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

### Circuits II: Parallel Circuits

- Ohm’s Law is an empirical law which states that the potential drop across a resistor is equal to the product of the current through the resistor and the resistance.
- Electrical power is the rate at which electrical energy is transformed into other types of energy.
- Conventional current flows from high potential to low potential in a circuit. Electron current flows in the opposite direction.
- Current only flows in complete paths.
- Kirchhoff’s Voltage Law states that the sum of all potential drops around a close loop must be zero. This is a restatement of the law of conservation of energy.
- Kirchhoff’s Current Law states that the sum of all current entering any point in a circuit must equal the sum of all current leaving that point. This is a restatement of the law of conservation of charge.
- Voltmeters are connected in parallel with the element to be measured. They have a very high resistance.
- Ammeters are connected in series with the element to be measured. They have a very low resistance.
- Real batteries have some amount of inherent resistance, known as the internal resistance of the battery. The terminal voltage is slightly lower than the battery’s emf due to the potential drop across the internal resistance of the battery in an operating circuit.

### Circuits II: Parallel 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
- Parallel Circuits
- Multiple Current Paths
- Removal of a Circuit Element May Allow Other Branches of the Circuit to Continue Operating
- Draw a Simple Parallel Circuit
- Basic Parallel Circuit Analysis
- Example 1
- Example 2
- Example 3
- Example 4
- Combination Series-Parallel Circuits
- Circuit Doesn't Have to be Completely Serial or Parallel
- Look for Portions of the Circuit With Parallel Elements
- Lead to Systems of Equations to Solve
- Analysis of a Combination Circuit
- Example 5
- Batteries
- Electromotive Force
- Pump for Charge
- Ideal Batteries Have No Resistance
- Real Batteries and Internal Resistance
- Terminal Voltage in Real Batteries
- Ideal Battery
- Real Battery
- Example 6
- Example 7
- Example 8
- Example 9

- Intro 0:00
- Objectives 0:16
- Parallel Circuits 0:38
- Multiple Current Paths
- Removal of a Circuit Element May Allow Other Branches of the Circuit to Continue Operating
- Draw a Simple Parallel Circuit
- Basic Parallel Circuit Analysis 3:06
- Example 1 5:58
- Example 2 8:14
- Example 3 9:05
- Example 4 11:56
- Combination Series-Parallel Circuits 14:08
- Circuit Doesn't Have to be Completely Serial or Parallel
- Look for Portions of the Circuit With Parallel Elements
- Lead to Systems of Equations to Solve
- Analysis of a Combination Circuit 14:51
- Example 5 20:23
- Batteries 28:49
- Electromotive Force
- Pump for Charge
- Ideal Batteries Have No Resistance
- Real Batteries and Internal Resistance
- Terminal Voltage in Real Batteries
- Ideal Battery 29:50
- Real Battery 30:25
- Example 6 31:10
- Example 7 33:23
- Example 8 35:49
- Example 9 38:43

### 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: Circuits II: Parallel Circuits

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

*I’m Dan Fullerton, in this lesson we are going to continue to try *0003

*and make the world a better place by helping folks understand how circuits work,*0006

*specifically focusing on parallel circuits and ideal and real batteries in this lesson.*0010

*Our objectives begin to understand the behavior of series in parallel combinations of resistors *0017

*to find current voltage resistance and power.*0023

*Applying ohms law and Kirchhoff's rules to DC circuits.*0025

*Understanding the properties of voltmeters and ammeter, and ideal and real batteries, understanding how they work and what the difference is.*0029

*Let us talk about parallel circuits.*0038

*Parallel circuits have more than 1 current path.*0040

*The removal of a circuit element from a parallel circuit may allow the other elements to continue operating.*0043

*Think of the lights in a room in your house, if 1 light bulb goes out, typically everything in the house does not go off.*0049

*Instead, you go switch the 1 light bulb, it comes back on but everything else continues operating.*0055

*That is a feature of a parallel circuit.*0060

*To draw a simple one, we will start off with a source of potential difference and *0062

*then we will draw some resistors but we will make sure that we now have multiple current paths.*0067

*There is an example of a parallel circuit.*0081

*Before, we found an equivalent resistance for series elements, we added up those resistances.*0084

