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 I: Series 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 I: Series 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
- Ohm's Law Revisited
- Example 1
- Example 2
- Example 3
- Example 4
- Electrical Power
- Example 5
- Example 6
- Example 7
- Electrical Circuits
- Closed-Loop Path Which Current Can Flow
- Typically Comprised of Electrical Devices
- Conventional Current Flows from High Potential to Low Potential
- Circuit Schematics
- Complete Conducting Paths
- Voltmeters
- Measure the Potential Difference Between Two Points in a Circuit
- Can Remove Voltmeter from Circuit Without Breaking the Circuit
- Very High Resistance
- Ammeters
- Measure the Current Flowing Through an Element of a Circuit
- Very Low Resistance
- Put Ammeter in Correctly
- Example 8
- Example 9
- Example 10
- Example 11
- Series Circuits
- Kirchhoff's Laws
- Example 12
- Example 13
- Basic Series Circuit Analysis
- Example 14
- Example 15
- Example 16
- Example 17

- Intro 0:00
- Objectives 0:08
- Ohm's Law Revisited 0:39
- Relates Resistance, Potential Difference, and Current Flow
- Formula
- Example 1 1:09
- Example 2 1:44
- Example 3 2:15
- Example 4 2:56
- Electrical Power 3:26
- Transfer of Energy Into Different Types
- Light Bulb
- Television
- Example 5 3:49
- Example 6 4:27
- Example 7 5:12
- Electrical Circuits 5:42
- Closed-Loop Path Which Current Can Flow
- Typically Comprised of Electrical Devices
- Conventional Current Flows from High Potential to Low Potential
- Circuit Schematics 6:26
- Three-dimensional Electrical Circuits
- Source of Potential Difference Required for Current to Flow
- Complete Conducting Paths 7:42
- Current Only Flows in Complete Paths
- Left Image
- Right Image
- Voltmeters 8:25
- Measure the Potential Difference Between Two Points in a Circuit
- Can Remove Voltmeter from Circuit Without Breaking the Circuit
- Very High Resistance
- Ammeters 9:31
- Measure the Current Flowing Through an Element of a Circuit
- Very Low Resistance
- Put Ammeter in Correctly
- Example 8 10:24
- Example 9 11:39
- Example 10 12:59
- Example 11 13:16
- Series Circuits 13:46
- Single Current Path
- Removal of Any Circuit Element Causes an Open Circuit
- Kirchhoff's Laws 15:48
- Utilized in Analyzing Circuits
- Kirchhoff's Current Law
- Junction Rule
- Kirchhoff's Voltage Law
- Loop Rule
- Example 12 16:58
- Example 13 17:32
- Basic Series Circuit Analysis 18:36
- Example 14 22:06
- Example 15 22:29
- Example 16 24:02
- Example 17 26:47

### 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 I: Series Circuits

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

*I'm Dan Fullerton and in this lesson, we are going to start talking about series circuits.*0003

*Our objectives include understanding the behavior of both series*0009

*and parallel combinations of resistors in order to find the current voltage resistance and power dissipated.*0011

*We will continue to work with parallel circuits into our next lesson.*0017

*Applying Ohm’s law in Kirchhoff's rules to DC circuits.*0020

*Determine unknown circuit characteristics.*0024

*Understanding the properties of voltmeters and ammeters.*0027

*Understanding the properties of both ideal and real batteries and we that will carry into our next lesson as well.*0030

*Let us start by revisiting Ohm’s law.*0037

*Ohm’s law relates resistance potential difference and current flow.*0040

*V = IR or you may see it written as R = V/ I or I = V/ R.*0044

*A material that obeys this is known as in Ohmic material.*0052

*It is an empirical law but you can do a pseudo derivation that comes close to approximating what and where it really comes from.*0056

*Although it is not a pure law of physics, it is extremely useful in analyzing circuits.*0063

*Let us start with a couple of simple examples to get us going.*0071

*The current on a wire is 24 amps when connected to a 1.5V battery, find the resistance of the wire.*0073

*R = V/ I which is 1.5 V/ 24 amps or 0.06250 ohms or we could write that as 62.5 milliohms.*0082

*Nice, straightforward, quick example.*0099

*Or slightly differently, in a simple electric circuit at 24 ohm resistor is connected across a 6V battery, what is the current in the circuit?*0101

