For more information, please see full course syllabus of AP Physics 1 & 2

For more information, please see full course syllabus of AP Physics 1 & 2

### Circuit Analysis

- The values of currents and electric potential difference in an electric circuit are determined by the properties and arrangement of the individual circuit elements.
- Kirchhoff's Current Law (KCL) states that the sum of the current entering any point in a circuit is equal to the current leaving that point. This is also known as the junction rule, and is a restatement of the law of conservation of charge.
- Kirchhoff's Voltage Law (KVL) states that the sum of the potential drops in any closed loop of a circuit has to equal zero. This is also known as the loop rule, and is a restatement of the law of conservation of energy.
- Real batteries and voltage sources have some finite amount of internal resistance. The terminal voltage of a real battery is equal to the battery's emf - the voltage drop across the internal resistance.

### Circuit Analysis

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
- Series Circuits
- Series Circuits Have Only a Single Current Path
- Removal of any Circuit Element Causes an Open Circuit
- Kirchhoff's Laws
- Tools Utilized in Analyzing Circuits
- Kirchhoff's Current Law States
- Junction Rule
- Kirchhoff's Voltage Law States
- Loop Rule
- Example 1: Voltage Across a Resistor
- Example 2: Current at a Node
- Basic Series Circuit Analysis
- Example 3: Current in a Series Circuit
- Example 4: Energy Expenditure in a Series Circuit
- Example 5: Analysis of a Series Circuit
- Example 6: Voltmeter In a Series Circuit
- Parallel Circuits
- Parallel Circuits Have Multiple Current Paths
- Removal of a Circuit Element May Allow Other Branches of the Circuit to Continue Operating
- Basic Parallel Circuit Analysis
- Example 7: Parallel Circuit Analysis
- Example 8: Equivalent Resistance
- Example 9: Four Parallel Resistors
- Example 10: Ammeter in a Parallel Circuit
- Combination Series-Parallel Circuits
- Analysis of a Combination Circuit
- Internal Resistance
- In Reality, Voltage Sources Have Some Amount of 'Internal Resistance'
- Terminal Voltage of the Voltage Source is Reduced Slightly
- Example 11: Two Voltage Sources
- Example 12: Internal Resistance
- Example 13: Complex Circuit with Meters
- Example 14: Parallel Equivalent Resistance

- Intro 0:00
- Objectives 0:07
- Series Circuits 0:27
- Series Circuits Have Only a Single Current Path
- Removal of any Circuit Element Causes an Open Circuit
- Kirchhoff's Laws 1:36
- Tools Utilized in Analyzing Circuits
- Kirchhoff's Current Law States
- Junction Rule
- Kirchhoff's Voltage Law States
- Loop Rule
- Example 1: Voltage Across a Resistor 2:23
- Example 2: Current at a Node 3:45
- Basic Series Circuit Analysis 4:53
- Example 3: Current in a Series Circuit 9:21
- Example 4: Energy Expenditure in a Series Circuit 10:14
- Example 5: Analysis of a Series Circuit 12:07
- Example 6: Voltmeter In a Series Circuit 14:57
- Parallel Circuits 17:11
- Parallel Circuits Have Multiple Current Paths
- Removal of a Circuit Element May Allow Other Branches of the Circuit to Continue Operating
- Basic Parallel Circuit Analysis 18:19
- Example 7: Parallel Circuit Analysis 21:05
- Example 8: Equivalent Resistance 22:39
- Example 9: Four Parallel Resistors 23:16
- Example 10: Ammeter in a Parallel Circuit 26:27
- Combination Series-Parallel Circuits 28:50
- Look For Portions of the Circuit With Parallel Elements
- Work Back to Original Circuit
- Analysis of a Combination Circuit 29:20
- Internal Resistance 34:11
- In Reality, Voltage Sources Have Some Amount of 'Internal Resistance'
- Terminal Voltage of the Voltage Source is Reduced Slightly
- Example 11: Two Voltage Sources 35:16
- Example 12: Internal Resistance 42:46
- Example 13: Complex Circuit with Meters 45:22
- Example 14: Parallel Equivalent Resistance 48:24

### AP Physics 1 & 2 Exam Online Course

### Transcription: Circuit Analysis

*Hi folks. Welcome back to Educator.com*0000

*I am Dan Fullerton and today we are going to talk about circuit analysis.*0002

*Our goals are going to be to draw and interpret schematic diagrams of circuits, to solve both series and parallel circuit problems using VIRP tables, to calculate equivalent resistances for resistors in both series and parallel configurations, and finally to calculate power and energy used in electric circuits.*0007

*Let us start by talking about series circuits.*0025

*Series circuits have only a single current path. If you remove any circuit element you cause an open circuit.*0029

*For example, think of a Christmas light.*0035

*If you have one go bad, you have seen the whole strand go out -- that is a series circuit.*0037

*The way we would draw a series circuit is we must have a source of potential difference, so we will put a battery here, where the long side is positive and the short side is negative.*0043

*And then we will draw some resistors.*0051

*That could be (R1), another resistor, a couple of more resistors, and then we complete the circuit.*0054

*Current wants to flow from positive high potential to negative potential, meaning electrons would go the other way.*0070

*Now, it is a series circuit, which means if any one of these elements breaks, the entire circuit stops working.*0077

*For example, if this resistor burnt out and became an open, all of a sudden nothing lights up anymore.*0082

