVT on.0337

I’m going to draw an asymptote in here at CVT, when it is fully charged.0341

It is going to have an exponential increase toward that value, getting really close to it about 5 τ 5 time constant.0350

The voltage across our capacitor started at 0.0357

We know after long time, we just determined that, that was going to be equal to VT.0361

We are going to follow that same asymptotic relationship getting very close when we get to about 5 τ.0369

How about if we were to discharge an RC circuit?0377

We have pulled the τ supply or the source of potential difference out of here, our capacitor starts out charged0380

and at time T equal 0, we are going to close the switch.0386

When we do that, let us go and we will go around our circuit this way for Kirchhoff’s voltage law.0390

The first thing I see, if I start down here at this blue point, I see - VC + IR, that brings me back to my initial point = 0.0398

We also know that I then is VC/ R with a little bit of rearrangement.0410

We also know that charge is going to be CVC.0417

Initially, current is VC/ R so I can draw to that in my graph down here, we start at VC/ R for our current flow.0423

Our charge is going to be C × VC, which initially CV is 0 for this problem which was CVT.0436

We are going to start at a fully charged capacitor, we will call that CV0 for this specific problem.0444

The voltage across our capacitor also has to start fully charged at whatever V0 happens to be.0451

We let some time elapse after a long time.0459

Now a capacitor as it becomes uncharged, it is going to act like a wire.0464

At that point, as we go around and write our Kirchhoff's current law equation.0473

Starting here, I have -VC + IR equal 0, but since it is acting like a wire, VC equal 0 that implies then that I must be 0 and Q equal 0.0478

We are going to get these exponential decay.0493

Current is going to DK something like that.0496

Our charge is going to decay with that 0 line as an asymptote.0502

Our potential is going to do the same basic thing in exponential decay.0508

It gets pretty close to about 5 τ so that is what happens as we look at this from a steady state looking at right when the switch is open or closed.0515

And right at the other end, after a long time.0524

Let us talk a little bit about this time constant τ, squiggly lowercase t or τ.0528

The time constant in an RC circuit is equal to the resistance × the capacitance indicates that time0538

at which the quantity under observation has achieved 1 – E⁻¹ or roughly 63% of its final value.0545

Why is it important?0552

After 1 time constant, you are closing in on 2/3 to your final value.0554

By 5 time constants, the quantity under observation is within 1% of its final value.0559

What you are talking about a long time with asymptotes that can be kind of confusing.0564

If you pick 5 τ, you are pretty much all the way to your final value.0569

You are within 99% of it.0574

We are going to use this time constant quite regularly, τ = RC for an RC circuit.0577

Let us take a look at a little bit of an RC analysis here with an example problem.0583

What is the current through R2 when the circuit is first connected and we got our capacitor down here C1 of 5 µf.0588

What is the current through R2 a long time after the circuit has been connected?0596

When it is first connected, that capacitor acts like a wire so we can almost pretend it is not even there.0600

And I see a combination circuit where we have got a series portion and a parallel portion.0606

I could redraw that if I wanted to, to find the current through our circuit.0611

Finding the equivalent resistance for these 2 to make an equivalent series type circuit.0616

I’m going to do that, let us call that our 20V.0621

There is our 200 ohm R1 and over here we have R2, 3 equivalent.0627

That is what our circuit looks like from an electrical perspective, when that capacitor is acting like a wire, when it is first turned on.0635

The value of R2,3 is going to be R2 × R3, 400 × 300/ 700, which is at about 171 ohms.0645

That allow me to find the total current in the circuit, I = V/ R which is going to be our 20V ÷ our total resistance.0658

If this is 171, that is 200, we divide by 371.0667

171 + 200 to come up with a current of about 0.0539 amps.0673

That means we have got 0.0539 amps walking through there.0681

We have that current 0.0539 amps and if this is what I'm going to call my ground here, 0 V.0687

0.0539 × 200, that means we are going to drop just over 10 V, about 10.78 V.0699

Meaning our potential up here is going to be about 9.22 V.0707

We can find the individual current through R2.0713

I2, the current through R2 then, is going to be V2/ R2, which will be 9.22 V/ 400 ohm, which is going to come out to be about 0.0231 amps.0719