*For a parallel circuit 1/the equivalent resistance is equal to 1/your first resistor + 1/your second resistor and so on.*0090

*So that your equivalent resistance is always smaller than your smallest individual resistor.*0100

*You could also look at the total current flow is going to be equal to the sum of the individual currents through each branch of the circuit.*0106

*So if we call this R1, R2, R3, and R4, the current flow through here I1, I2, I3, I4, and so on.*0116

*These all have to come together to give you your total current flow.*0131

*That is really just restatement of Kirchhoff's current law .*0135

*The potential is going to be the same across each of these elements *0139

*because they are connected opposite sides of that source of potential difference.*0145

*The voltages are the same in a parallel circuit, for however many elements you might have.*0149

*Now this formula can get a little cumbersome when you are dealing with a lot of resistors.*0157

*A quick way, a little cheer formula if you only have 2 resistors in parallel to find the equivalent resistance, *0161

*you can say that the equivalent resistance is equal to your first resistor × your second resistor/ the sum of those 2 resistances.*0167

*It only works for 2 resistors in parallel.*0176

*You got more than 2, you have got to go back to your original formula here.*0179

*Let us take a look at basic parallel circuit analysis and we are going to do this with a VIRP table again.*0184

*We have a 12 V source of potential difference and three 2000 ohm resistors.*0189

*We will make our VIRP table and label our elements.*0195

*Let us call this R1, R2, and R3.*0201

*We will have a line for R1, R2, or R3, and our total.*0206

*Maybe might not need this table but I think it will play.*0221

*We will start by filling in what we know again.*0225

*We know our total voltage is 12V, we know our 3 resistances 2000, 2000, 2000.*0228

*And because it is a parallel circuit, right away I can look at it and tell the potential drop across each of these resistors is also 12V.*0240

*This ensures we have 12V on either side of our battery at any point on the wire has the same potential.*0247

*Therefore, these are all 12.*0253

*Once I have done that, I can find my current flow I = V/ R.*0257

*In that case, 12/ 2000 is going to be about 0.006.*0262

*R2 will be the same and the current through R2 will be the same and R3.*0267

*Because all of these currents are going to add together, our total current flow is going to be 3 × that *0272

*or the sum of those which is 0.018 amps or 18 milliamps.*0282

*We can find our equivalent resistance.*0288

*Before I do that though, we could have used the 1/ R equivalent = 1/ R1 + 1/ R2, that would work.*0290

*And I can also look by inspection and say you know I have to be certain my total resistance*0298

*because it is a parallel circuit has to be less than 2000 ohms,*0304

*because it has to be smaller than the smallest individual resistor in that parallel configuration.*0307

*R = V/ I 12/ 0.018 is going to give me a resistance of 667 ohm which is less than 2000.*0313

*That checks out.*0323

*To find our power again, any of your favorite power formulas V × I, I² R, V²/ R.*0325

*If you happen to be in that kind of mood to fill in your power.*0331

*0.072 W or 72 mw, same here, same here.*0334

*Find our total power, I could add all those up or I can do the same thing with the information from these rows.*0341

*Any of my favorite power formulas to get the same answer, 216 W or 0.216 W.*0347

*Basic parallel circuit analysis.*0355

*Let us do another one, a 15 ohm resistor R1 and a 30 ohm resistor R2 *0360

*are to be connected in parallel between points A and B, in a circuit containing a 90V battery.*0363

*Complete the diagram to show the 2 resistors connected in parallel.*0369

*Let us do that first, we want to connect these in parallel.*0373

*We need to make 2 current paths.*0378

*We will call this one R1, this will be R2.*0382

*Determine the potential difference across resistor R1 and calculate the current resistor R2.*0387

*Et us go to our VRIP table and our elements are R1, R2, R total.*0392

*Our total potential 90V.*0412

*We know that R1 is 15 ohms, it tells us R2 is 30 ohms.*0415

*As we look at this, because the battery is connected directly to either side of those resistors, *0420

*we can tell that the potential across each one is also 90V.*0427

*To find current here, that is V/ R 90/ 15 that is just going to be 6 amps, 90/ 3 that is going to be 3 amps.*0430