*This time we are given resistance and potential and we asked to find current flow.*0111

*I = V/ R which is 6V/ 24 ohms or 0.25 amps, which we can also write as 250 ma.*0117

*A constant potential difference is applied across the variable resistor held at constant temperature,*0135

*which graph best represents the relationship between the resistance of the variable resistor and the current through it?*0141

*V = IR or I = V/ R, if we wanted to put that in a form that is similar to Y = NX the equation of the line.*0147

*It does not really work that way.*0158

*If we look at our Y, it is equal to a constant in this case.*0161

*A constant potential difference divided by resistant.*0165

*We are looking for an inverse relationship.*0168

*The correct answer here must be number 1.*0170

*And example 4, the graph shows the relationship between the potential difference V across a resistor and the current I, through that resistor.*0178

*Through which interval does the resistor obey ohms law?*0186

*Where it obeys Ohm’s law, where it is known as an ohmic material is where we have this linear relationship.*0190

*Here from B to C, because that is where our IV graph is linear.*0196

*Current flowing through a circuit causes a transfer of energy into different types.*0208

*The rate at which electrical energy is transforming the other types is the electrical power dissipated.*0211

*For example of a light bulb, electrical energy is converted to light and heat.*0218

*When a television electrical energy becomes light, sound, and heat.*0222

*All examples of energy conversions.*0226

*We can look at this analytically, we take a look at a toaster.*0229

*A 110V toaster oven draws a current of 6 amps at its highest setting converting electrical energy into thermal energy.*0233

*They can use some nice tasty toast.*0240

*What is the toaster’s maximum power rating?*0242

*Power is current × potential which will be 6 amps × 110V or 660W.*0245

*And we have also talked about power = I² R or power = V²/ R.*0257

*An electric iron operating 120V draws 10 amps of current, how much heat energy is delivered by the iron in 30 seconds?*0267

*Let us start off by finding the power that is going to be current × voltage or 10 amps × 120 V, which is 1200 W.*0276

*But we do not want the power, we want to know how much energy.*0289

*Energy is power × time which will be 1200 W, × that 30 seconds which is 36,000 J or 36 kJ.*0293

*The potential drop of 50V is measured across a 250 ohm resistor, what is the power developed in the resistor?*0314

*We will use V² / R because we are given potential and resistance, that will be 50V²/ 250 ohms which is 10 W.*0321

*Let us try and put this all together to start making some electrical circuits.*0339

*An electrical circuit is a closed loop path through which current can flow.*0343

*If you do not have a closed loop path, you are not going to get any current flow.*0347

*It has to be at closed loop.*0350

*It can be made up of most any materials but typically what we are talking about are things like wires, batteries, resistors, switches.*0352

*We will talk about capacitors in a little bit, inductors towards the end of the E and M course.*0358

*Conventional current flows from high potential to low potential, the direction a positive charge would move.*0363

*Keep in mind, in most circuits it is actually electrons that are moving which are going in the opposite direction from low to high potential.*0371

*Therefore, electron current is in the opposite direction of what we call positive conventional current.*0377

*As we talk about circuits, we are going to talk about symbols.*0386

*We are going to represent them with schematics on paper,*0389

*because trying to draw 3 dimensional circuit on paper and making it look really good is very difficult.*0392

*We have come up with these symbols to help us diagram these.*0397

*You can see much more easily how they operate.*0401

*Lines represent wires and these symbols over here are fairly standard symbols for different elements in the circuit.*0404

*Things like the voltaic cells, batteries, cells, voltage sources, often× use the same symbols.*0410

*A switch whether it is open or closed, a voltmeter and ammeter, a resistor look like Charlie brown shirt.*0419

*A variable resistor has an arrow that goes to that Charlie brown squiggly line thing.*0425

*A lamp is just a resistor that also gives off light and heat, as do all resistors but it is designed to do that.*0431

*Sometimes, you will see a lamp portrayed as a resistor.*0438

*Other times, you will see the lamp symbol like this in the circuit schematic.*0441

*In order for current to flow, the first thing you need is a source of potential difference.*0446

*Those voltaic cells, batteries, power supplies, any of those are required in order to have a circuit with current flow.*0451

*If you do not have a potential difference, you are not going to have any current flow.*0457