There is our first answer.0735

What is the current through R2 when the circuit is first connected?0737

0.0231 amps.0740

After a long time, or at least 5 τ as an approximation.0744

Then, our capacitor acts like an open, it is fully charged.0750

Our circuit starts to look really simple then, we have our source of potential difference or 20 V.0757

We have R1 over here, 200 ohms.0763

Nothing is going to flow through this right branch because we have got an open.0768

The only thing we have to worry about here is our 400 ohm R2.0771

Now, current = V/ R which is going to be 20 V our total resistance 600 ohms, or about 0.0333 amps.0780

There is our current after a long time.0794

An example where we are using this steady state analysis of an RC circuit, to solve a slightly more complex problem.0797

Let us do another one, here we have a circuit that has a source of potential difference.0806

We have got 3 resistors and another capacitor, what is the current through R3 when the circuit is first connected?0812

What is the current through R2 a long time after it has been connected?0818

Let us look when it is first connected.0823

This is uncharged so it acts like a wire.0825

Our circuit is going to look something like this.0827

We got our source of potential difference, we have got our 100 ohm R1, we have got 200 ohm here, and we have got 200 ohm here.0830

I’m going to find the equivalent resistance of these two.0850

200 × 200/ 200 + 200 or by inspection I know that that is going to be half the value, since they are identical 2 capacitors in parallel.0854

What we are going to have is our source of potential difference,0864

our 100 ohm resistor and our equivalent resistance right over here and each is going to be 100 ohm.0867

If this is 10 V, pretty easy to see that the voltage is going to be divided evenly between them0877

because they got an equivalent equal resistance, that point must be 5 V.0883

If that is the case, then that means when over here, this must be 5 V when the circuit is first turned on.0891

We can find the current through R3, when it is first turn on.0896

It is just going to be V/ R or that 5 V ÷ 200 ohm which is 0.025 amps.0898

So there is our current through R3 when the circuit is first connected.0910

Let us take a look what happens after it is a long time after it is been connected, to find the current through R2.0917

After a long time, this is going to act like an open current through R2, a long time after the circuit has been connected.0924

We will have 10 V, we will have R1, and we will have our 200 ohm, 100 ohm R1, we have a 200 ohm R3.0932

This is open so it is not even in play at the moment.0952

What do we have here?0955

Our current is going to be V/ R which is going to be 10 V/ 300 ohm, which is 0.0333 amps, that is our current through R1 and through R3.0958

But it asks us, what is the current through R2 a long time after the circuit has been connected?0972

A long time after the circuit has been connected through R2, there is nothing because we have got an open branch.0978

Its capacitor is open, nothing flows through R2.0984

The current through R2 after long time is going to be 0.0987

Taking a look at some equivalent capacitances.0995

What is the equivalent capacitance of the capacitor network shown here?0999

You have got a 5 µf, got 210 µf in parallel, and a 5.1003

I can redraw that to try and make something a little bit simpler for me to see as 5 µf,1008

we will replace these 2 in parallel with their equivalent.1017

The capacitors in parallel add up so that is the equivalent of 20 µf capacitor.1020

And down here, we have got our other 5 µf capacitor.1026

Now I can use my equivalent capacitance equation for capacitors in series.1030

1/ the equivalent capacitance is 1/5 µf + 1/ 20 µf.1036

Keeping in mind that micro × to 10⁻⁶ + 1/ 5 µf or 1/ the equivalent capacitance is going to be 450,000, 1/ F,1045

which implies then that the equivalent capacitance is going to be 1/ 450,000, which is 2.22 × 10⁻⁶ F or 2.22 µf.1059

How about another equivalent capacitance problem?1083

When you look at things like this, you usually tend to be a little tricky when you first look at them.1086

See if you can simplify and put in a format that looks a little bit more easy on your eyes.1089

As I look at this one, 45° angle start to just mess with my brain.1095

Let us see if we can straighten this out a little bit.1099

I could redraw this, it looks down on the left hand side we have got C to another capacitor C,1104

to another capacitor C, and then we come back to some point.1112

On the right hand side, we just have 1 capacitor that is the equivalent circuit to that.1117

That looks a whole lot easier.1125

We have got these 3 C that are in series.1128

Let us find that they are equivalent, 1/ C equivalent is going to be 1/ C + 1/ C + 1/ C which is 3/ C.1131