*Which means our total current is the sum of those in a parallel circuit or 9 amps.*0438

*Now, I can find my equivalent resistance R = V/ I or 10 ohms .*0443

*Which checks out, it is less than my smallest individual resistor in the parallel configuration.*0448

*All I have to do is find my powers to finish up my table.*0454

*V × I, this will give me 540W, 270W, and 90 × 9, 810W.*0458

*Or I could just added does up.*0466

*Once I have my VIRP table, I can answer the question that is asking me.*0468

*Determine the potential difference across resistor R1.*0471

*There it is, 90V.*0475

*V R1 = 90V.*0478

*And calculate the current in resistor R2.*0482

*Current through R2, I look up right on my table, there it is 3 amps.*0485

*That is straightforward.*0492

*Looking more at this equivalent resistance.*0494

*Let us take an example with 3 identical lamps connected in parallel with each other.*0497

*If the resistance of each lamp is X, what is the equivalent resistance of this parallel combination?*0501

*Right away, we know the answer has to be less than X because they are in parallel.*0506

*We can get rid of that one right away.*0510

*1/ our equivalent resistance is going to be 1/ X + 1/ X + 1/ X which is going to be equal to 3/ X, *0518

*which implies then that 1/ R equivalent of 3/ X then R equivalent must be equal to X/ 3 ohms.*0528

*The correct answer must be B.*0538

*How about a circuit that has an extra resistor?*0544

*Let us look at one that has 4 parallel resistors.*0547

*We got a 12V battery, we want to know the current measure by ammeter A which is the current flowing through this resistor,*0551

*the circuit’s equivalent resistance and how much power is dissipated in the 36 ohm resistor.*0559

*Let us label our resistors R1, R2, R3, R4 and we can make our VIRP table.*0564

*Those are R1, R2, R3, R4, and total.*0579

*Now that masterpiece of artwork is on the paper, we can start filling in the things that we know.*0602

*Our total potentially we have 1 battery here 12V.*0607

*We know R1 is 6, R2 is 12, R3 is 36, and R4 is 18 ohms.*0611

*We also can tell just by looking at these that our equivalent resistance is going to be less than 6 ohms, *0619

*has to be less than our smallest resistor.*0626

*Now it is a parallel circuit so all of the potentials have to be the same.*0630

*We can fill those in 12, 12, 12, 12.*0634

*And since I know 2 things in these rows, I can solve for the others.*0639

*Current I = V/ R, that is going to be 2 amps through R1.*0642

*By the way, there is the answer.*0648

*What is the current measured by ammeter A?*0650

*It is the current through R1 which is going to be 2 amps.*0651

*I through R2 is going to be 12/ 12 or 1, 12/ 36 that is going to be 1 /3 .33, 12/ 18 2/ 3 0.67.*0657

*And all of our currents add up to our total current here which is going to give us 4 amps.*0667

*The equivalent resistance then, R = V/ I, 3 ohms it is less than 6.*0672

*Everything is good.*0678

*Let us calculate our powers while we are here, it is good practice.*0681

*12 × 2 gives us 24 W, 12 W that will be 4 W.*0683

*12 × 2/3 8 W, and we can at all those up or just 12 × 4 should give us 48 W.*0689

*I will check, 24, 12, 4, and 8 is 48 W.*0695

*We got our VIRP table, we can answer the rest of the questions.*0699

*What is the circuits equivalent resistance?*0702

*There it is, 3 ohms.*0705

*And how much power is dissipated in a 36 ohm resistor?*0707

*That would be 4 W.*0710

*Let us take a look at an ammeter in a parallel circuit.*0716

*The circuit diagram shown below, ammeter A1 reads 10 amps, what is the reading of ammeter A2?*0719

*To analyze this, it is a parallel circuit, try and vision those ammeters are not there and you can see it is simple, 2 resistor parallel circuits.*0726

*Let us make our VIRP table and we have R1, R2, and total.*0735

*Let us fill in what we know.*0756

*We know our total current flow must be 10 amps because that is what the ammeter tells us.*0758