*You will no longer have current flow and your circuit no longer functions, so that is a series circuit.*0089

*One of the tools we are going to use to analyze circuits is known as Kirchhoff's Laws.*0096

*Kirchhoff's Laws are tools utilized in analyzing these circuits.*0101

*We are going to focus on two laws: Kirchhoff's Current Law, or KCL, states that the sum of all current entering any point in a circuit equals the sum of all current leaving that point in a circuit.*0105

*It is really just a restatement of conservation of charge -- What goes in, must come out.*0115

*It is also known as the Junction Rule.*0121

*The other law we are going to look at is Kirchhoff's Voltage Law, or KVL, and that says that the sum of all the potential drops in any closed loop in a circuit must sum to zero.*0125

*That is really a restatement of the conservation of energy and it is also known as the Loop Rule -- KVL, Kirchhoff's Voltage Law.*0135

*Let us take a look at our first example here.*0145

*A 3-ohm resistor and a 6-ohm resistor are connected in series in an operating electric circuit.*0147

*At some point in the circuit we must have a 3-ohm resistor and a 6-ohm resistor, and there has to be more to the circuit, but we will just draw that piece for now.*0153

*If the current through the 3-ohm resistor is 4 A -- so we have 4 A going through the 3-ohm resistor.*0164

*What is the potential difference across the 6-ohm resistor?*0170

*To answer this I have to figure out the current going through the 6-ohm resistor, and if I have 4 A going through the 3-ohm resistor, Kirchhoff's Current Law says I must have that same current here and same current here. -- what goes in, must come out.*0173

*So I must have 4 A going through the 6-ohm resistor; current does not get used up in a circuit.*0186

*Now that I know the resistance and the current flow, I can find the voltage drop.*0193

*The potential difference from one side to the other is going to be I × R, which is 4 A × 6-ohms or 24 volts.*0198

*So if the current is going this way, we must drop 24 volts from the positive to the negative side as the current goes in that direction.*0209

*What is the potential difference across the 6-ohm resistor? 24 volts.*0218

*Let us take another look here.*0226

*The diagram below represents currents in a segment of an electric circuit. What is the reading of ammeter A?*0227

*We have an ammeter here, and here we have a junction where all of these come together.*0233

*This is going to be a great place for us to apply Kirchhoff's Current Law, or the Junction Rule.*0237

*The current coming into that junction must equal the current leaving that junction, so coming in we have 2 A, we have 3 A -- so in, we have 2 + 3 and let us just assume for now that we will call that current coming in...*0242

*...so +A -- and if we get a negative we know it is going the opposite direction -- must equal the current leaving and the current leaving -- well, here we have our 4 A, we have 1 A, and we have 2 A.*0260

*When I put this all together, I have 5 + A = 7, therefore, A must equal 2 A, and yes, it is going in, so our ammeter would read: 2 A.*0277

*As we talk about basic series circuit analysis, we can use Kirchhoff's Current Law and Kirchhoff's Voltage Law along with Ohm's Law to find a bunch of unknowns.*0294

*But one of the most helpful tools, especially when you are starting out with simple circuits is what I call a VIRP Table.*0303

*What we are going to do is we are going to analyze the voltage, (V), the current, (I), the resistance, (R), and the power dissipated, (P), for every element in the circuit as well as for the total circuit.*0308

*Here is what I am talking about.*0320

*We have here a power supply and 3 resistors; I will call this R1, that one R2, and that one R3.*0322

*Now I am going to make a table. I will make a row for every circuit element: R1, R2, R3, and one for the total.*0331

*What I am going to analyze for each of these is the potential difference (V), the current flow (I), the resistance (R), and the power dissipated, (P).*0344

*I am just going to make this look like a nice pretty table -- to help us organize our thoughts.*0355

*Then we will fill in what we know about the circuit already.*0372

*Well, I know R1, R2, and R3, each of those is 2 kilo-ohms or 2,000-ohms so I will fill those in.*0375

*I know the total voltage must be 12 volts because that is the extent of my power supply, so that is 12.*0386

*I could also figure out the equivalent resistance.*0393

*Because it is a series circuit, so I just add up my resistors: 2,000 + 2,000 + 2,000, which will give me a total of 6,000.*0395

*Now what is really neat here is anytime you know any two things in a row, using Ohm's Law or your power definitions, you can figure out the other two.*0404

*Once you know two items in any row, you can find the other two, so current -- right here -- I is going to be V/R --Ohm's Law, so total current, I = V/R, is 12 volts/6,000 ohms, which is going to be .002 A, so I will fill that in right here.*0414

*That means I have .002 A flowing through my circuit this way, and since it is a series circuit, I must have the same current everywhere in the circuit.*0436

*That means that I have .002 A flowing through R1; I have .002 A flowing through R2, and of course, .002 A flowing through R3.*0445

*Now I know two things in all of these rows.*0457

*If I want the voltage here, through R1, the voltage drop across R1 -- that I can find by Ohm's Law V = RI, so that is 0.002 × 2,000, which is going to give me voltage of 4 volts -- potential difference of 4 volts from one side to the other is 4.*0460

*It is pretty easy to see R2 and R3 are going to have the same values.*0479

*The sum of the potential drops here all add up, so the potential drops 4 volts here, 4 volts there, 4 volts there, and that brings us back to 12 volts, which is what we would expect.*0483

