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Lecture Comments (2)

1 answer

Last reply by: Professor Dan Fullerton
Tue Feb 2, 2016 8:58 AM

Post by Shehryar Khursheed on February 1 at 06:25:33 PM

On the 2008 FRQ, part b, can you please explain why the final value of the current with the inductor and the initial value of the capacitor current are both higher than the resistor current?

RL Circuits

  • When a circuit is first turned on, an inductor opposes current flow and acts like an open circuit. After a time, the inductor keeps the current going and acts like a short (like a wire).
  • If the battery is removed after the circuit has been on, the inductor acts like an emf source to keep the current going for a time. As the circuit dissipates power, the current will decay exponentially to zero.
  • To analyze an RL circuit, use Faraday’s Law. You cannot correctly use Kirchhoff’s Voltage Law (the loop rule) since the magnetic flux in the circuit is changing.

RL Circuits

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

  • Intro 0:00
  • Objectives 0:11
  • Inductors in Circuits 0:49
    • Inductor Opposes Current Flow and Acts Like an Open Circuit When Circuit is First Turned On
    • Inductor Keeps Current Going and Acts as a Short
    • If the Battery is Removed After a Long Time
    • Resister Dissipates Power, Current Will Decay
  • Current in RL Circuits 2:00
    • Define the Diagram
    • Mathematically Solve
  • Voltage in RL Circuits 7:51
    • Voltage Formula
    • Solve
  • Rate of Change of Current in RL Circuits 9:42
  • Current and Voltage Graphs 10:54
    • Current Graph
    • Voltage Graph
  • Example 1 12:25
  • Example 2 23:44
  • Example 3 34:44

Transcription: RL Circuits

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

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

Our objectives include, applying Faraday’s law to a simple RL series circuit to obtain a differential equation for current as a function of time.0011

Solving the differential equation for the current as a function of time 0020

using that separation of variables technique that we have used a couple of times.0023

Calculating the initial transient and final steady state currents through any part of that simple series 0027

in parallel circuit containing an inductor and a couple of resistors.0033

Finally, being able to sketch graphs of the current through or the voltage across the resistors0036

or the inductor in these simple series and parallel circuits.0042

All about circuits that have resistors and inductors in them.0045

Let us start off with our analysis of the circuits.0049

When the circuit is first turned on, the inductor opposes the current flow and it acts like an open circuit.0053

Contrast that to the capacitor, where when you first turn it on, it acted like a wire.0059

After a long time, the inductor keeps current going, it acts as a wire acts, as a short.0064

Contrast that with the capacitor which after a long time it acted like an open.0070

They are opposites.0074

After a long time, if you remove the battery, the inductor is going to act as an EMF source to keep the current going.0076

You can almost think of an inductor is something that likes to oppose change.0082

If there is no current flow in the circuit, it does not want current flow.0087

But after you eventually get it going for a while and you stop the current flow, it wants to keep it going.0090

As the resistor dissipates power, the current will decay exponentially to 0 as you use up that energy.0096

Standard RL circuit diagram or source of potential difference, resistor and our inductor.0104

Something important to note here.0110

As we go through these analyses, the electric field inside that inductor is 0.0112

Let us take a look at the current in the RL circuit.0120

The first thing I'm going to do is I'm going to define a couple things.0124

Let us talk about our current flow.0127

Let us say it is in that direction.0128

Let us also note our electric fields in here.0130

If this is our source of potential difference, positive and negative side, we know the electric field points that way in there.0133

Our electric field is going to go in this direction through our resistor.0141

And inside our inductor, keynote, electric field is 0.0145

When analyzing these sorts of circuits, lots of people like to use Kirchhoff’s voltage law.0151

They like to say, as you go around here - V + IR + the voltage across the inductor equal 0.0157

Although you can get the right answer doing that, it is not technically correct because Kirchhoff’s voltage law 0165

is really just looking at items where you have a non changing magnetic flux.0170

Because you have a non changing magnetic flux, we have to go back to some of these first principles.0176

We are going to start there and do this the more correct way.0179

Let us start by writing our equation, the integral / the closed loop of E ⋅ DL is going to be the opposite of D φ B DT.0186