*We have got 10 amps there, we will call that R1 and that R2.*0764

*R1 is 20 ohms, R2 is 30 ohms.*0769

*Let us see, what else do we know here?*0775

*We have got 10 amps through the entire circuit, 20 and 30.*0778

*Let us do this one a little bit differently.*0787

*If this is 20 and 30, let us find out our equivalent resistance just calculating it the old fashioned way.*0789

*Our equivalent is going to be R1 R2/ R1 + R2 which will be 20 × 30/ 20 + 30 which is 50, that is going to be 600/ 50.*0795

*For a total equivalent resistance of 12 ohms.*0809

*V = I × R, that means it is going to be 120V.*0815

*This will also be 120V, it is a parallel circuit.*0818

*I = V/ R, 6 and 4 amps.*0822

*The question we are asked is what is the reading of ammeter A2?*0827

*Ammeter A2 is the current flowing through R1 which is 6 amp.*0831

*The answer to the question that is asking, 6 amps.*0836

*Not all circuits are purely series or parallel, some× we have combination circuits.*0842

*Let us take a look at an example there.*0847

*A circuit can be series in parallel and have portions of each.*0850

*First, look for portions of the circuit that have parallel elements.*0855

*Replace the parallel resistors with the equivalent single resistor to simplify the circuit down and down, *0858

*until you get something it is a little bit more recognizable that you can deal with.*0863

*You can analyze with a VIRP table if you want, to help keep track of things.*0868

*But it is not quite as straightforward with a VIRP table anymore.*0871

*Then, work back to your original circuit using Kirchhoff's laws until you know the current voltage and *0875

*resistance of all the individual circuit elements.*0881

*Oftentimes, you are going to end up with a system of equations to solve those.*0882

*Sometimes the math can be a little bit involved with these combination circuits.*0886

*Let us take a look here.*0891

*We have a combination circuit here, R1, R2, R3, and R4.*0893

*I'm going to start by making my VIRP table just to help track of what is going on.*0899

*We will start off VIRP and we have our 4 resistors, R1, R2, R3, R4, and our total.*0904

*I always like to start by filling in what I know, what we are certain about.*0933

*Only 1 battery, so our total potential is 10V.*0937

*We know our resistances, we have 20 ohms, 30 ohms, R3 is 50 ohms, and R4 is 20 ohms.*0941

*And I think that about covers what we know to begin with.*0952

*As I look at the circuit, I see we have almost a series circuit except we have these 2 elements in parallel.*0956

*What I'm going to do is, I'm going to redraw the circuit replacing these 2 resistors *0962

*with their equivalent single resistors so that I have a series circuit to analyze, to help me understand what is going on.*0966

*The way I will do that is, we still have our battery 10V, we have our 20 ohm resistor up here R1.*0974

*We have our equivalent 2 and 3, which we are going to call R2, 3.*0985

*We do not know its value yet.*0990

*And we have R4 which is 20 ohms.*0993

*To find R2, 3, I’m going to find the equivalent resistance of R2 and R3.*0998

*R2, 3 is going to be equal to the product of the resistances.*1004

*30 × 50 ohms/ the sum of their resistances, 30 + 50 is 80.*1009

*This trick only works when you have 2 resistors in parallel.*1015

*And I come up with about 18.75 ohms.*1019

*I can find our total resistance here if I know R1 and R4, 20 and 20.*1025

*The equivalent of these two is 18.75.*1032

*Our total will be 20 + 20 + 18.75 or 58.75 ohms.*1035

*Current is V/ R so 10/ 58.75 that is going to be about 0.170 amps.*1043

*Which means we have 0.170 amps here, that has to be our current flowing through R1.*1051

*There is no other place for it to go by Kirchhoff's current law.*1058

*That will be 0.170.*1061

*We also have to have that 0.170 going through R4.*1065

*When we get here, the current is going to branch and the total between R2 and 3 has to equal 0.17.*1070

*We do not know exactly what those values are yet.*1076

*We will see what else we can learn from the circuit for now.*1079

*I know 2 things in these rows, I can find the potential drop across R1 and R4.*1083

*Let us do that, V = IR, 20 × 0.17 that is going to be a potential drop of 3.4V.*1088

*The same math down here for R4, 3.4.*1095

*That means that we are left with, that is 6.8 V so we are left with 3.2 V to drop across these.*1100