*For the power, I could get that a bunch of different ways.*0497

*I could use power equals IV, I could use power equals I ^{2}R, or I could use power equals V^{2}/R, since I know all of those things.*0501

*We will take the easy one right now -- power equals V × I, that will be 0.008 watts dissipated in R1 or 8 milliwatts.*0510

*We will have the same for R2, the same for R3, and the total power dissipated is just going to be the sum of those or I could find it by multiplying V times I down here, but either way 0.008 + 0.008 + 0.008 or 12 × 0.002 -- regardless, I am going to get the same answer of 0.024.*0519

*I am dissipating 0.024 W in my entire circuit.*0539

*That is basically how we are going to use these VIRP tables to help us organize our information.*0544

*As I write these down you will notice that I am skipping putting units anywhere.*0548

*I am assuming that V potential difference is in volts, current is in amps, (R) is in ohms, and power is in watts.*0552

*Let us take a look at another problem -- current in a series circuit.*0562

*A 2-ohm resistor and a 4-ohm resistor are connected in series with a 12-volt battery.*0565

*Let us draw that first. We have a 12-volt battery and we have in there a 2-ohm resistor, and a 4-ohm resistor.*0570

*If the current through the 2-ohm resistor is 2 amperes, so we must have 2 A flowing through the 2-ohm resistor, what is the current through the 4-ohm resistor?*0586

*I do not even have to do a VIRP table because this is just Kirchhoff's Current Law.*0595

*I must have 2 A flowing through there as well due to KCL, Kirchhoff's Current Law -- I = 2 A -- very straightforward there.*0599

*Let us talk about energy expenditure in a series circuit.*0614

*In the circuit diagram below, we have two 4-ohm resistors connected to a 16-volt battery as shown.*0617

*Fill in a VIRP Table for the circuit and determine the rate at which electrical energy is expended in the circuit.*0623

*Let us call this R1 and this is R2, so that when we do that now and make our VIRP table, we have R1, R2, and a row for the total, and we have V, I, R, and P.*0629

*We will start with what we know. Our total voltage must be 16 volts, R1 is 4, and R2 is 4.*0656

*And if we have two resistors in series then that means our total resistance is the sum of those, and that is going to be 8-ohms.*0665

*We know two things in the row: I = v/r or 16/8 and that is going to give us a current of 2 A.*0673

*If we have a current of 2 A here, we must have a current of 2 A through R1 and a current of 2 A through R2, so we can fill those in.*0680

*Now, it is just an exercise in math. V = I × R, 2 × 4, and that will be 8 volts, and 8 volts there, and the sum of (R), individual potential drops, gives us our total potential drop.*0689

*For power, we will just go right to power equals V × I or 8 × 2 = 16 W and 8 × 2 = 16 W, and I could add those together or 16 × 2 = 32 W.*0704

*What is our total power dissipated here? 32 watts -- right from our VIRP Table.*0716

*Taking a look at another series circuit, we have a 50-ohm resistor, an unknown resistor, (R), a 120-volt source, and an ammeter connected as shown.*0727

*The ammeter reads half an amp. Calculate the equivalent resistance of the circuit, the resistance of resistor (R), and the power dissipated by the 50-ohm resistor.*0739

*Let us do this with a VIRP Table again.*0749

*If we call this R1, we will make that R2, then our table will have R1, R2, and total.*0751

*We do not need to add the ammeter in there because our ammeter should have a negligible effect on the circuit. 0767 All the ammeter does is to provide us information.*0761

*V, I, R, P -- And we will start by filling in what we know: R1 is 50-ohms; we know that we have a 120-volt source.*0770

*We know the current over here because the ammeter says it is half an amp and that means we must have half an amp going through R1 and R2, so, 0.5 A, 0.5 A, so we must have a total current of 0.5 A, so that must be the same.*0790

*Now, let us start down here. We know two things in this row: if R = V/I or 120/0.5, that is going to be 240, so our total resistance must be 240.*0806

*If our total resistance is 240 and R1 is 50-ohms -- they are in series -- what do we add to 50 to get 240?*0820

*Well, that must be 190-ohms.*0828

*We can figure out the voltages now -- V = I/R, so half of 50 is going to be 25.*0831

*Half of 190 is going to be 95, and of course, those potential drops will add up to our total potential drop.*0837

*And if we want the power here too, one more step -- V × I is half of 25, which is 12.5 W; half of 95 is 47.5 W, and half of 120 is 60 W, or add up 12 1/2 and 47 1/2.*0845

*Now, I know everything I need to about this circuit, so I can go answer the questions.*0862

*Calculate the equivalent resistance of the circuit.*0867

*That is easy. The equivalent resistance is our total resistance there, 240.*0870

*The resistance of resistor (R) is R2 and its resistance is 190-ohms. We have answered that as well.*0875

*The power dissipated by the 50-ohm resistor, or by R1 = 12.5 W.*0883

*By filling out our VIRP table we have answered all these questions effectively.*0890

*Let us take a look at a voltmeter in a series circuit.*0896

*In the circuit represented in the diagram, what is the reading of voltmeter V?*0899

*It looks like we have 2 resistors, and if this looks confusing at first, pretend the voltmeter is not there. 0909 If it is hooked up correctly, it should not have any significant effect on your circuit.*0903

*As I take a look at this, let us call this R1, we will call that R2...*0913

*...and we will make our VIRP table -- R1, R2, total -- V, I, R, P.*0919

*Let us go and let us fill in what we know already -- R1 is 20-ohms, R2 is 10-ohms and our source is 60 volts.*0939