*Current only flows in complete conducting paths.*0463

*Here in the circuit to the left, we have a source of potential difference.*0466

*We have a switch which is open at the moment and we have a lamp.*0469

*No current will flow because we do not have a closed path.*0473

*Over on the right here, once we have closed that path, close that switch,*0476

*current is going to flow from high potential, the positive side through our lamp.*0480

*And that is the direction of conventional current flow.*0486

*Note that the longer side is the positive side.*0490

*The way I remember that, is if you draw this, it takes more ink to make the longer side.*0493

*It takes more ink to make a + , the short side looks like a minus.*0497

*Let us take a look at voltmeters, use to measure the potential difference between two points in a circuit.*0504

*Voltmeters are always connected in parallel with the element to be measured.*0509

*The two lids on the voltmeter go to either side of the object you want to measure the potential difference across.*0513

*What you do not want to do is have a voltmeter in the circuit, where all of the current goes through the voltmeter.*0519

*It is very bad for the voltmeter and your circuit.*0524

*If the voltmeter is connected correctly, you can always remove it from the circuit without breaking the circuit.*0528

*And voltmeters have a very high resistance and the reason they do that, is that they have a negligible effect on the circuit.*0533

*Although they take the tiny amount of energy in order to make them operate, you really do not want it affecting the circuit as a whole.*0539

*The way you do that is by using them in parallel and they have a very high resistance.*0545

*You know it is connected correctly because in this diagram, we can actually go*0549

*and we could disconnect it and our circuit would still function.*0554

*Our voltmeter would not do anything but it would still work as a whole.*0557

*If we wanted to put that voltmeter back in, now it is connected in parallel on either side of the element we want to measure the voltage across.*0561

*Looking at the ammeters, they measure the current flowing through an element of the circuit.*0571

*Unlike voltmeters, ammeters are designed to go right in line with the circuits.*0576

*Any current going to the element also goes to the ammeters so we can measure the entire current going through it.*0580

*Ammeters have very low resistance so that they have a negligible effect on the circuit.*0586

*That they do not cause a large potential difference, a large voltage drop.*0591

*Therefore, affect the rest of the circuit.*0594

*They are designed to have a negligible effect.*0597

*In order to put an ammeter in correctly, if you break the circuit, the circuit is not going to function correctly.*0600

*If you put an ammeter in a circuit in parallel, because it has such a low resistance, chances you are going to break it.*0607

*You are going to smoke it and burn it up.*0613

*Be very careful when inserting ammeters.*0615

*Ammeters always go in series with what you are trying to measure.*0618

*Let us take an example where we are looking at ammeter and voltmeter placement in a circuit.*0624

*In the circuit diagram here, possible locations of an ammeter and voltmeter are indicated by 123 and 4.*0629

*Where should we place an ammeter to measure the total current and the voltmeter to measure the total voltage?*0636

*As I look at this, it seems to me if we want the total current flow, we would have to put our ammeter here at 1.*0643

*Here is our source of potential difference, we are going to have current flowing this way and around that way.*0652

*The only place where we get all of the current is if we place the ammeter here at 1 and that is in series with the rest of the circuit.*0661

*That eliminates choice B and C.*0667

*And the voltmeter, we are left with choice 4 or 2.*0671

*If we put the voltmeter at 4, that would be bad, that is in series.*0674

*Voltmeters go in parallel so it cannot be that when you are in a circuit.*0678

*Instead, if we put it here at 4, our voltmeter right there is going to measure*0683

*the potential difference across our voltage source which should be just about perfect.*0688

*I'm going to pick choice A.*0693

*Which circuit diagram below correctly shows the connection of ammeter A and*0700

*voltmeter V to measure the current through and potential difference across resistor R.*0704

*Let us take a look at each of these here.*0710

*In number 1, we have an ammeter connected in parallel with the resistor.*0712

*That is not going to be good at all, that is going to get nice and smoked and crispy.*0719

*We do not want one.*0723

*As we look at 2, our ammeters now in series but our voltmeters in series is not parallel, that is not going to work.*0726

*As we look at 3 here, we have our voltmeter in parallel with R but our ammeter is connected*0733

*on the same portion of the circuit as the voltmeter.*0740

*It is not going to measure the current going through R, it is going to measure the current going through*0742

*the voltmeter which we are trying to make as small as possible.*0747

*Although that is probably not going to hurt anything, it is not going to give you the information you want.*0750