We have been doing that for a while, Faraday’s law.0200

When we do this, let us look at our E ⋅ DL side.0203

We are going to have - V + our potential drop across R IR, so - V + IR =, and we know that D φ B DT is just going to be - L DI DT.0209

We showed that in our last lesson.0228

That is how we set up our equation to get going here.0230

Let us start to rearrange this a little bit.0234

We have to I - V/ R must equal - L/ R DI DT, as we work to separate our variables and our differential equation.0237

Which implies then that all we have as we separate these DI/ I - V/ R must equal - R/ L DT.0250

The integral from I = 0 to some final I must equal the integral from T equal 0 to some final value T.0267

Starting to solve this, DU/ U that form the integral of that is going to be the log of U.0278

The left hand side is going to look like the nat log of I - V / R evaluated from I = 0 to some value I =, 0284

and the right hand side just becomes - R/ LT.0296

Expanding out this left hand side, substituting in our limits.0303

That implies then that the log of I - V / R - the log of - V / R must be equal to - R/ L × T and the difference of the logs is the log of the quotient.0306

Therefore, we could write the left hand side is the log of I - V/ R divided by -V/ R = - R/ LT.0327

Once again, this nat log is pestering me.0341

Let us raise everything, E raised to those powers on both sides so we can get rid of our natural log.0344

And we can write the left hand side as I - V/ R divided by - V/ R must be equal to E ^- R/ L × T.0351

Or doing a little bit of algebra here, we will multiply both sides by that - V/ R so that the left hand side becomes I - V/ R = - V/ R E ^- R/ LT.0365

Or going to step further, I =, if we add that V/ R to that side, we are going to get V/ R.0383

If I factor that out of the right hand side, it would be 1 – E ⁺_R/ L × T.0394

I = V / R, that current that you would get if you do not have the inductor × (1 – E ^- R/ LT).0405

There is our current flow as a function of time.0415

When we talk about RC circuits, we talk about a time constant τ which is equal to R × C.0419

And we are talking about RL circuits, our time constant which we still call τ is now L/ R.0426

We can write this then in the form I = V/ R 1 - E ^- T/ τ. 0434

It fits that same basic form of the solution that we saw previously when we are dealing with RC circuits.0447

Once again, though we are dealing with something 1 – E ^- T/ the time constant or E × – E ^- T/ the time constant.0453

You are going to see that form again and again and again.0464

There is the current, let us take a look at the voltage in RL circuits.0468

The voltage across our inductor is a function of T, is just L DI DT which is going to be L × the derivative with respect to time of V/ R × 1 – E ^- R/ LT.0475

And that is going to be, we can pull our V/R, that is a constant.0498

That will be L V/ R × the derivative of 1 - E ^- R/ LT.0502

Taking that derivative, we have LV/ R.0515

We have - E ^- R/ LT × - R/ L.0519

It complies then that the voltage across our inductor as a function of time is just going to be VL / R, R/ L, make a ratio of 1.0529

Our negatives cancel out so V × E ^- R/ LT.0539

If you want that to put that in terms of τ again, that is a VE ^- T/ τ.0547

Those are equivalent statements, as long as τ is L/ R.0557

There is the voltage in RL circuits.0564

Let us also take a look at the rate of change of current in RL circuits.0566

Of course, we could just come back here with our DI DT and look at that because we know what that is.0570

But instead, I think it is worthwhile to go to the math once more and get a little bit more practice.0576

Let us take a look at the rate of change of current in the RL circuit.0580

DI DT is going to be the derivative with respect to time of our V / R × 1 - E ^- R/ L × T, 0586

which we can pull out V/ R our constant, V/ R × the derivative with respect to T of 1 - E ^- R/ LT.0601

Or DI DT = V/ R - E ⁻R/ LT × - R/ L which just gives you V/ L E ^- R/ LT.0613

Or in terms of τ, our time constant V/ L E ^- T/ τ.0633

Both of those are equivalent again.0646

When we only take a look at the graphs of these, we already talked about the general shape of them.0650

But looking at current in voltage graphs, let us start with current up here.0655

Current is a function of time T, we said we started with 0 current and 0659

we are going to approach some maximum level that is called Imax and just V/ R.0675

If we start at 0, we are going to approach that asymptote and we are going to do that 0677

in such a manner that it start to get mighty close to it, right about the time we hit 5 τ, 5 time constants or 5 × L/ R.0681