*Let us just draw the same here.*1107

*I'm going to call this point, ground.*1108

*Usually not a physical connection, just a point we are going to call 0V to simplify our analysis.*1111

*That means this side of the battery must be 10V higher.*1118

*We dropped 3.4 across R1, that means we are left with 6.6 V here.*1122

*We also had dropped 3.4V across R4 so that means if this side is 0, this side must have been 3.4.*1130

*6.6 V to 3.4, we must have dropped 3.2V across R2 and across R3.*1139

* We know their potential difference.*1146

*We can find the current through those.*1148

*I = V/ R so 3.2/ 30 is going to give us 0.107 amps, going through our 30 ohm resistor.*1151

*3.2/ 50 is going to be about 0.064 amps which does not quite add up to 0.17 just by a rounding error, *1163

*but that is really has to do with tracking our significant figures and how accurately we are calculating these.*1176

*Roughly these two add up to 0.17.*1182

*The current splits and then it comes back to be 0.17 amps again.*1185

*For our powers, your choice of formulas.*1190

*Just going down, I get 0.578 W, 0.342 W, 0.205 W, 0.578 W.*1193

*Again, 10 × 0.17 is going to give me 1.7 W, which should be the sum of all of those, and it is.*1202

*Analysis of a combination circuit. *1212

*The VIRP table did not tell us everything we needed to know but it was kind of useful for tracking some of the characteristics of our circuit.*1215

*Let us start getting into some circuits and add a little bit more challenge to them.*1224

*Here we have 2 voltage sources and we are asked to find the current flowing through R3, if R3 has a value of 6 ohms.*1231

*What is the power dissipated in R3?*1239

*R3, let us label that, we know that is 6 ohms.*1243

*I'm going to start by picking a point on my circuit to call ground or 0V.*1249

*I’m going to do that right here.*1253

*That is 0V and if I do that that means the side must be 12V.*1256

*That is also 0V so this must be 16 V over there.*1261

*I’m also going to go and start defining some of my currents.*1267

*I’m going to pick a direction, if I'm wrong we will just a negative value.*1270

*It really does not matter what direction you write.*1273

*Let us assume that we have this current through R1, I'm going to call I1 and call that direction positive.*1275

*I'm going to call this direction positive 4 I2, the current through R2.*1282

*We will label this is I3, the current through R3.*1287

*If I analyze the circuit right at this point here, I can use Kirchhoff's current law to give me an equation.*1293

*It says I1 + I2 is going to equal I3.*1299

*That can be handy.*1305

*I'm going to go and I'm going to start making loops using Kirchhoff's voltage law to see if I can get some other equations that will help me solve the system.*1307

*If I start down here and I make a loop this way, I see the negative sign of the 12V battery first, as I start.*1316

*I'm going to write -12V + the voltage drop across R1 is 8 I1, 8 I1 + 6 I3, brings me back to 0.*1323

*That is going to equal 0.*1338

*I also know that I1 + I2 = I3, which implies then because I1 + I2 = I3, I can write this as -12 + 8 I1 + 6.*1341

*I’m going to replace I3 with I1 + I2 = 0.*1358

*Or 14 I1 8 I1 + 6 I1 + 6 I2 = 12.*1365

*I'm going to call that my equation number 1.*1379

*14 I1 + 6 I2 = 12.*1382

*I got that from applying Kirchhoff's voltage law around this loop and also throwing in Kirchhoff's current law *1384

*from that point where all my currents are coming together.*1387

*Let us take a different loop and maybe I will start here and we will go around this way.*1394

*As I go past the battery, I see the negative side first.*1400

*I'm going to write - 16 + the drop across R2 which is going to be 12 I2 + 12 I2 + 6 I3 *1403

*brings me back to my starting points so the entire sum of voltage drop across a closed loop must be 0.*1417

*Then we are going to use what we learn up above I1 + I2 = I3 again.*1423

*In order to rewrite this as -16 + 12 I2 + 6 ×, replacing I3 with I1 + 2 = 0, *1434

*which implies then as I multiply this out, that is 6 I1 + 18 I2 is going to equal 16.*1451

*We will call that equation 2.*1467

*We got to find some way to combine those.*1471

*A bunch of different things you can do but the easiest one I see is as I look at 1, if I happen to multiply the left side and the right side by -3, *1474

