AP Physics C: Electricity & Magnetism 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 series0009

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 go0549

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 measure0683

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 and0700

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 connected0733

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 through0742

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 mw1305

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