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Lecture Comments (15)

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

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


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


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?

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 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

Transcription: Circuits II: Parallel Circuits

Hello, everyone, and welcome back to

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 resistance0298

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

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