*I keep that equality but then I have got some manipulations I can do pretty easily with number 2.*1483

*I'm going to come over here to the left and I'm going to multiply this side by -3 and that side by -3.*1490

*The reason I'm doing that is I want to cancel out the I2.*1496

*14 × -3 is going to be -42.*1500

*I want to make sure that I track what I'm doing.*1504

*I’m going to take 1 and multiply it by -3 to get 14.*1507

*That is going to be -42 I1 -18 I2 = 12.*1513

*Right down below it, I'm just going to rewrite my equation 2.*1521

*6 I1 + 18 I2 = 60.*1526

*A little bit of math trick here.*1536

*If the left side = the right side for this equation, the left side = the right side for that equation, which they do that.*1539

*It is what the equal signs means.*1545

*That means we should be able to add both left sides, add both right side and maintain an equality.*1547

*That is what I'm going to do realizing that I'm going to get rid of I2 in this equation.*1551

*When I do that, the left hand side I end up with the -36 I1 must be equal to.*1557

*I did not multiply this side by -3.*1567

*-3 × 12 will be -36.*1574

*-36 I1 = -20.*1577

*I can solve for I1 then to find that I1 is 20/ 36 or 0.556 amps.*1581

*Now that I have got that, I can go plug that back into my equation number 2 again, in order to find out what I2 is.*1591

*Let us plug that in the 2 and I will do that over here if I have 6 I1 + 18 I2 = 16, that implies then that 6 × I1 0.556, + 18 I2 = 16.*1598

*A little bit more math there and I come up with 18 I2 = 12.664, *1621

*which implies then that I2 must be 12.664 divided by 18 or about 0.703 amps.*1634

*We have got I2, we have got I1, to find I3, that is just the sum of I1 and I2.*1644

*I3 = I1 0.556 + I2 0.703 and I find that I3 is equal to 1.26 amps.*1651

*Find the current flowing through R3.*1665

*We did that, that answer is right here 1.26 amps.*1668

*Finally, what is the power dissipated in R3?*1673

*To find the power dissipated in R3, a couple of ways we can do that but the simplest that I see at the moment *1676

*is the power through 3 is going to be I² R just 1.26² × that 6 ohms, or 9.52 W.*1683

*As you can see, these couple of voltage source problems can get a little bit messy with systems of equations and you can have more or less.*1704

*And there is a lot of other ways to solve these besides what I did with this little math trick.*1711

*Regardless though, you are just doing algebra once you have written these Kirchhoff's current law and *1716

*Kirchhoff's equivalent equations, eliminating variables and solving for unknowns.*1720

*Let us take a look at batteries.*1729

*A solar battery which is a combination of cells, provides a potential difference often× referred to as an electromotive force or EMF.*1732

*It is important to note, electromotive force is not a force, it is a potential difference.*1739

*A battery can be thought as a pump for charge raising it from a lower potential to a higher potential.*1744

*Ideal batteries, these magic batteries we use in so many of our circuits have no internal resistance.*1750

*In practicality, it is really not that simple.*1757

*Real batteries have some amount of resistance to the flow of charge within the battery itself *1759

*which we call the internal resistance given the symbol R sub I.*1767

*In real batteries, the terminal voltage is going to be slightly lower than the battery’s EMF because you *1772

*lose some of that potential difference due to a voltage drop across the internal resistance of the battery.*1778

*You love to have that internal resistance as low as possible but it always exist there to some extent.*1784

*Let us take a look at an ideal battery.*1789

*We have current flowing from A to B through our battery, positive and negative.*1793

*There is a resistance, the voltage across our battery we call Δ V which is going to be the voltage of B - the voltage of A, *1797

*which implies that the voltage of the battery is going to be equal to its EMF script E which is also terminal voltage VT.*1810

*Pretty straightforward, same sort of battery we have been dealing with.*1821

*The real battery gets a little bit more complicated because we have this internal resistance inside the battery.*1825

*Potential difference of our battery here is still Δ V or the voltage at B - the voltage at A.*1833

*The voltage on our battery is equal to IR as well, which now we have to worry about the EMF *1843

