Circuits C

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Из курса от партнера University of Houston System

Preparing for the AP Physics 1 Exam

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University of Houston System

Preparing for the AP Physics 1 Exam

36 оценки

This course is designed for high school students preparing to take the AP* Physics 1 Exam. * AP Physics 1 is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.

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Introduction to Circuits

Topics include elective charge, electric force, and DC Circuits. You will watch 3 videos, complete 3 sets of practice problems and take 2 quizzes. Please click on the resources tab for each video to view the corresponding student handout.

Circuits A26:18

Circuits B14:51

Circuits C22:25

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Dr. Paige K. Evans
Clinical Associate Professor
Math
Mariam Manuel
Instructional Assistant Professor
University of Houston

So in order to solve a complex circuit you need to keep your series rules and

your parallel rules in mind.

In other words, when does voltage add up, when does voltage stay the same,

when does current add up, when does current stay the same.

So you want to keep those things in mind.

In order for me to solve for this, let's look at what it is I'm solving for.

My unknowns here are voltage one, voltage five, and voltage six.

Those are the values I need to solve for.

So I can go ahead and

notice over here that I've got these two circuits over here that are placed in

parallel with one another, voltage four and voltage five, and voltage one and

voltage two, these resistors over here are also placed in parallel with each other.

Well, my rules for parallel tell me that the voltage is going

to be the same in each branch, which means that if voltage for

resistor four is five volts, then voltage five is also going to equal five volts.

Similarly for voltage two, for resistor two, that's two volts,

which means voltage one would also equal two volts.

But now let's look at voltage six.

This is directly in series with the battery.

So the voltage isn't going to be the same.

However, if I treated all of this as if it were one big series circuit,

in other words, here's my battery, here's voltage six,

this one could be voltage four and five combined.

Then I have voltage three, and then I have voltage or resistors one and two combined.

Well, what this tells me is that if I combined the voltages across all of these,

they should add up to the total, which is 12 volts.

So let's go ahead and look at that now.

I've got five volts over here.

Remember, I'm not going to add those.

They are both each getting the same voltage, which is five.

I've got one volt over here.

Two volts over here.

And an unknown amount of voltage over here,

all of which needs to add up to 12 volts.

This tells me that my unknown voltage, or

voltage six, it's going to equal four volts.

When I add up four plus five plus one plus two, I end up with 12 volts.

Since I solved this question, again, like I said with the other complex circuits

problem, we want to keep out rules for series and parallel in mind.

If you notice the server, it gives me both the total voltage, and the total current.

Now let's look at what they're asking me to solve for.

So, I'm going to go ahead and solve for,

what the current is over here, as well as here.

For current four and current six I'm going to solve for voltage five.

I'm going to solve for the current and voltage for resistor three.

Similarly, for resistor two, I'm going to solve for the current and voltage.

So basically, solving for any missing current or voltage at each resistor.

Now let's go ahead and start.

There's several different points of view you may want to start from.

Where I'm going to start from is looking at my battery.

My battery gives me the total current which is five amps.

I notice that I have resistors that are in series.

Remember that in series, the current stays the same.

What this tells me is that I've got this resistor over here,

resistor six, while the current over here is going to be five amps.

I can go ahead and continue from here.

I can notice that these two resistors, four and

resistor five, each are parallel with one another.

However if I combine them as one resistor their total currant would be five amps.

What that tells me is current five plus current four need to equal to five amps.

Because again remember that would place them combined in series with the battery,

having the same current as the battery, which would be five amps.

So if current five is four amps, then that means current four must be one amp,

adding up to a total of five amps.

Then I notice over here I've got a similar situation with current one or

resistor one and resistor two.

Now for resister one I know what my current is.

It's three amps, and I know what my voltage is.

It's two volts.

Resister one and two are in parallel with each other.

If you think back to the problem we worked through just before this one,

if they are parallel, these resistors to each other,

then that means that their voltage is going to be the same.

Which means that v2 = 2v.

Now keeping in mind, if I combine resistors one and twp together,

place them in series with the battery, the total current both of these together or

combined would get would be the total current of the battery,

which is five amps.

Which means, if current one is three amps,

then current two must be two amps, adding up to a total of five amps.

Then I can go ahead and look at resistor three.

For resistor three, again, looking at it as it,

it is in series with the battery which it would be.

It gets the same current, five amps.

Then I can go ahead and look at what the voltage would be.

Well keep in mind what my rules for voltage are in series.

Our rules for voltage and series is that the tool voltage, which in this case is

ten, is going to be equal to the total voltage, that's at this point over here,

with the resistor three, as well as resistors four, and five, and six.

In other words, total voltage.

The voltage is going to drop at each point as it goes across in series.

When I look at the total voltage that I have,

having the two volts over here, between resistors one and two.

Having three volts over here for resistor four.

And then if I look at resistor five, and I want to solve for

what my voltages over there,

I know that that would also be three volts because they're parallel with one another.