*As we look again, this is a series circuit again.*0949

*Current is going through R1 and R2, and we can assume very little current through (V) -- pretend it is not there -- it has a negligible effect on the circuit.*0953

*Therefore, we could find that for the total equivalent resistance, we can just add them for a series circuit of 20 + 10 = 30.*0961

*Once I have that, of course, I can figure out the current -- I = V/R or 60/30 = 2 A.*0970

*If we have 2 A going here, we must have 2 A through R1 and R2, so we can fill those in.*0978

*And now, it is pretty easy to go find the potential drop across R1 -- V = I/R by Ohm's Law or 2 × 20 = 40 -- 2 × 10 = 20, and of course, those add up to our total.*0984

*And while we are here, let us figure out the power -- 40 × 2 (Power = V × I) = 80 W; 20 × 2 = 40 W, and 60 × 2 = 120, or I could have added those up.*0997

*Now, to answer the question -- What is the reading of voltmeter V?*1010

*Voltmeter V is reading the potential drop across R1, so the potential drop across R1, I can just look up on my table and it is 40 volts, so our answer must be 40 volts.*1015

*Let us take a look now at some parallel circuits. Parallel circuits have multiple current paths.*1030

*Removal of a circuit element may allow other branches of the circuit to continue operating.*1035

*This is kind of like the wiring in your house.*1040

*If you looked at a circuit in a bedroom, you have a power supply, a resistor, maybe another resistor, like a lamp, an alarm clock, a stereo, or whatever it happens to be.*1042

*And with your power supply here -- (+/-)V -- if one of these gets interrupted or broken, the other ones continue operating because you still have a complete path here for the electricity to follow.*1058

*It just does not go through the one element that is broken, which happens when a light bulb goes out in your house, you do not lose everything in that room, you typically lose just that light bulb.*1070

*Now, basic parallel circuit analysis works very similar to what we did for series circuit analysis, and we are going to use VIRP tables again.*1100

*Here I have 3 resistors and let us call them R1, R2, and R3. I will make our VIRP Table -- R1, R2, R3, and a row for a total and our data: V, I, R, and P.*1108

*Our first step again is always filling in what we know: R1 is 2,000, R2 is 2,000, R3 is 2,000, and our total voltage is 12 volts.*1135

*Now, as we do this, there are a couple different ways we could go about doing this, or different ways to start. 1155 One is we could calculate the equivalent resistance for 3 resistors in parallel and put it down here.*1150

*You could do that, but there is another way to do this without having to go through that math.*1161

*Anywhere on a wire has the same potential, so, if this is 12-volts difference here, these are connected to the same wire, so that must be a 12-volt drop here, here, and here.*1166

*We have the same potential drop across all 3 resistors of 12 volts.*1176

*In a series circuit, the potential drops are all the same.*1181

*Now we could figure out the current through each of these -- I = V/R -- the current through here is going to be 12-volt drop/2,000-ohms or about .006 A; R2 is going to be the same and R3 the same.*1185

*As we look at these, if this is .006 A, we must have .006 A going that way, and if that is also .006, now we must have .012 at this branch in the circuit plus this .006 over here gives us .018, so all of the currents add together for the total.*1201

*Now, I could find R over here and I have V/I = 12/.018 is going to give us 667, which is the same thing we would have gotten if we calculated an equivalent resistance for the resistors in parallel using 1/R-equivalent = 1/R1 + 1/R2 + 1/R3. -- Try it, it works out.*1222

*For the powers -- well, that is just going to be V × I, or .072 W, which is the same math for all three of these, and for the total, we could add these three up or 12 × .018 = .216 W or 216 mW for the parallel circuit.*1245

*Let us take a look here at another parallel circuit problem.*1265

*We have a 15-ohm resistor, R1 and a 30-ohm resistor, R2 that are to be connected in parallel between points (A) and (B) in the circuit that contains a 90-volt battery.*1269

*Complete the diagram to show the two resistors connected in parallel between (A) and (B), so let us do that first. 1284 If they are in parallel, we must have two current paths, so let us make that one of them, and we will make that the other, so now current has two different directions to go here, R1 and R2.*1278

*Now, determine the potential difference across resistor R1 and calculate the current in resistor R2.*1300

*Well, the potential difference across R1 has to be the same as the battery here.*1307

*If this is a 90-volt difference between the two, then let us make this simple -- let us call the negative side of the battery 0 volts.*1312

*If we call that 0, then this side must be 90 higher, so that is 90.*1319

*We have 90 volts here on the same wire and over here we have 0 volts.*1323

*The potential difference across R1 is 90 volts, so V1 = 90 volts.*1327

*And the current in resistor R2 -- well, that should be easy.*1334

*I2 is going to be V2/R2, where the voltage drop across R2 is the same as it is in R1, 90 volts, so 90 volts/R2 = 30 ohms, and that is going to give us a current of 3 A -- That easy.*1338

*Equivalent resistance -- Three identical lamps are connected in parallel with each other.*1360

*If the resistance of each lamp is x-ohms, what is the equivalent resistance of this parallel combination?*1365

*1/R-equivalent is going to be 1/x + 1/x + 1/x which is equal to 3/x and if 1/R-equivalent is 3/x, then R-equivalent must be x/3-ohms - Answer B*1372