*of our battery - I × our internal resistance which is what we call the terminal voltage for a real battery.*1851

*The terminal voltage is the EMF - the voltage drop across that internal resistance of the battery.*1860

*Let us take a look at a problem involving these.*1868

*The terminal voltage of a real battery is 15 V.*1873

*If the battery is an EMF of 18V and supplies 10 W of power to resistor R, find the value of R and RI, the internal resistance.*1876

*I can analyze this with a VIRP table because really it looks like a simple 2 resistor series circuit.*1885

*The only difference is RI, one of the resistor is inside the battery.*1893

*Let us make our VIRP table and our resistors are RI, the internal resistance R, and we have a ρ for our total.*1897

*Filling in what we know.*1918

*We know that we have a total EMF of 18V.*1921

*Let us see, we know that the voltage drop across R, the terminal voltage of the battery is 15V which means the voltage drop across RI must be 3V.*1926

*Let us see, what else do we know in here.*1936

*We have 10W of power supplied to resistor R so that must be 10W.*1939

*Once we know that is 10W, power = V × I, that means our current must be 0.67 amps.*1944

*It is a series circuit so we have the same current everywhere throughout our circuit.*1953

*We can start to find our resistances.*1957

*R = V/ I 3/0.67 is going to give us 4 ½, 15/ 0.67 will give us 22 ½.*1960

*Their total resistance, we can add these up or 18/0.67.*1968

*Either way, we get 27.*1972

*Just finishing our VIRP table to powers.*1975

*V × I, 3 × 2/3 is going to be 2, and 18 × 2/ 3 is going to give us 12.*1978

*Of course, our powers add up.*1984

*The question, find the value of R and RI.*1986

*I have no problem, I did that a little bit ago.*1990

*For R and RI, we found that we have 22.5 ohms and 4.5 ohms.*1993

*Looking at another internal resistance problem.*2003

*A 50 ohm and 100 ohm resistor are connected as shown to a battery of EMF of 40V and internal resistance of R.*2006

*Find the value of R if the current in the circuit is 1 amp.*2013

*What is the battery’s terminal voltage?*2017

*Let us see here, as we start I'm going to take a look and see if I can simplify this a little bit.*2021

*I’m going to redraw this with an equivalent resistance here, we have got 100 and 50.*2027

*Let us draw this as our potential difference of 40 V.*2032

*We have our R there and we will put in a our R1, 2 there which is going to be equal to 100 × *2038

*50/ their sum 150 which is about 33 ohms, which works that is less than either one of those.*2050

*There is R.*2056

*What I can do is, I can take a look and use KVL as we go around the loop, the current going this way.*2059

*As I apply Kirchhoff's voltage law, I see I will get -40 starting at this point, + IR + 33 I, gets me back to my starting 0.*2068

*Which implies then as I look at this, that 33 + R × I = 40.*2084

*We know the current in the circuit is 1 amp.*2095

*If our total is 1 amp, that means I is 1, that means 33 + R = 40.*2098

*R, our internal resistance must be 7 ohms.*2107

*We found that is 7 ohms.*2111

*What is the battery’s terminal voltage now?*2114

*Our voltage across R is going to be IR which is 1 amp × 7 ohms or 7 V.*2120

*Our terminal voltage which is the EMF - IR is going to be 40V -7V or 33 V for the terminal voltage of our battery.*2129

*Let us take a look at another circuit.*2145

*This one that has some meters in it.*2150

*We have got 2 voltage sources and given the schematic, determine the reading of both the ammeter and the voltmeter.*2152

*Let us start by taking a look at our circuits and see what we can define here.*2160

*I'm going to set this point down here, I'm going to call that my ground or 0V, which means this must be 30, that point must also be 30, it is on the same wire.*2164

*The ammeter should not affect it.*2173

*If this is 0 down here too, then this point must be -5V.*2176

*We can do that.*2181

*Let us start defining a couple of our currents.*2183

*Let us call that current 1 through R1, that is going to be.*2185

*Voltmeter does not affect our circuit, you should have a very high resistance so no current flowing up there.*2190

*They are minimal.*2194

*This is I1 as well.*2196

*This we could call I2 through there, which means I3 heads down through there.*2198

