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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來自 University of Houston System 的課程

Preparing for the AP Physics 1 Exam

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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.

- Dr. Paige K. EvansClinical Associate Professor

Math - Mariam ManuelInstructional Assistant Professor

University of Houston

There are two rules called Kirchhoff's Laws that are very helpful when analyzing

circuits.

The first law we will study is called Kirchhoff's junction rule,

sometimes referred to as Kirchhoff's current rule.

Fundamentally, this rule is based upon the concept of conservation of charge.

When three or more wires meet at one point, all the current

coming into the junction must be equal to the current leaving the junction.

Since charge cannot be created or

destroyed, all of the electrons entering must also leave.

Here's an example where Kirchhoff's junction rule can be applied.

So, in order to solve this question,

I am going to use Kirchhoff's law that we discussed.

It tells me that the wire is carrying a total of 10 milliamps and

that two of the wires carry 3 milliamps.

So here's how I'm going to look at this.

My total current is going to equal

current 1 + current 2 + current 3.

I know that my total current is this 10 milliamps.

Say that branch 1 has 3 milliamps,

branch 2 has 3 milliamps and I need to solve for

the current going through this third bridge.

Well, 6 + 4 = 10, so

the current there is going to be 6 milliamps.

The second rule is called Kirchhoff's loop rule or Kirchhoff's voltage rule.

This physics principle is based upon the concept of conservation of energy.

Since voltage is related to the energy of the charges as they move around a circuit,

energy must be conserved and total charge in voltage for a complete loop must be 0.

If a charge were to gain or

lose voltage each time it completes a loop around a circuit,

it would either gain or lose energy and would not fit energy conservation.

Let's look at an example in a simple circuit.

So, for this question I'm going to use my Kirchhoff's rules, and

in this case they tell me that the total voltage of the battery is 6 volts.

They tell me that the first resistor is measured to have 2 volts of voltage drop

and they want me to figure out the voltage drop in the second resistor.

So the way that I can go ahead and solve for

this, Here would be a sketch of the picture.

6 volts is in the total battery over here.

I lost 2 volts over here.

So my combined voltage has to equal to 6 volts,

which means that the voltage drop across the second resistor is 4 volts.

>> The example we just worked is what is called the series circuit.

Just like a series of books, a series circuit refers to a scenario in which

circuit elements are placed directly one after another.

Because of Kirchhoff's junction rule, we know that all parts of a series circuit

must have equal current since there are no junctions or splits.

All of the charge from the battery must go through all of the other elements,

because charge is conserved.

Because of Kirchhoff's loop rule, we also know that the voltage dropped across

all the circuit elements must equal the voltage across the battery.

When solving problems with series circuits,

these three steps are helpful guidance.

1, find the equivalent resistance of the entire circuit.

2, use Ohm's law to find the total current, I, through the battery.

3, apply Ohm's law to each resistor

to find the voltage drop across each.

What do we mean by equivalent resistance here?

Suppose that three resistors are connected in series to a battery.

It is possible to buy one resistor to replace the three separate resistors so

that the current through the battery would not change.

To find the equivalent resistance in series,

you only have to add them together mathematically so

that the equivalent resistance equals R1 + R2 + R3.

Here's an example.

In this question, it tells me we have three resistors,

one that's 2 ohms, one that is 3 ohms and one that is 4 ohms,

and they're connected in series to a 9 volt battery.

It wants us to calculate the voltage across and

the current through each resistor.

My first step in a problem like this is to always draw a picture

to help me visualize what's going on.

I have three resistors and when they are connected in series that means that they

are connected end to end to one another.

So, I would call this Resistor 1, this Resistor 2, and this Resistor 3.

This is our symbol for our battery.

Remember that the long side of the diagram is the positive side of the battery.

The small side is the negative side.

That means that current through the circuit

will leave the battery going this way.

We talk about conventional current, positive current flow.

That means it goes clockwise around this loop and back again.

So we know the direction.

Also since they're in series they all have the same amount of current because

whatever electrons leave the battery must go through each of those resistors.

There's no other forks or splits for them to go anywhere else.

So, the first step in a series circuit like this is to find your equivalent

resistance.

And since they are in series,

I can find my equivalent resistance by adding them together.

They're end to end, like this, so literally, my equivalent resistance,

I can go to the store and buy, instead of a 2, 3 and

4 ohm resistor, I could go and buy a 9 ohm resistor instead,

and redraw the [INAUDIBLE] what it looks like at this point.

I have my same 9 volt battery, but

instead I have an equivalent resistance of 9 ohms.

Okay, the next thing it wants us to know is the current and

voltage through each resistor.

Well, now that I have the equivalent

resistance I can find the equivalent curve.

V = IR.

The voltage that this equivalent resistance is connected to is 9 volts.

The resistance, 9 ohms.

So, I know that there is 1 amp going through this equivalent resistance.

That means that there must be 1 amp going through the battery.