*Let us take a look at a circuit that has four parallel resistors in it.*1396

*The diagram below represents an electric circuit consisting of four resistors and a 12-volt battery.*1399

*What is the current measured by ammeter A, what is the circuit's equivalent resistance, and how much power is dissipated in a 36-ohm resistor?*1404

*A lot of questions, but again, if we make a VIRP table we will probably answer all of them just in filling out that table.*1412

*Let us call these R1, R2, R3, and R4.*1418

*We will make our table and we have elements: R1, R2, R3, R4, and total V, I, R, and P.*1426

*We will fill in our lines here before we fill in what information we know right away from the circuit.*1440

*What do we know right away? R1 is 6-ohms, R2 is 12, R3 is 36, and R4 is 18, and we also know our total voltage is 12.*1453

*You can also look at this battery -- remember if we called this the 0 side that means we have 0 volts here, here, here, here; they are all connected together and this is 12, 12, 12, 12; they are all on the same wire, so the voltage drop across each of these is 12 volts: 12, 12, 12, 12.*1467

*We can now fill in our currents -- I = V/R or 12/6 = 2, 12/12 = 1, 12/36 = 1/3 (.333), and 12/18 = 2/3 (.667).*1487

*And the total current in a parallel circuit adds together, or we could have used the equivalent resistance formula to figure out what that is going to be, but if I add these currents together, 2 + 1 + 1/3 + 2/3, it is going to give me 4 A, so, R = V/I or 12/4 and we get 3-ohms there.*1500

*Now, let us stop for a second and check.*1518

*We know the equivalent resistance of any parallel circuit has to be less than the smallest resistor in that configuration, so is 3-ohms less than all of our other resistors?*1519

*It most certainly is. We probably did something right.*1534

*Power is just V × I, so let us work our way right through here: 24, 12, 4, 12 × 2/3 = 8, and add them all together, or 12 × 4 = 48.*1535

*Let us answer the questions it asked to begin with.*1549

*What is the circuit's equivalent resistance? R _{(eq)} right from our table is 3-ohms.*1552

*What is the current measured by ammeter A? Well, A is measuring the current through R1, so, let us say that I1 must be 2 A, and finally, how much power is dissipated in the 36-ohm resistor?*1561

*That is the power through resistor 3, which is going to be 4 watts -- All right from our table.*1576

*Looking at an ammeter in a parallel circuit again -- In a circuit diagram shown below, ammeter A1 reads 10 amperes, so 10 A right there. 1595 What is the reading of ammeter A2?*1588

*Well, let us try a VIRP table solution again.*1599

*If we call this R1, and this one is R2, then we have R1, R2, and total, and V, I, R, and P.*1601

*We will make our table and start by filling in what we know: R1 is 20-ohms, R2 is 30-ohms, and our total current is 10 A because that is the combination current of what is going through our battery here.*1618

*As I look at this now, what is my next step going to be?*1639

*Well, there is a lot of different things I could do here, but what I am going to do first, since we have not done it this way yet, is I am going to find the equivalent resistance of the circuit.*1643

*I am going to say that R-equivalent with only two resistors is going to be R1R2/R1 + R2, so that is going to be 20 × 30/20 + 30, which is 50, which is going to give me an equivalent resistance 600/50 or 12-ohms.*1652

*My potential drop then, V = IR must be 120 volts -- parallel circuit, which means we have the same voltage drop across these other elements -- 120.*1673

*If I want the current flow, I = V/R, that is 120/20 or 6 and 120/30, that is 4.*1685

*What is the reading of ammeter A2?*1693

*The current through A2 is the current through resistor 1, which is going to be 6 A, so the reading of ammeter A2 reads 6 A, and did not even have to fill in the power on this one to keep going.*1702

*If we wanted to though it would be pretty easy to say that the power here is 720 W, 480 W, and a total of 1200 W.*1718

*What happens if your circuit is not completely series or parallel but it has a combination of these different elements?*1730

*Well, first thing I like to do is look for portions of the circuit that have parallel elements and see if you can replace those by an equivalent resistance until you get it into a series circuit configuration and analyze with a VIRP Table.*1736

*Or, work back to your original circuit using KCL and KVL until you can find all of those unknowns.*1748

*It is a little bit easier to show than it is to explain.*1755

*Here we have a combination circuit -- 10 volts -- we have R1, and now we have two resistors, R2 and R3, which are in parallel and then back to series for R4.*1761

*What I would probably do first thing is draw an equivalent circuit where I am going to put R2 and R3, and I am going to replace those by an equivalent resistor.*1770

*I would draw this circuit as -- we have our 10 volts here, we have R1, which is still 20-ohms; we have our equivalent resistance between R2 and 3, and we still have R4 down here which is 20-ohms.*1780

*To find out R2-3 -- since it is in parallel -- it is going to be R2 × R3 = 30 × 50/30 + 50, which is 80...*1805

*...gives us 1500/80 and that is going to be about 18.75-ohms -- That is a much easier circuit to analyze.*1817

*As we look at this, we will go back to our VIRP Table now.*1830

*We have R1, R2, R3, and R4, and V, I, R, and P...*1833

*...R1 is 20-ohms, R2 is 30, R3 is 50, R4 is 20, and our total voltage is going to be 10 volts.*1856

*We know our total resistance because these are all in series -- 20 + 18.75 + 20 will give us our total, so 40 + 18.75 gives us a total resistance of 58.75.*1876