At that point we are more than 99% of our way to our final value.0690

If we do the same thing, taking a look at voltage here, we are going to start at our maximum voltage, 0695

Vmax or the voltage of our source of potential difference B for battery.0702

Overtime we are going to decay to 0.0710

As we do that, our voltage across our inductor, let me bring that down just a little bit better.0714

That kind of matches our graph up above.0723

We start to get pretty close to that final value around 5 τ.0727

There are some more here, why do not we do a couple example problems.0735

We will pull out some old AP exams and look at some problems from their that involve these RL circuits.0738

Starting with the 2011 exam, let us take a look at E and M problem number 2.0746

You can find that at this link or google it.0752

Take a minute, print it out.0756

Look it over, see if you can solve it yourself.0757

Hit the pause button, check it out, then come back here and we can compare solutions.0759

We have got a circuit that has a resistor, a capacitor, and inductor in the switch in there.0766

It looks like it says an experiment, when the switch is close to position S1 at time T1 and it is left there for a long time.0772

If I want to redraw my circuit when we are in that configuration A1, we have got a 9V battery, 0779

we have a 500 ohm resistor, and we have a 25 mf capacitor.0789

With the switch in that position, it looks like that is all we have to worry about and it is been on a long time.0802

Calculate the value of the charge on the bottom plate of the capacitor a long time after the switch is closed.0807

Key thing to note, usually we are solving for the magnitude of the charge.0815

Since, we are looking at the bottom plate of the capacitor, we are looking for the negative charge.0819

We have to be careful with your answer, we get our sign right.0822

We can use C = Q/ V to solve this, knowing that the charge on one of the plates is CV, which is going to be 0.025 F, 0825

25 mf × the voltage across our capacitor, the potential which after a long time when it is charged up, there is no current flowing.0838

No potential drop across the resistor, all 9V must be spread across that capacitor.0846

9V is going to give us then that Q our charge = 0.225 C.0853

We want the charge on the bottom so that is going to be -0.225 C.0861

Let us take a look at part 2 then, part 2 gives us a graph and it wants us to 0884

sketch a graph of the magnitude of the charge on the bottom plate of the capacitor as a function of time.0891

We are looking at the bottom plate of the capacitor but we are only looking at magnitudes,0899

we do not have to worry about signs and drawing our graphs upside down or anything like that.0903

Let us sketch out our axis nice and neatly here.0910

It looks like this is our time axis in seconds, this is the magnitude of our charge and it gives us a point here T1.0920

Before we close that switch, we cannot have any charge on the capacitor, it is initially uncharged.0933

The initial part should be really straightforward 0.0939

After that, we are going to have rise to our final value of 0.225 C.0944

We are going to do that, approaching that asymptote of our maximum charge.0950

It should look something like that, where we have an asymptote here of about 0.225 C.0955

It gets to this point, and if we want to sketch everything in their, right around T is approximately 5 τ.0968

Which in this case, it is going to be 5 RC, which is going to be 5 × R 500 × RC 0.025.0976

Or I come up with about 62.5s.0988

After that, you get just a touch over a minute you are almost here to your final value.0992

I would use that for my graph for part A2.0996

Moving on, let us give ourselves some more room on the next page.1002

For A3, we are given another graph.1006

Sketch a graph through the resistor of the current through the resistor as a function of time.1012

Let us make our axis again, our Y and our X.1022

Now we have our current flow through the resistor as a function of time.1036

We have got to our T1 labeled here.1041

As we start off again, our current flow before we turn anything on, that is an open circuit.1045

We are not going to have any current flow.1050

That piece is pretty straightforward.1052

Then, once we turn it on, we are going to initially have our maximum current flow as the capacitor acts like a wire, 1055

we are going to start at some high value.1061

It is just going to be V/ R, 9V/ 500 ohms is going to give us 18 ma.1064

We are going to have that exponential decay of which, maybe not the prettiest graph there 1074

but you get the idea and we start to get really close to our 0 point at around 5 τ.1081

A3, let us move on to our next page which says part B.1089

An experiment to the capacitors on charge and the switch is close to position S1 at T1 1100

and it is looped at position S2 at time T2 when we have 105 mc on the capacitor, allowing electromagnetic oscillation in the LC circuit.1106