*We want a reading of both the ammeter and the voltmeter.*2208

*As I run KVL around the loop, I'm going to start here and I see -30 first + 10 ohms × I1, the voltage drop across R1, + 20 I1.*2211

*I see a -5 and I get back to 0.*2233

*I have 30 I1 must be equal to 35 or I1 = 35/ 30 or 1.17 amps.*2236

*There is the current I1 and let us see here.*2249

*If I have got 30V here and I1 is 1.17 amps, that means that I’m dropping 10 × I1, which is dropping 11.7V across here.*2253

*If I had 30 and I dropped 11.7, that means I have got 18.3V here at this point.*2265

*If I have got 18.3V here and I know I have 0V here, because it is attached to what I called 0, *2273

*my ground over there, the voltmeter reading must be the difference from 1 side to the other or 18.3V.*2283

*There we go, that is a little bit near.*2305

*This was actually a lot more straightforward than solving the entire circuit.*2307

*Terminal reading of the ammeter and voltmeter.*2311

*The ammeter must read I1 1.17 amps.*2313

*The voltmeter reads 18.3V.*2317

*Let us take a look at one last problem here.*2322

*3 resistors 4 ohms, 6 ohms, 8 ohms, are connected in parallel in an electric circuit.*2324

*The equivalent resistance of the circuit is, we know with parallel resistors, *2329

*the equivalent resistance is always smaller than the smallest individual resistor so it has to be less than 4 ohms.*2334

*There is our answer right there, A.*2342

*Thank you so much for watching www.educator.com.*2345

*We will see you again real soon, make it a great day.*2347

1 answer

Last reply by: Professor Dan Fullerton

Sun Jan 24, 2016 3:24 PM

Post by Shehryar Khursheed on January 24 at 10:22:41 AM

If you had three lightbulbs connected in a parallel curcuit and then you remove one bulb, will the other two get brighter or will they remain with the same brightness?

1 answer

Last reply by: Professor Dan Fullerton

Mon May 18, 2015 10:33 AM

Post by Arjun Srivatsa on May 18, 2015

Hi,

How do you determine whether to add or subtract voltages when using Kirchhoff's voltage law. I cant seem to determine when to add the element or subtract it

1 answer

Last reply by: Professor Dan Fullerton

Sun Apr 19, 2015 6:03 PM

Post by Patrick Jin on April 19, 2015

Hi!

For the analysis of the combined circuits, the voltage does not quite add up to total of 10. 3.4 + 3.4 + 3.2 + 3.2 = 13.2 Is there a reason that these numbers do not match?

1 answer

Last reply by: Professor Dan Fullerton

Tue Apr 7, 2015 7:45 PM

Post by Thadeus McNamara on April 7, 2015

at 38:01, when you got that the voltmeter measured a drop of 18.3 volts, does that include the drop from the 5 V battery?

Would an alternate way to find the voltmeter measurement be to find the individual voltage drops of the Resisto 2, Resistor 3, and the 5 V battery? and then find the summation of those 3 voltages?

2 answers

Last reply by: Thadeus McNamara

Tue Apr 7, 2015 2:39 PM

Post by Thadeus McNamara on April 7, 2015

for example 5, how do you how many I variables you need? For example, you used I1 I2 and I3, but I'm no sure why you wouldn't need I4 and I5 at the bottom of the circuit, I4 going to the left and I5 going to the right

1 answer

Last reply by: Professor Dan Fullerton

Wed Feb 11, 2015 5:53 AM

Post by Jingwei Xie on February 10, 2015

Hi,

I am also confused by #8. Doesn't R3 use the voltage of the 30V battery as well? Wouldn't I2 be more than 0.25A? It seems like the voltmeter is measuring the potential difference of R2 (23.3V using V=IR) plus that of the 5V battery, since R3 is in parallel with the 5V battery. Did I misunderstand anything? Thank you so much!!

1 answer

Last reply by: Professor Dan Fullerton

Wed Oct 15, 2014 2:11 PM

Post by Ram Manohar Oruganti on October 15, 2014

Hi Dan,

I have a question about Example 8. If we compute I2 it becomes 0.25A and I1 is 1.17A. This means the KCL is not satisfied at the junction between R2 and R3. How come?