Therefore, C equivalent for these 3 is just going to be C/ 3.1140

I will redraw this now as, I got a C/ 3 here and that still that my C from over here on the right.1146

Capacitors in parallel, we just add them up.1159

Our total equivalent capacitance, C + C/ 3 is going to be 4/ 3 C.1162

Let us take a look and do a couple of practice AP problems.1177

We are going to take a look at the APC E and M free response number 2 problem.1181

You can find a link to it here at the top.1185

Search it on, it is pretty easy to find.1188

Take a minute and print out the questions, see if you can do it and then follow along.1190

If you get stuck, use that to keep you going or if you get through the whole thing, excellent, you can check your answer.1194

Let us take a minute here and look at this problem.1200

We have got a circuit that has a couple capacitors in it with a switch.1203

Initially switch S is open and the battery has been connected for a long time.1207

What is the steady state current through the ammeter?1212

As we look at part A here, after a longtime, the current through the capacitor is going to be 0.1216

Therefore, you have 0 current through the ammeter.1224

Let us write that here.1226

After a long time, the current through the capacitor equal 0.1230

Therefore, you have 0 current through your ammeter.1238

Because at that point, if you were to draw the circuit, it kind of looks like this.1252

Ignoring that switch because nothing is going through it while it is open.1255

You have got a resistor, a capacitor, and your ammeter.1258

You have another resistor, another to capacitor, and the whole thing comes back to your source of potential difference.1261

But if it is fully charged up, no current flows.1275

Therefore, this one is pretty easy.1279

Part B says, calculate the charge on the 10 µf capacitor, this capacitor up here.1283

To do that, C = Q/ V which implies then the charge = C × V, which we know is 10 µf, 10 × 10⁻⁶ F × the potential difference across it.1292

As we look at that, that is got be 30 V.1308

We end up with 300 µc.1312

We got out 30 V power supply here, 40 ohm, 20 ohms, 5 µf, 10 µf.1322

For part C, calculate the energy stored in the 5 µf capacitor.1331

The potential energy there is going to be ½ CV² which is ½ × our capacitance, 5 × 10⁻⁶ F × the square of our potential difference 30 V² to be 2250 mJ, or 2.25 mJ.1337

Just highlighting our answers.1362

We had 0 current 300 µc and 2.25 mJ.1368

Next, it is going to ask us to close the switch and after we come into a new steady state,1374

calculate the steady state current through the battery.1380

Let us go to a new page, give ourselves some room here.1383

I'm going to redraw the circuit now that we have that switch closed, see if we can make that a little bit clearer.1386

If I were to draw it, I’m going to draw it in a way that is a little bit more familiar to me.1392

We will that our 30 V source of potential difference.1396

We have our 5 µf capacitor here.1400

We have a 10 µf capacitor here, and we also have over here our resistor network 40 ohm, 20 ohm,1407

and we have a connection right there where the switches to be.1425

Rearrange a little bit with an equivalent circuit.1430

We are trying to find what the current is through the battery.1434

As I look at that then, I see we have got 30 V across our 2 resistors here, that should be pretty straightforward.1440

I = V/ R which is going to be 30 V/ 60 ohm or 0.5 amps.1447

Moving on to part E, calculate the final charge on the 5 µf capacitor.1462

We already said charge is capacitance × voltage which will be 5 µf, 5 × 10⁻⁶ F.1470

Our voltage here is going to be 20 V across our 5 µf capacitor because we have got a basic voltage divider here.1479

We are going to drop 20 V across our 40 ohm resistor and 10 V across our 20.1490

That is going to be 100 µc.1496

Let us take a look now at part F, calculate the energy dissipated as heat in the 40 ohm resistor in 1 minute, once a circuit has reached steady state.1505

It sounds like we need to find the power first and then multiply that power by the time, in order to find the energy expended.1515

Power is I² R is going to be our ½ amp² × 40 ohm resistance or 10 W.1522

The energy expended is going to be power × time, that is the electric field here.1535

Energy which is 10 W × 60 s or 600 J.1542

I guess 10 W is not really an answer, it is part of the way to an answer.1554

That takes care of the 2010 free response number 2.1557

Pretty straightforward, once you realize that an uncharged capacitor acts like a wire and a fully charged capacitor acts like an open.1562