We are going to charge up that capacitor and then close our circuits.1117

All we have is a capacitor and inductor.1120

The capacitor discharges, sending current in the inductor.1121

When it does that, it create a magnetic field in the inductor.1125

It does that like that it opposes that.1128

As that runs out, it starts putting that magnetic field energy back into current flow.1130

You are going to have this oscillating system.1135

Let us draw our circuit at this point, let us call it LC circuit because it has an inductor and a capacitor.1138

There is our inductor and this is 25 mf here.1149

We have got a 5 H inductor over there and their initial charge Q0 when our capacitor is 0.105 C.1157

What are we trying to find?1169

The energy stored in the capacitor at time T2.1170

When that happens, our energy stored in the capacitor we know the charge on it.1175

If ½ CV² is the energy of the capacitor, we know C = Q/ V.1181

Therefore, V = Q/ C so that can be rewritten as ½ × our C.1191

Instead of V², I’m going to have Q² /C².1197

Let us put Q/ C².1201

Then, UC = ½, we will have Q² /C as we simplify that, which is going to be our 0.105 C² / our capacitance 0.025 F.1204

I come up with about 0.221 J stored in that capacitor at that point in time.1224

Taking a look at B2, calculate the maximum current that will be present during the oscillations.1236

We are going to have the maximum current flow when we have depleted that capacitor.1244

When the charge on the plate of that capacitor = 0, max current when QC is equal to 0.1249

Therefore, the energy stored in the inductor is going to be 0.221 J.1252

No energy stored in the capacitor, it is all in the inductor that point.1263

The energy stored in the conductor is ½ LI².1267

Then our I² is just going to be 2 × 0.221 J/ our inductance of 5 H.1272

We want just the current not the current², so if I take the square root of both sides, I determine that the current is going to be about 0.297 amp.1284

There is our maximum current flow in the circuit.1300

Let us take a look at part B3, calculate the time rate of change of the current DI DT when the charge on the capacitor plate is 50 mc.1305

We are trying to find DI DT when the charge on our capacitor is 0.05 C .1317

I’m going to go to Faraday’s law here and state that the integral around the closed loop of E ⋅ DL 1330

is - the time rate of change of our magnetic flux, which by the way is - L DI DT.1339

Then, as we go around our E ⋅ DL, we have Q/C from our capacitor.1350

Its contribution to the electric field Q/ C must equal - L DI DT, which implies then that DI DT is going to be equal to -Q/ LC, 1356

which implies that DI DT is –Q, which is -0.05 C/ 5 H our inductance × our capacitance 0.025 F,1381

which implies then that the time rate of change of current at that point is going to be about -0.4 amps/ s.1397

I think that covers the 2011 question number 2.1414

Let us take a look at our next question moving to 2008.1419

The 2008 exam, we are going to look at the E and M portion free response question number 2,1425

where again we have a couple configurations of circuits.1431

Taking a look at part A1 here, giving us a circuit A and B are terminals to which we can attach different circuit components.1437

We are going to calculate the potential difference across R2 immediately after the switch is closed.1448

When first off, we have a 50 ohm resistor connecting A and B.1453

Let us draw that circuit first.1457

It looks like we are going to have our source of potential difference 1500 V.1460

We are going to go over here to R1, which is 200 ohms.1466

We have got a 300 ohm resistor there, that brings us back to the power supply.1472

We have another branch now that has a 100 ohm resistor.1480

We are also going to insert for this part of the problem a 50 ohm resistor.1489

200 ohms, 300 ohms.1502

At this point, we want to know the potential difference across R2.1506

I suppose I would label all these here so it is a little easier to see which one is which.1511

There is R1, R2, R3, and our special resistor.1515

What I would do here, if I wanted to find the potential across R2, 1525

is recognized that I have 2 resistors here in series that have an equal resistance of 150.1530

We just have to add them up and these are in parallel.1535

300 in parallel with 150, let us find an equivalent resistance for these 3 resistors.1540

I’m going to call that R2,3 equivalent, is going to be, our first resistance 300 × the second 150.1546

Adding those up and divided by the sum of those which is going to be 450 1558

which shall give us an equivalent resistance of 100 ohms for these 3 in that configuration.1563