*That means our total current, (I) must be V/R, which is going to be about .170 A.*1893

*If that is .170 A, right there -- let us think about it -- On this version that is .170 A and it all goes through there.*1901

*It is going to split here, but as it comes back through R4, we have the total current again .170 A.*1910

*We can fill in .170 here, and we can fill in .170 here.*1917

*We will figure out what else we can do right now -- V = IR, so the voltage drop across R1 or 20 × .17 is going to be about 3.4, and R4 -- same math -- 20 × .17 = 3.4.*1923

*Now, as I look at this, if we start here at 10 volts, I am going to call that 'bottom side,' the negative side zero, and that means this is 10, and across R1 we drop 3.4.*1940

*If we had 10 and we dropped to 3.4, that means we must have 6.6 over here.*1951

*Down here at our fourth, this is zero side and it dropped to 3.4, so this side must be 3.4.*1960

*So, what is the potential difference across R2 and R3?*1967

*Well that is going to be 6.6 - 3.4, which is 3.2, so R2 and R3 have the same voltage drop.*1970

*Current flow now I can gets from Ohm's Law: I = V/R, 3.2/30 is going to be .107 and 3.2/50 is going to be .064 and with the exception of a little bit of rounding error there, if we add these two together we are going to get .17...*1980

*...which we should because if we have .17 coming this way through R2 now, we have 0.107 and that gets most of it, the remainder 0.64 comes through here and then they re-combine again to give you .170 -- they add up.*1997

*Now, to go finish this off, let us calculate our powers: V × I (0.578), VI down here, 0.341 watts; 3.2 × 0.064 = 0.205 W; 3.4 × 0.17 = 0.578 W again...*2011

*...and our total is 10 × .170 = 1.7 W, which is what I should get when I add all of these up -- and I do.*2031

*So that is how you could analyze a combination circuit.*2040

*Simplify it until you can figure out some more of your unknowns and use your VIRP table to help organize your thoughts.*2043

*So far we have been dealing with ideal voltage sources, batteries and so on, but in reality, voltage sources typically have some amount of real internal resistance.*2052

*Hopefully fairly small, but they have some amount of that.*2061

*The terminal voltage of a voltage source is actually reduced slightly by the potential drop of current flowing through this internal resistance, so if we wanted to look at a real voltage source, we could model it as a source of emf (electromotive force).*2065

*That is not really a force, it is another term for potential, but what it really is is that is the maximum voltages could put out and then you have this internal resistance.*2080

*Your terminal voltage (VT) is what you actually measure across the terminals of the battery when you have some current flowing through here, and because you current flowing, you are going to drop a little bit of your potential, your emf is going to be reduced a little bit by this internal resistance.*2090

*Terminal voltage is going to be E - I times your internal resistance.*2106

*Let us take a look at a problem with two voltage sources here. We have not done one of these yet.*2117

*Find the current flowing through R3 if R3 has a value of 6 ohms. What is the power dissipated in R3?*2121

*Well to do this, the way I am going to start is I am going to set up my circuit and think for a minute.*2131

*I am going to call that point 0 volts because that is on the negative side of the battery, so that is going to be 12 and we will say that we have some current flowing, so let us call that I1 flowing through R1.*2136

*Over here we have 0 volts on this side of the battery, which are connected by a wire, so they have to be the same.*2147

*The positive side, we will call 16 volts and we will say that the current flowing this way through R2, is what we will call I _{2}.*2153

*That means that if I _{1} is flowing this way and I_{2} is flowing this way -- let us call this current3, and it should be pretty easy to see by Kirchhoff's Current Law at this point that I_{3} = I_{1} + I_{2}.*2160

*Now that we have established that, let us see if we can apply Kirchhoff's Voltage Law around some of these loops to get us some more information.*2177

*If I apply Kirchhoff's Voltage Law around just this loop here, what that says is that some of the potential drops around that loop must be equal to 0.*2185

*So what I am going to do is I am going to list the positive and negative sides of each of these -- this will be the positive side of the battery of the resistor -- the current is flowing and dropping potential this way -- plus, minus, plus, minus, plus, minus.*2194

*Now as I make my loop around here, if I start down here I always look at what sign I see first -- if I am going this way up, I see the negative sign of that battery first.*2210

*So I am going to write -12 volts plus -- the next potential drop is going to be V = IR, or our current flow times our resistance by Ohm's Law, so this will be 8 ohms × I _{1}, so plus 8I_{1} and as I come around to this part of the loop, I have I_{3} × R_{3}.*2222

*That is going to be plus 6I _{3}, and at that point, I am back to where I started -- it must be equal to 0.*2245

*I can do the same thing over here with this loop. Let us say we go this way around the loop to get us another equation.*2253

*If I start down here, I see the negative side of this power supply first so I could write -16 + 12I _{2}, is what I come to next, + 6I_{3}, and now I am back to where I started, which equals 0.*2260

*So now I have a system of equations with several different unknowns.*2276

*A lot of different ways to solve them, but let us just work through this one in a sort of the brute-force method.*2281

*Over here, let us start with this: -12 + 8I _{1} + 6I_{3}, well I_{3} = I_{1} + I_{2}...*2287

*...so that will be 6 × (I _{1} + I_{2}) = 0, or -12 + 8I_{1} +6I_{1} + 6I_{2} = 0...*2297