*Number 4, here now we have our ammeter connected so that all the current has to flow through our ammeter.*0756

*That is what we want, to measure the total current and the current that is going to resistor R.*0762

*Our voltmeter is connected in parallel with each lead connected to either side of R to measure the potential difference across it.*0766

*EA 4 must be our answer.*0774

*Looking at voltmeters again, a student uses a voltmeter to measure the potential difference across a resistor.*0780

*To obtain a correct reading, the student must connect the voltmeters go in parallel with the element you want to measure.*0785

*Which statement about ammeters and voltmeters is correct?*0797

*The internal resistance of both meters should be low.*0801

*That is not good, if you did that with the voltmeter, you will pull a lot of current and change the way your circuit works.*0804

*It cannot be A.*0810

*Both meters should have a negligible effect on the circuit being measured.*0811

*Yes, that is the whole point.*0815

*We want to get information from our circuit, about how it is operating without affecting the way it operates.*0817

*B is going to be our correct answer there.*0822

*A little more on series circuits.*0828

*Series circuits have only a single current path.*0830

*You do not have multiple current paths.*0833

*The removal of any single circuit element causes an open circuit.*0835

*Think of it like Christmas or holiday lights.*0838

*Have you ever seen where 1 ball goes out and they all go out?*0840

*That would be a great example of the series circuit.*0844

*A diagram of one, if we started with a source of potential difference and we will put a couple of resistors here.*0846

*There is one, there is another, and let us say one more resistor here.*0854

*We have one current path in which our current can flow.*0865

*Remove any single element and all of the current stops, that is a series circuit.*0870

*What we can do is we can find the equivalent resistance for resistors in series, if we have a bunch of resistors all lined up,*0875

*we could replace them with a single resistor where the equivalent resistance value is just the sum of all your individual resistors.*0880

*Now in a series circuit, anywhere in the circuit the current is the same.*0892

*Whatever current enters a point must leave a point, by the law of conservation of charge.*0896

*The total current anywhere in the series circuit is going to be the same.*0901

*The current through resistor 1, let us make it smaller here.*0905

*R1, R2, R3, R4, total current the current through I1 must be equal to the current through R2 must be equal to I3,*0909

*the current through R3 and so on, for however many resistors you have in that series circuit.*0919

*The potential drop is going to be equal to the drop across R1 + R2 + R3 + R4.*0925

*The potential drops across each of these elements all add up to give you your total.*0933

*That will be V1 + V2 + V3, for however many resistors you happen to have.*0938

*I did this a moment ago but we have a couple laws or rules that will help with analyzing circuits.*0949

*Kirchhoff’s laws are tools utilized in analyzing circuits and there are 2 of them.*0954

*Kirchhoff’s current law, which you will often× hear me abbreviate as KCL.*0959

*It is also known as the junction rule, states that the summit enters any point of the circuit,*0963

*the current that enters any point in the circuit also has to leave that point, by the law of conservation of charge.*0968

*Kind of like if you pour water into a cup, you keep pouring water in.*0975

*Whatever you pour in eventually is going to come out.*0978

*Same thing with current, whatever you put into a point has to come out.*0981

*Restatement of the law of conservation of charge.*0986

*Kirchhoff's voltage law which I call KBL, states that the sum of all the potential drop in any closed loop of the circuit has to equal 0.*0991

*It is a restatement of the law of conservation of energy.*0998

*Basically, what goes up must come down.*1001

*As you go around any closed loop in a circuit, you have to come back to the exact same energy or the same potential.*1004

*And that is oftentimes called the loop rule.*1010

*And you will see how we apply these here in just a moment.*1013

*Taking a look at voltage across a resistor, you have a 3 ohm resistor and a 6 ohm resistor connected in series.*1018

*Let us draw those, there is a 3 ohm resistor, there is a 6 ohm resistor.*1025

*If the current through the 3 ohm resistor is 4 amps, so we have 4 amps going this way.*1033

*What is the potential difference across the 6 ohm resistor?*1038

*Kirchhoff's current law KCL says is that whatever comes into a point in the circuit has to leave that.*1042

*Which means we must have 4 amps going through the 6 ohm resistor.*1047

*The potential drop then across the 6 ohm resistor, using ohms law is the current × the resistance or 4 amps × 6 ohms which gives us a potential drop.*1053