Our total current flow, treating it as a 200 ohm resistor and a 100 ohm resistor in series,1570

is going to be V/ R or 1500 V/ our equivalent total resistance 300 ohms, which is going to be 5 amps.1575

Now that we have done that, that means that our potential across R2 is going to be the potential that goes through our 100 ohm resistor.1589

Equivalent resistor there so V R2 is going to be equal to our 5 amps, the current flow, × our 100 ohms or 500V.1598

That is a little better.1620

For part 2, we are going to have a 40 mh inductor in place of that 50 ohm resistor.1624

Let us take a look at that circuit.1630

We have our 1500 V source of potential difference.1634

We have our 200 ohm R1 resistor.1642

We have our 300 ohm R2.1650

Our way over here, we have our 100 ohm R3 and we have our inductor which is a 40 mh inductor.1660

We want to know the potential difference across R2 immediately after the switch is closed.1683

Immediately after the switch is closed, an inductor is going 2 act like an open.1689

It does not want any change in that state.1696

The current through the inductor at time T = 0 is 0.1699

If this entire circuit here is open, so what we have effectively, we can pretend that this is all gone 1702

and all we have is this nice simple series circuit.1711

If we want to know the voltage drop across R2, we can just do the ratio as a voltage divider1715

and I can see pretty easily that that is going to be 900V.1721

3/5 of the resistance is R2, therefore, 3/ 5 of the voltage will be dropped across R2.1725

Or they can do this mathematically, recognizing that the total equivalent resistance at T equal 0 1730

is just going to be R2 resistors in series or 500 ohms.1737

Therefore, the current is going to be V/ R or 1500 V/ 500 ohms, which is going to be 3 amp.1742

3 amp through 300 ohms is going to give us, V = IR Ohm’s law.1753

V through R2 = I2 R2 is going to be 3 amps × 300 ohms or 900 V.1761

That was not so bad.1777

Taking a look at part of 3, now we are going to replace that inductor with an uncharged 0.8 µf capacitor.1780

We are getting good at drawing these circuits so let us do it again.1789

A3, we have our 1500 V supply, we have a resistor R1 200 ohms.1793

We have another resistor R2 300 ohms.1806

Moving further to the right, we have our next branch which is R3 which is 100 ohms and we have our capacitor.1817

Which we say it was 0.8 µf or 800 nf.1831

What we want to find out is what the potential drop is across R2 right after the switch is closed.1843

Right after the switch is closed, this is going to act like a wire.1849

All we have is another circuit now or we can pretend this does not exist here.1855

Let us even erase it out of our picture.1860

At T = 0, there is our equivalent circuit.1866

The equivalent resistance between the 300 and 100 over here, R equivalent for 2,3 is going to be 300 × 1869

100/ their sum 400 or about 75 ohms.1879

Our total effective resistance for our circuit is 200 + 75 in series or 275 ohms.1888

Our total current flow is going to be V/ R which is 1500 V/ 275 ohms or about 5.45 amps.1896

Then V R2 is going to be equal to the current flow × the equivalent resistance, our equivalent 2,3 1911

which is 5.45 amps × 75 ohms which is about 409 V.1923

We answered part A1, 2, and 3.1936

For part B, it looks like we are going to be get into the graphing game.1942

The switch gets close to time T equal 0, on the axis below sketch the graphs of the current, in R3 vs. Time for each of the 3 cases.1948

The case for we have got to resistor there, the inductor, and the capacitor.1957

Let me go to the next page to give us lots a room and we will make our graph of current vs. Time.1962

There is our time, there is our current, and let us start with the case for we have got that resistor.1983

When it is a resistor, we are not going to have any time dependents so our current flow is going to be a constant.1991

Let us do that here in blue.2000

Here is our current flow through our resistor R3.2005

In the case where we happen to have a resistor in that blank spot between A and B.2012

When we put the inductor in there, remember the current flow through R3 started at 0 and then it got bigger.2019

It follow that path up until it heads for that asymptote.2028

For that, when we are going to started 0 it is going to have a higher over all current.2031

It is probably going to look something like that.2036

There is the case where we have an inductor.2040

With the capacitor in there, we start at maximum current where it started looking like a wire.2043

After a long time as it becomes charged up, it becomes an open.2050

It is going to start in its maximum value of current and it is going to exponentially decay down to 0.2053