And then it has one volt left as well as resistor state.

If I want to solve for this and figure out what's left in order to add up to a total

voltage of ten, my answer comes out to four volts.

Feel free to check these and

make sure that your total voltages across each series adds up to ten volts.

And that solves for all of the unknowns in this question.

>> When you first approach a problem like this, it can look a little overwhelming.

We're going to follow the same rules we've been following for complex circuits, and

take it one step at a time.

First, I'm going to look for any resistors that are obviously in series or

obviously in parallel.

Then start to simplify the circuit to get one equivalent resistance and

then solve from there.

So I notice first of all that these two resistors on the end

are definitely in series.

And I can combine those two by adding their resistances.

The equivalent for those two would be, in this case, R2 + R3.

If I combine those two by adding them since they're

in series it'd be 19 + 5, I get 24 ohms.

A good practice in solving a problem like this is to then redraw your circuit each

time with the new information that you just picked up.

So I notice that I still have resistor five and resistor six.

I still have resistor one and resistor four.

But we've essentially gone to the store bought ourselves the new equipment

resistance here of 24 ohms.

I still have resistor four here though.

Which was eight ohms.

And I noticed that those two are definitely in parallel.

Their ends are connected together.

To combine resistors in parallel,

my equation is a little bit more complicated.

one over R4 plus one over what I would call R2,3,

the combination of two three that we got from above.

This number here.

So we begin subbing in one over one over eight plus one over 24.

Don't forget the one over.

A common mistake is to forget to take the inverse when you're combining them

like this.

The one over 24 plus one over 28 gives me six ohms.

So far pretty nice numbers that we're dealing with.

I'm going to again, let me draw my circuit so

that we can keep track of what we've learned.

I still have this resister, and this resister.

I have this resister here, but

now we've combined those two into a nice six ohm resistor.

I notice then that our six ohms resistor, and I'm going to go back up and

notice our one here which is 15 ohms

is now definitely in the series.

With our new equivalent resistance.

So I can combine those and notice that it's a process of kind of going back and

forth, back and forth.

The equivalent resistance there would be R1, and in variables what we have

now is something more like R2, three, and four, combined together.

That means that my equivalent resistance

would be 15 plus 6 or 21 ohms.

Again drawing my new equivalent circuit I still have these two resistors, and

now I have the 21 ohm resistor that we've just located.

I notice that R5, however, this one here, let me scroll up to see what that was,

a nine ohm resistor.

Was R5, those two are end to end and

in parallel, so we've got one more parallel combination to set up.

The equivalent resistance.

one over one over R5 plus one over, and

if you wanted to name this appropriately I supposed I'd call it one through four.

Those are where this is one through four added together.

one over, that means R5 was nine ohms.

one over nine Plus the one over 21 we just found

gives mean equivalence resistance of 6.3 ohms.

Not as nice as some of our other numbers, but

that means that we have one last circuit to draw.

We've got the 6.3 ohms we just found.

And one little resistor left, our R6, just two ohms.

And this is still our battery.

That means the last equivalent resistance would be our R6.

Plus, these are in series so there's just one added to the other.

What we had, which was R1 through five.

Which means we would have 2+6.3 or 8.3 ohms.

So what did we just find?

By combining all of this, we now know that it would be perfectly

acceptable to go to the store by one equivalent resistor of eight

point three ohms and the battery could not tell a difference between this or

that whole network of resistors we had a moment ago.

At this point the steps usually the same as we've been doing.

V equals IR, ohms law, for this equivalent circuit this 8.3

ohm resistor is connected to a battery of voltage 17 volts according to our problem.

That means the current, and I'm going to try to go out a few decimal places here,

to not round too much and introduce too much error.

I get a current of 2.048 might have a slightly different length than myself.

Now, that I know that.

That's the current that goes through this circuit.

That is the current that would go through this real battery here.

The battery can't tell the difference between our two circuits, so

this is 2.048 amps.

And at this point, it becomes a little bit more of a puzzle.

You have to be able to break down what you know in the problem and

what you can solve for using the two rules, Kirchhoff's Junction and Loop Rule.

The first thing I notice and you'll get better at this as you practice it more.

Is this resistor and the battery are in series.

That means they have to have the same current.

So I automatically know the current through this resister because

they're in series, 2.048 amps.

Once you know two things, you can always find the third.

Because I'm going to use Ohm's Law for this little resistor here.

I know the current, 2.048, the resistance, 2 ohms.

So I can find the voltage across this resistor, which I'll mark here in red.

Voltage six across resistor six.

4.096 volts.

And now you have to start looking around.

With what we know now, how does that help us get to the next step?

Well, let's keep going with current and see how that helps us.

We just said that 2.048 amps comes from our battery.

And it's this junction.

There's a split.

That means some current will go this way.

That will be in I5.

It also means some current goes this way.