*Potential difference of 24 V across the 6 ohm resistor.*1066

*Taking another look at Kirchhoff's current law.*1073

*We have a diagram that shows current in the segment of an electric circuit.*1076

*What is the reading of ammeter A?*1079

*To do this, I'm going to take a look at everything that is entering this point and everything that is leaving this point.*1082

*It looks like we have 3 amps and 2 amp coming in, so we have 5 amps coming in.*1087

*We also have coming out 4 + 1 + 2 or 7 amps out.*1095

*Whatever comes in, must come out so we are missing 2 amps in.*1101

*That means that this must be 2 amps entering the circuit.*1106

*The ammeter reading 2 amps.*1111

*Let us take a look a little bit of basic series circuit analysis.*1117

*As we do these, I’m going to start off for simpler circuits using what I call A VIRP table.*1120

*What that is, is just the way of tracking different characteristics of different elements in the circuit.*1126

*I'm going to label the column for potential current flow resistance and power and then do that for the different elements in my circuit.*1132

*Where here we have resistor 1, resistor 2, resistor 3, and a make a line for total.*1140

*We will just make ourselves a nice little table.*1151

*R1, R2, R3, R4, and the nice thing about these VIRP tables is as you fill them in,*1156

*any time you know 2 things in a row you can always solve for the other items in a row using ohm’s law or a power equation.*1166

*Let us start off by filling in what we know.*1174

*Right away, I know since I have a 12V battery that my total and there is only one of them that might total voltage must be 12.*1177

*Let me make that line there to make it a little meter.*1184

*I know my resistance is R1 is 2000 ohms, 2 kl ohms.*1188

*R2 is 2000 and R3 is 2000.*1193

*Now we also know for a series circuit that the total resistance, the equivalent resistance is the sum of the individual resistances.*1197

*My total resistance here must be 6000 ohms.*1205

*Once I know that, I can start filling in the other items because that 6000 and that is 12,*1208

*I can find the current here using ohm’s law I = V/ R, 12 /6000 gives me a current of 0.002 amps.*1215

*That means that I must have 0.002 amps flowing through that portion of the circuit.*1224

*By Kirchhoff's current law, that means I have to have the same current through R1, through R2, and R3.*1232

*These all must be 0.002 amps as well.*1238

*When I know the current and the resistance, I can then use ohm's law to solve for the potential drop V = IR, 2000 × 0.002 is just going to be 4V.*1244

*Same there and same there.*1255

*I'm dropping 4V here, 4V here, 4V here, that makes sense because I'm lifting up the source of potential difference is raising our potential 12V.*1257

*We drop 4, we drop 4, we drop 4, we come back to 0, we raise 12 again.*1268

*If you are possible to draw, you make a closed loop in the circuit, you come back to where you start.*1274

*The sum of all the potential drops in a closed loop must be 0.*1278

*To find the power, we have our choice of equations.*1282

*We could use power = V × I, power = I² R, power = V²/ R.*1285

*For any of these, we will get the same result which will be for R1 0.008, 0.008, 0.008.*1290

*And of course, we are expanding 8 mw.*1300

*In each of these, the total will have to add up those powers to be 0.024 or 24 mw*1305

*which by the way we can also get using V × I, I² R, or V²/ R.*1312

*There is our VIRP table which tells us just about anything we could ever want to know about this simple series circuit.*1317

*Let us take a look at some more sample problems.*1326

*A 2 ohm resistor and a 4 ohm resistor are connected in series with a 12V battery.*1329

*If the current through the 2 ohm resistor is 2 amps, what is the current through the 4 ohm resistor?*1334

*That is a trick question because right away if it is a series circuit, we know the current is the same everywhere.*1339

*It is just going to be 2 amps.*1344

*Here, we have got a circuit connected below with two 4 ohm resistors connected to a 16V battery.*1350

*Fill in the VIRP table for the circuit and determine the rate at which electrical energy is expanded in the circuit.*1356

*That is what we call power.*1361

*Let us call that R1, R2, and make our VIRP table.*1364

*Our circuit elements, we have R1, R2, and total.*1373

*Let us start to fill in what we know.*1385

*Our total potential is 16V, we know R1 is 4 ohms, we know R2 is 4 ohms, and because it is a series circuit our total resistance is the sum of those or 8 ohms.*1387

*I = V/ R, 16/8 tells us that is 2 amps.*1400

*That current must be 2 amps and anywhere in the series circuit you must have the same current so those must be 2 amps as well.*1406