Keep in mind that the battery can't tell the difference between this equivalent

resistance and those three separate resistors.

That means that there is 1 amp going through this battery,

1 amp going through this resistor, 1 amp going through this resistor, and

1 amp going through this resistor.

And now we have all of the currents in the problem.

Our next thing it would like us to find is the voltage across each resistor.

Well, remember that we just found the current.

That's I.

We know the resistance, R1 is 2 ohms,

R2 is 3 ohms and R3 is 4 ohms.

So, we know the current and the resistance for each.

Then you come along and use R Ohm's Law for each individual circuit element,

V = IR, to find the voltage.

In this case, the current being 1, the resistance being 2,

R voltage for resistor 1 would be 2 volts.

Same thing here for this resistor.

V = IR for Resistor 2.

The current is 1 amp like we found earlier, 3 ohms,

we have a 3 volt drop on that resistor.

And for the last one here, V = IR.

1 amp x 4 ohms gives me 4 volts.

So, notice then what we have.

We dropped 2 volts, and then 3 volts, and then 4 volts.

So, as your check, go back and follow your loop role.

I know that going from the negative to the positive of

a battery means we go up 9 volts.

We then dropped 2, we then dropped 3, we then dropped 4.

That means we back down to 0 volts.

And the loop continues.

But all the rises in voltage minus all the falls should bring you back to 0.

It's based upon conservation of energy.

>> A parallel circuit is one in which the ends

of the circuit elements are connected together as seen here.

Notice how that a junction exists.

The current from the battery must split.

Because of Kirchhoff's junction rule, the total current through the battery must

be equal to the sum of the current in the three branches.

In this case, total current equals current 1 + current 2, + current 3.

Because there are now three separate loops, Kirchhoff's loop

rule shows us that all three resistors must have equal voltage drop across them.

In this case voltage 1 = voltage 2 = voltage 3,

which equals the voltage of the battery.

When solving problems with parallel circuits,

these three steps are helpful guides.

Step 1, find the equivalent resistance of the circuit.

Step 2, use Ohm's law to find the total current through the battery.

Step 3, apply Ohm's law to each resistor to find the current in each

branch using the total voltage.

So, for this question, if you recall, are rules for parallel circuits.

I've got three resistors connected in parallel.

So, it looks something like this.

They tell me that I have resistance of 2 ohms, 3 ohms and 4 ohms.

They also tell me that they are all connected to a battery that is

total 9 volts.

They want me to calculate the voltage across and

the current through each resistor.

So, in order for me to do this, I can solve for

the voltage across each by knowing my rules which is that my

total voltage is going to equal the voltage going down each branch.

So for instance, I have 9 volts running through here,

9 volts running through here and 9 volts going down this bridge as well.

In order to solve for my current, say this is 1, 2, 3.

I can solve by saying V1 = current 1 x R.

V is 9, current is what I'm solving for and R is 2.

This gives me current equaling 9/2,

which equals 4.5 amps.

Similarly I can solve for

the rest of them by saying current 2 equals voltage

divided by resistance 2, which is 9/3.

This gives me 3 amps.

Current 3 equals voltage divided by

resistance 3, which is 9/4,

which equals 2.25 amps.

And those are all of the currents, and

the voltage as mentioned was 9 volts across each branch.

>> So, now that we are looking back at the previous problem which was

a series circuit, recall that we found the voltage and

the current for each of these three resistors.

Now, what it wants us to do is calculate the power dissipated in each resistor.

A question that might come afterward is that if these were instead lightbulbs

rather than resistors, which is the brightest bulb and

which one is the dimmest bulb?

To do that we need to compare power, and in circuits,

the base equation for power is current multiplied by voltage.

And of course, you could sub n, V = IR, Ohm's law,

into this equation to get other versions.

But since we know everything at this point that we need, I x V works great.

And I can do this calculation for each of the three resistors.

For instance, for resistor 1 here, the current was one amp, and

the voltage across it, the voltage dropped 2 volts.

So, there's a 2-watt power dissipation over resistor 1.

If I wanted to do this calculation again, but this time for resistor 2,

the current was 1, the voltage this time 3 volts, the power is 3 watts.

And for the third resistor, P = I times V, the current again was 1,

they all had the same amperage, because they were all in series.

The voltage, 4 volts, I get 4 watts.

This means that if these were lightbulbs, that third resistor,

the one we just calculated, the 4 watt resistor, would be the brightest by volt.

Quick check to see that we haven't made any mistakes here

is that the power dissipated for each of these resistors

should add up to the power dissipated by the battery overall.

So, I can actually come over here to the battery itself,

multiply the current of the battery times the voltage of the battery, and

see that overall, the circuit is using from the battery, 9 volts.

And when I add these three to the right, together, 4 + 3 + 2, that also gives me 9.

And so I can see the power given off by the battery is then used up

by the resistors, which is what we expect.