*...or if I put that together, we could say that that is 14I _{1}, adding 8 and 6I_{1} + 6I_{2} is going to be equal to 12.*2317

*I got that from my first loop combined with Kirchhoff's Current Law up here.*2330

*Now let us take a look at this equation and see what we can do with that.*2335

*We will start by writing it as -16 + 12I _{2} + 6, and I am going to replace I_{3} again with I_{1} + I_{2} = 0.*2339

*Therefore, -16 plus -- well we are going to have 6I _{1} plus we will have 6I_{2} + 12I_{2} is going to give us 18I_{2} = 0...*2355

*...so I could write that then as -- well we have 6I _{1}, we have 18I_{2}, and all that must be equal to 16.*2370

*All right. So I have two equations, two unknowns. How am I going to solve this?*2383

*Well I see a nice and easy way to do that right here.*2388

*What I am going to do is I am going to take and I am going to multiply that whole equation by -3 and rewrite it down here.*2391

*If I multiply both sides by -3, I keep that equality -- that is a fair thing to do algebraically.*2398

*So -3 × 14 is going to be -42I _{1}; 6 × -3 is going to give us -18I_{2}, and -3 × 12 will be -36 on that side.*2404

*Why did I do that? Well if you look here now, if I add these two equations, if I add up the left-hand sides and I add up the right-hand sides, I can keep that equality.*2419

*When I do that, here is what I get: 6I _{1} - 42I_{1} -- well that is going to give me -36I_{1}, 18I_{2} and -I_{2} gives me 0, and that must be equal to 16 + (-36) = -20, so now I can easily solve for I_{1}, which is going to be -20/-36 or about .556 A.*2429

*I know I _{1}. I was able to get my first variable.*2455

*Now what I can do is I can take that value and I can plug it back into one of my previous equations over here.*2459

*What I am going to do is take that and let us come over here and I am going to start with this 14I _{1} + 6I_{2} = 12, but now 14I_{1} is .556 + 6I_{2} = 12.*2466

*14 × .556 + 6I _{2} = 12 -- fairly easy to solve that to say then that I_{2} is going to be equal to .703 A.*2486

*I know I _{1}, I know I_{2}, and I_{3} is just I_{1} + I_{2} -- so I_{3} = .556 + I_{2} (0.703), or I find that I_{3} is going to be equal to 1.26 A.*2499

*Find the current flowing through R3? That is I _{3}, so we have done one of those.*2520

*Let us make sure we box our answers so we do not lose it.*2525

*Now what is the power dissipated in R3?*2528

*Well for the power dissipated in R3, power3, well I know (I) and I know (R), so I can use power = I ^{2}R, where my I is 1.26 A^{2} times the resistance of R3 (6 ohms) or 9.5 W, so there is my other key answer.*2532

*The current flowing through R3? -- 1.26 A; and the power dissipated by R3? -- 9.5 W.*2555

*Let us do a problem with internal resistance now.*2566

*We have a 50-ohm and a 100-ohm resistor, connected as shown, to a battery with an emf of 40 volts and an internal resistance of (R).*2570

*So there it is. Here, this entire thing is our battery.*2578

*Find the value of (R) if the current of the circuit is 1 A. What is the battery's terminal voltage?*2580

*Well over here I have two resistors and parallels, so the first thing I am going to do is I am going to re-draw this as a nice series circuit, or I am replacing these by their equivalent resistance.*2589

*So, we will have our 40 volts here, we still have our internal resistance (R), and our R-equivalent between 100 and 50?*2600

*Well R-equivalent is going to be 100 × 50/100 + 50, so 100 × 50 = 5000/150 -- R-equivalent is about 33.3 ohms.*2614

*I will put that in here, 33.3 ohms, and our circuit just became a whole lot simpler.*2631

*Now we know the current flowing through here is 1 A, and if that is the case, the current flowing through here is 1 A as well.*2640

*So as I run through and use Kirchhoff's Voltage Law around this loop -- positive side and negative side of my battery -- the first thing I am going to see is -40, then I have 1 A × R, so that is going to be +1R or +R, + 33.3 × 1 A, which is 33.3, so that must equal 0.*2649

*It is pretty easy to solve for this -- R is going to be equal to 40 - 33.3 or 6.7 ohms. We found the value of R.*2672

*What is the battery's terminal voltage?*2685

*Well if we want the terminal voltage of our battery, Vterminal is the battery's emf - IR.*2688

*That is going to be our 40 volts - I (1 A) times the internal resistance (6.67 ohms), so 40 - 6.7 is going to be 33.3 volts.*2695

*Because our battery has some internal resistance, you do not get the whole 40 volts out when the circuit is in operation, you get 33.3 volts across the terminals of our battery.*2709

*Let us take a look at a complex circuit with meters.*2723

*Given the schematic diagram below, determine the reading of both the ammeter and the voltmeter.*2726

*Well what I am going to do to start this off -- we have a couple of power supplies here -- is just look at this for a minute and see what we can figure out.*2732

*We will call this the zero side of our battery, so that is 30 volts, that must be 30 volts here, we have current flowing, we will call this I _{1} flowing through R1.*2739

*Over here we have +5 volts -- well I am going to call this side 0 volts again because it is on the same wire, and this must be 5 lower so that must be -5 volts over there.*2753

*If this is 0, we must have 0 volts over here as well, and we are trying to find the reading of both the ammeter and the voltmeter.*2760