*We can now find our potential difference V = I × R, which is 8 and 8, and those add up as we expect.*1413

*And for the powers, we have our choice of formulas again.*1421

*V × I, will be simply 16 and 16 is 32.*1424

*The rate at which electrical energy is expanded is the total power, power = 32 W answers the question that we were asked.*1429

*Another problem, we have got a 50 ohm resistor, some unknown resistor R, 120V source and the ammeter connected as shown.*1444

*The ammeter reads 0.5 amps so we know the current right there is 0.5 amps.*1452

*Find the equivalent resistance of the circuit, the resistance of resistor R, and the power dissipated by the 50 ohm resistor.*1459

*The way I find out all of these things is I make the VIRP table and then just look it up.*1466

*Once we have a VIRP table made, everything else we should be able to look up right on there.*1470

*We will write our VIRP, I'm going to call this R1 and this R2.*1476

*R1, R2, total, we do not have to make a line for the ammeter because it is supposed to have a negligible effect on the circuit.*1487

*We will start filling in what we know.*1504

*We have a total of 120 V, we know the current is 0.5 amps, and that is the same anywhere in the circuit because it is series.*1506

*We know R1 is 50 ohms and I think that will get us started.*1517

*Our voltage through R1 is I × R, half of 50, that is just going to be 25V.*1523

*Since we know that potentials add up in a series circuit, we know that this must be 95V.*1530

*The resistance R = V/I, 95/ ½ means that this is going to be 190 ohms and resistance is add in a series circuits or total resistance must be 240 ohms.*1538

*We can find our power as well with any of the formulas that you are most happy with.*1551

*V × I, 25 × ½ will be 12.5, 95 × 1/2 will be 47 ½, and 120 × 0.5 will be 60 W.*1556

*Of course, the total power is the sum of the individual powers from those elements.*1565

*What did it ask us?*1570

*Find the equivalent resistance in the circuit, right from our table that is going to be 240 ohms.*1572

*Find the resistance of resistor R, that is what we call R2.*1580

*R2 we said was 190 ohms and our power dissipated by the 50 ohm resistor.*1585

*The power dissipated by R1 is 12.5 W, right from my VIRP table.*1589

*Let us take a look at 1 more simple series circuit here.*1604

*In the circuit represented in the diagram, what is the reading of the voltmeter V.*1608

*First thing, let us label some of these.*1614

*Let us call this R1 and this R2, make our VIRP table.*1616

*We have resistor 1, resistor 2, and total.*1628

*Start filling in what we know.*1643

*We have a 60V source so there is our total.*1646

*We know that R1 is 20 ohms, we know that R2 is 10 ohms.*1649

*Once we have done that, we know that the total resistance in our circuit is the sum of the individual resistances because it is a series circuit.*1655

*30 ohms I = V/ R, this must be 2 amps.*1663

*That means that anywhere in our circuit, because it is a series circuit we have the same current 2 amps and 2 amps.*1670

*We can find our voltage using Ohm’s law, I × R so that is 40.*1677

*The potential drop across R1 this must be 20, we drop 20V across R2 and those add up to the total, that make sense.*1681

*And for the powers, take your choice of equations V × I, 80, 40, and 120.*1689

*And of course, our 2 individual powers expanded add up to a total power expanded.*1696

*What was the question asks?*1701

*What is the reading of voltmeter V?*1702

*If we look up here at V, V is measuring the potential difference across R1.*1705

*We come down to my VIRP table and I find that the potential difference across R1 is 40V.*1711

*Hopefully that gets you some good fleeting, a start on series circuits and I know this is a calculus based physics course.*1720

*And by all means, these are simple problems that we are starting with.*1727

*We are just laying the groundwork for as we get to more complex circuits and more complex topics.*1731

*We got to have these basics down first.*1736

*We will start getting to those into the next lesson and further on as well.*1738

*Thank you for your patience, thank you for watching www.educator.com.*1742

*We will see you again real soon, make it a great day.*1746

0 answers

Post by Professor Dan Fullerton on November 6, 2016

Yes, that would be an acceptable answer in that circumstance.

0 answers

Post by Mohsin Alibrahim on November 5, 2016

Hello,

In Example 8, would it be a right answer if there were Ammeter at 1 and Voltmeter at 3 ?