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

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

### 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
- Objectives
- Inductors in Circuits
- 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
- Voltage in RL Circuits
- Rate of Change of Current in RL Circuits
- Current and Voltage Graphs
- Example 1
- Example 2
- Example 3

- 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

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

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

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

Electric Fields | 1:19:22 | |

Gauss's Law | 52:53 | |

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

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

Conductors | 20:35 | |

Capacitors | 41:23 | |

II. Current Electricity | ||

Current & Resistance | 17:59 | |

Circuits I: Series Circuits | 29:08 | |

Circuits II: Parallel Circuits | 39:09 | |

RC Circuits: Steady State | 34:03 | |

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

III. Magnetism | ||

Magnets | 8:38 | |

Moving Charges In Magnetic Fields | 29:07 | |

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

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

The Biot-Savart Law | 21:50 | |

Ampere's Law | 26:31 | |

Magnetic Flux | 7:24 | |

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

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

Inductance | 6:41 | |

RL Circuits | 42:17 | |

LC Circuits | 9:47 | |

V. Maxwell's Equations | ||

Maxwell's Equations | 3:38 | |

VI. Sample AP Exams | ||

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

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

### Transcription: 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 resistors*0036

*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 divider*1715

*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

1 answer

Last reply by: Professor Dan Fullerton

Tue Feb 2, 2016 8:58 AM

Post by Shehryar Khursheed on February 1, 2016

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?