Let us take a look at the 2004 exam now, free response number 2.1574

Take a minute, find it, look at the problem.1579

Download it, give it a shot, and we will come back here and see what we can do with it.1582

As I look at this one, we are given a circuit and we are also given a graph of voltage vs. Time.1586

The first thing it asks us to do is determine the voltage across resistor R2 right after that switch is closed.1593

It usually helps me to draw the circuit again.1599

Let us put that up here.1602

For part A, we have our voltage source.1604

Over here, we have that is R2 according to their notation.1609

Here is R1, and as we go a little further in the circuit, we come to our switch S and our capacitor which is C 20 µf.1615

I'm going to call that voltage C +- VC.1635

At time 0, we are closing that switch.1639

At time T = 0, VC equal 0.1646

There is no charge on our capacitor.1651

Therefore, V across R1 if that is 0, that is got to be 0.1653

Therefore, V R2 must have the entire 20 V of them, source across it.1662

The voltage across R2 right after the switch is closed must be 20 V.1669

Part B, determine the voltage across resistor R2 a long time after the switch is closed.1676

For part B, we are looking as T gets really big.1683

In that case, VC is going to start approaching the voltage that is across R1.1686

The way we could figure that out is that is going to be, we have got 15 kilo ohms here, and we need to figure out what R2 is going to be.1693

We can look right at our graph though, and see that the capacitor voltage is 12 V after a long time.1705

If this is 12 V here, that means the voltage across R1 here has to be 12 V.1711

VC is 12 V and V R1 is 12 V, which means V R2, the remaining of the 20 V must be 8 V.1717

That graph just came in mighty handy.1730

For part C, calculate the value of resistor R2.1734

To do that, I'm going to look again as T approaches infinity, we said that V R1 was 12 V,1740

which implies that I R1, you have to have the same voltage drop which is V R1/ R1 is 12 V/ 15,000 ohms.1748

Therefore, we get a current of about 0.0008 amps.1767

V R2 is 8 V which is I R2 × R2.1775

Therefore, R2 must equal V R2/ I R2 or I R1 because there is nothing flowing through your capacitor.1785

That is going to be 8 V/ 0.0008 amps from up above, or 10,000 ohm which is 10 kilo ohms.1797

We found that resistance 10 kilo ohm.1816

What else do we have here?1822

Find the energy stored in the capacitor a long time after the switch is closed.1824

The energy stored in the capacitor after a longtime.1830

U is ½ CV² is just going to be ½ × our capacitance 20 µf.1833

20 × 10⁻⁶ F × 12 V² or 0.00144 J, which is 1.44 mJ.1840

It looks like we are going to be making a graph.1861

Graph current R2 from 0 to 15 s.1863

Let us give ourselves some more room here.1866

I will make a graph for part E, our axis first.1869

It looks like we are graphing current and amps on the Y.1885

We got our time in seconds over here.1889

We will break it up, 5, 10, 15.1892

Graph the current through R2 from 0 to 15 s.1899

At T = 0, current through R2 is V R2 ÷ R2, or 20 V/ 10 kilo ohms which is 0.002 amps.1903

We are going to have to start up here at 0.002.1917

After a long time, as T approaches infinity and I R2 is V R2/ R2, which now is going to be 8 V/ 10 kilo ohms or 0.0008 amps.1923

Let us see if I make it thick there at about 0.001, we are really targeting this asymptote down here of 0.0008 amps.1945

I would expect our graph to looks something like, time constant looks like that happens pretty quickly.1961

We are probably looking at something kind of like that, as we get closer and closer to the longer and longer time.1969

Part F, indicate whether the energy stored in the capacitor is greater than, less than, or equal to its value when R2 was in the circuit, explain.1982

The potential energy is ½ CV² but the voltage across the capacitor goes up if R2 decreases.1994

Therefore, U, our potential energy must be increasing.2010

It is got to be greater than.2016

Hopefully, that gets you started on RC circuits in a steady state.2025

In the next lesson, we will get to RC circuits, talking about their transient analysis.2028

What happens between time equal 0 at a long time, trying to put some quantitative numbers around that.2034

Thank you so much for watching, make it a great day.2041