I'll call it I1. And

I'm giving them those names based on the resistors.

You could call them anything else you'd like.

Now I don't know exactly how that splits.

The current doesn't split evenly, necessarily,

it depends on the resistance that's there.

Less current will go through an area of more resistance, and vice versa.

So what I do know, though, is my voltage loop rule.

Notice, for example, that I started off going up 17 volts.

Now I'm going to pick a path current can flow.

That means it drops some amount of voltage across this resistor I.

The remaining voltage across this resistor, and

now I need to be back at zero.

That's what my loop rule tells me.

All the ups minus all the downs brings me back to zero voltage.

So we went up 17 down some unknown,

down 4.096 volts, and that brings you back to zero.

That means that we could set up an equation based on that path.

You wanted to set up a nice neat equation of variables,

I'd probably do something like this.

It went up the voltage of the battery.

Minus the voltage on five, minus the voltage on six, and

that should bring me back to zero.

And I could sub in both the voltage of battery, which is 17.

And the voltage of resistor 6, 4.096 volts, equals 0.

In the end, what I learned is that the voltage across resistor 5,

is going to be 12.904 volts, and again, I'm going outside my here.

Now that I know two of the three things.

I know the voltage, I know the resistance there.

I can use V equals IR for resistor 5.

I know the voltage which is are 12.904.

I'm looking for current and I know the resistance was 9.

So again I find that the current that went in this branch which is 1.434 amps.

I know all three.

Now I'm going to use Kirchhoff's junction rule.

And it's a series of steps

you'll notice as we go through that it looks very similar.

We do voltage and then current.

Voltage and then current.

Now that I'm at this junction again, I know how much current came into this

junction, which we said was this number here, 2.048.

I know some went down, and

I know exactly how much that was, that was this value, 1.434.

So I can find the remaining by finding the total current.

I know that the current from the battery minus the current that went I

5 equal how much current goes I 1.

If it's easier for you to imagine it this way, think of it this way.

The current's coming into the junction which is the current through the battery

would equal I1 plus I5 both leave.

Either way I'm going to sub in for the current from the battery and

the current through I5.

And I solve that I1 equals the remaining

which is 0.6145 amps.

Now that I know that I1 and R1 I'm going to multiply using V equals IR.

I'm going to go ahead and change colors here.

And the current we just got .614 amps, the resistance 15,

so the voltage across the first resistor

there 9.217 Volts.

And you'll notice it starts to get kind of messy.

So you might need to space out your work or however you need to keep track of that.

Again, now I know that this amount of current comes into this junction and

again I don't know exactly how it splits.

I know there's an I4.

I know there's an I2.

So instead I'm going to use my loop rule again.

Again, I pick any loop that current could travel, and it should still apply.

We go up 17 volts, we said we went down across resistor one, 9.217.

We would go down some unknown voltage, down the remaining 4.096 and

I'm going to use that equation to help me solve for voltage 4.

So my calculation would be 17

volts minus 9.217 minus 4.096.

Give me the remaining voltage that used to be across resistor 4.

Let me cleanup my board real quick.

So using that loop rule, by subtracting from voltage 4,

I get 3.687 volts.

Again, once I know two of the three things, I can use a law V=IR.

We know the voltage already, voltage 3.687,

the current is what I'm looking for.

I4 is .4608 amps.

Now that I know the current that went this way,

you'll notice it's a pattern.

We know how much current came in, that's this number here, .614.

I know how much current went in this section,

in this junction, which was 0.468.

I could subtract to find the remaining current, I2.

When I do that calculation, I get a current of 0.1536 amps.

Again I know 2 of the 3, so I'm going to use V equals I R.

The current, 0.1536, the resistance 19.

The voltage across resistor 2,

2.919 volts.

Now I'm going to get to the last section here.

I noticed that R2 here and R3 are in series,

they have to have the same current.

So I automatically know current's three, which is the same as

the current through number two, which is 0.1536 amps.

Again I know two of the three using Ohm's law, V equals I R,

I get the final voltage, which is 0.7681 volts.

The first thing you're probably thinking,

after all of that, how can you tell if you've made a mistake?

Well, one obvious place to check would be the power.

You could go through, calculate the power for each resistor.

Power being I times V and you know the power of all those circuit

added up should equal the power used up by the battery.

That would be one check.

Another possible check to show if I made a mistake,

would be to pick another voltage loop.

For instance, one we haven't checked yet is that current could leave the battery

going this route, go all the way around, down and back.

If that's the path, I know it needs to go up 17, down the voltage here, 9.217.

Down the voltage here, 2.919.

Down some voltage here, 0.7681.

And down the remaining, 4.096.

And should add up awfully closely to 0.

Just remember that overall the process is to first find an equivalent resistance as

we drew down here.

These diagrams will often help you not make a mistake or check your work.

And then you go through using Kirchhoff's loop rule and

junction rule to find the rest.

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