*Let us add I _{2} for the current flow through there and we must have I_{1} flowing through here as well.*2773

*That would be an I _{3}, so let us see what we have as we apply Kirchhoff's Voltage Law around the loop.*2781

*What we are going to do is we are going to start here as I apply Kirchhoff's Voltage Law, I am going to see -30 first, plus 10I _{1} + 20I_{1}, then I see the -5 before I get back to where I started, which equals 0.*2789

*Well I have one equation and one unknown: this just became relatively simple.*2810

*What I can do then is, with a little bit of Algebra here, and say that I have 30I _{1} must equal 35, therefore I_{1} = 35/30 or about 1.17 A.*2815

*I _{1} is what goes through the ammeter, so there is our first answer.*2834

*Let us try and find the reading on the voltmeter.*2840

*Well the voltage drop across this R1 is going to be I _{1} × R1, so V1 = I_{1} × R1.*2844

*That drop then, I _{1} (1.17 A) × r1 (10 ohms) is going to give us a drop of 11.7 volts.*2855

*If this side is 30 and we drop 11.7, that must mean we have 18.3 volts left over here and the voltmeter measures the difference from this point to that point, that is how it is attached.*2865

*The difference between 18.3 and 0 volts is just 18.3 volts, so the voltmeter reads 18.3 volts -- our second answer.*2879

*That was not nearly as bad as I expected when we first looked at it.*2896

*Let us take a look at one more nice, simple problem to round this out.*2901

*Three resistors: 4 ohms, 6 ohms and 8 ohms are connected in parallel in an electric circuit.*2907

*The equivalent resistance of the circuit is...?*2912

*Well right away if they are in parallel I know that my equivalent resistance must be less than 4 ohms.*2916

*Oh wait, there is our answer right there -- less than 4 ohms. All of our other choices are larger than 4 ohms.*2920

*Very little thinking required, it is just knowing that formula and a couple of facts.*2927

*Hopefully that gets you a great start with circuit analysis.*2931

*Thanks so much for your time in watching Educator.com. We will talk to you again soon.*2934

1 answer

Last reply by: Professor Dan Fullerton

Sat Jan 14, 2017 3:16 PM

Post by Vivian Ni on January 13 at 06:06:30 PM

At 33:02, I understand how you got 3.2 V, but how is V the same for both R2 and R3? The resistance is different for both, so I'm confused as to how the voltage is the same. Thanks!

1 answer

Last reply by: Professor Dan Fullerton

Thu Aug 20, 2015 8:05 AM

Post by Anh Dang on August 19, 2015

Sorry about this question, but can you remind me what exactly a potential drop is?

2 answers

Last reply by: Professor Dan Fullerton

Thu Jun 25, 2015 3:45 PM

Post by Derek Boutin on June 25, 2015

Professor Fullerton, the lecture was extremely helpful. However, I do have two questions. In Example 11, why do you use -12 and -16? Also, in Example 13, why do you start with 30 Volts and end with 0 Volts? Thanks!

1 answer

Last reply by: Professor Dan Fullerton

Mon Feb 23, 2015 9:03 PM

Post by Yahaira Leon on February 23, 2015

Why did you divide 50*30 by 80

1 answer

Last reply by: Professor Dan Fullerton

Sat Mar 29, 2014 7:39 AM

Post by Hoa Huynh on March 29, 2014

Dear Professor,

At 46:11, I do not understand why the voltage right below the emf 5V is 0. If we use the loop, the current should flow from the + side to -side. Please, explain me how can we know the current flow on what way. Thank you

1 answer

Last reply by: Professor Dan Fullerton

Wed Mar 5, 2014 5:42 AM

Post by ibrahim shawi on March 5, 2014

is it possible to use VIRP for this problem?

1 answer

Last reply by: Professor Dan Fullerton

Wed Mar 5, 2014 5:44 AM

Post by ibrahim shawi on March 4, 2014

for example 13 complex circuit with meters, You labeled the first resistor R1 and then the second one also as R1 is it because they are in series? also in the beginning of the equation you made the 30 negative.... is that because of equation..... kirchhoff's law?

1 answer

Last reply by: ibrahim shawi

Tue Mar 4, 2014 9:52 PM

Post by ibrahim shawi on March 4, 2014

for the basic parallel circuit analysis lecture at 20:53 you said that total R is equal to 667 and i see how you got it but when i solve for the answer using 1/R=1/R+1/R+1/R I get 1.5x10^-3....

1 answer

Last reply by: Professor Dan Fullerton

Thu Jan 30, 2014 7:38 AM

Post by Karpis Sanosyan on January 29, 2014

You really help me understand, but i don't understand how you get the voltage by putting one side 0

?

1 answer

Last reply by: Professor Dan Fullerton

Sun Oct 27, 2013 10:01 AM

Post by Yadira Perez on October 26, 2013

Excellent explanations, I was so lost in class but you have made a difference thank you!!

1 answer

Last reply by: Professor Dan Fullerton

Wed May 8, 2013 6:12 AM

Post by Nawaphan Jedjomnongkit on May 8, 2013

Thank you so much for great lecture and I like your teaching style that give a lot of example questions after theory part. Yet I still not so confident about circuit analysis with 2 voltage sources. Is it possible that the voltage from one sauce cancel the other like when I put batteries in wrong side will have no power? Take care , don't catch a cold ^_^