Something like that, that will be the case where we have our capacitor in there.2062

Those would be the sketches I would put in place for current vs. Time through that R3, which is right in line with our mystery element.2066

Let us take a look at 1 more free response problem.2077

Moving to the 2005 E and M exam free response 2.2084

We have got another inductor circuit and another RL circuit.2088

Here we have resistors 1 and 2 of resistance R1 and R2 and an inductor connected to a battery and we have got a switch, of course.2092

The switch is close at time T equal 0, find the current through resistor 1 immediately after the switch is closed.2100

I like to redraw the circuits, it helps me just feel better and get a feeling for what is actually in there to draw it myself.2107

We have got our source of potential difference, positive, negative.2114

Over here, we have our first resistor R1 ± , we have R2 ±, and we have a second branch coming 2122

down over here to our inductor ± and there it has some inductance.2140

At T equal 0, find the current through resistor 1.2151

At time T = 0, the current through our inductor is equal to 0.2155

As if this branch does not exist.2160

All we have here is a straightforward series circuit.2162

The current at that point is going to be V / R is going to be our electromotive force E/ our total resistance R1 + R2.2166

There is the current through resistor 1 right after we turn that on.2177

For part B, determine the magnitude of the initial rate of change of current DI DT in the inductor.2185

The initial rate of change in the inductor.2199

What we can do is take a look at Faraday’s law as we make the big loop around here.2201

We will skip R2 as we make our loop.2207

The integral/ the closed loop of E ⋅ DL = – D φ B DT or as I fill that in, as we go this way, we see -E first + our initial I0 × R1.2209

No electric field in the inductor so nothing there and that is going to be equal to R- D φ B DT which by now we know is - L DI DT for our inductor.2227

Solving for DI DT, DI DT is going to be equal to, we have E / L we divide that out, E/ L – I is 0 R1/ L, 2246

Which implies that DI DT is going to be equal to E/ L - we can replace I₀ with what we just found up here E / R1 + R2.2271

There is our I₀ × R1/ L.2289

With a little bit of simplification, I can factor out an E/ L to say that DI DT is equal to E/ L × the quantity 1 - R1/ R1 + R2.2296

We can simplify that further if you want to.2316

I think we can make this quantity look like R2/ R1 + R2, but that looks like it is puny simplified to make.2318

I think we will call that our answer for part B.2325

In many cases here, you are more than welcome to always go through and try clean these up further 2329

but that should meet the requirements of the problem.2333

Let us take a look at part C.2339

Let us give ourselves more room on the next page.2343

Part C says, determine the current through the battery a long time after the switch has been closed.2346

As time approaches infinity a long time, the potential across our inductor is going to be 0 because it is acting like a wire at that point.2353

The current through the battery is just going to be your total current V/ R which is going to be E/ R1.2365

That is going to completely bypass R2, it is got that short around it.2375

Part C, taking a look at part D.2383

On the axis below, sketch a graph of current through the battery as a function of time.2388

We have done all the hard work there, we only have to do is fill in our graphs.2392

Let us do that, let us put our axis up here.2395

We have our current through our battery I.2406

We have our time T and initially we already found that I 0 was E/ R1 + R2.2409

We found that after a long time, you are actually up here at E/ R1.2419

That is going to be our asymptote.2426

What we have to do is fill in our graph as we approach that asymptote, something like that.2429

Finally part E, let us give ourselves some more room again.2439

Part E says, determine the voltage across resistor 2 right after you open the switch.2445

When you do that, we are going to have a new circuit that looks kind of like this.2451

We have got our resistor R2, we have our inductor L, and because the switch is open, the rest of it does not come into play.2457

I R2 is going to be equal to the current through our inductor which is just E / R1.2472

Exactly what we had right before that because that conductor wants to oppose the change.2484

It is going to keep that going as long as it can.2489

It is going to keep that current for the time being and it is going to decay as that power is dissipated in the resistor.2493

Then V R2 is going to be equal to the current through × the resistance which is going to be our current through R2 × I R2 × R2, 2498

which we just said I R2/ R1, that is E/ R 1 R2 or E × R2/ R1.2514

That should finish them up.2528

Hopefully, that gets you a great start on RL circuits.2530

Thank you so much for watching www.educator.com.2533

Make it a great day everybody. 2536