AP Physics - Electric CurrentDid you ever imagine what your pathetic life, grim as it already is, would be like without those little wall socket thingees you have mounted on the wall just above the deck of your house? Know what the Physics Kahuna is talking about? Well, he’s talking about your basic electricity!

People have only had electricity for the past hundred years or so. What a difference it makes in how we live. Just about everything in the modern home works off electricity – we make light, wash dishes, clean the deck, drill holes, cook food, freeze pork chops, sew clothes, cut wood, listen to music, cool food, watch moving images, mash potatoes, dry our hair, tape TV shows for later viewing, heat air and water, manipulate bits of information . . . the list goes on and on.

When we studied electric fields, we mainly dealt with charges that weren’t doing much. You ginned up a bit of charge by rubbing a balloon with a rabbit fur, and then had fun waving the thing around, attracting bits of paper. Great sport, but what good is it? Can’t really do any proper work with it, can you? Well, truth be told, static electricity was not terribly useful (even though the Physics Kahuna loves it) - aside from a physics lab (where it was explored with great interest) it was mainly used to entertain people during lectures or as toys.

Still, static electricity, while cool and interesting in and of itself, was not terribly useful. For one thing it was a lot of trouble. You had to rub things or rotate things – this entailed a lot of work – and the payoff was rather pitiful – you get a lousy spark or attract a bit of fluff to a charged up rubber stick. Big deal.

Electricity didn’t really take off and become the huge, dynamic engine of change that transformed the planet until a way was found to have a continuous flow of charge through a conductor. What was needed was a voltage source.

The first voltage source was what we now call the battery.

Battery Business: In the 1790’s Luigi Galvani found that he could remove the leg of a frog (from the frog), place it on his lab table, and make it twitch by touching it with a scalpel. This actually happened by accident. Kind of scary don’t you think? A dead leg suddenly twitching?

Well, it quite caught the fancy of Galvani. How could this be? After a series of experiments, he found that the cause of the twitch was a flow of charge generated by two dissimilar metals – the scalpel and the metal lab tabletop.

The word galvanized remains as a tribute to Mr. Galvani. One is galvanized when the muscles suddenly contract.

Alessandro Volta was inspired by this and learned to manufacture the first wet cells (the thing that led to the modern battery). He used bits of dissimilar metals to generate electricity placed into a sequence of electrolyte bearing bowls. The bowls were placed in a line and he connected them together with strips of metal that were half copper and half zinc. He called this device the “crown of cups”.

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The device produced electricity continuously! As fast as you drew the electrons off they were replaced with new ones. Amazing!

Allesandro also developed an even better method – this one involved a stack of discs that alternated between two different metals. Between the metal discs was a cardboard disc that had been soaked in salt water. If the top metal disc was connected to the bottom disc with a conductor, a continuous flow of electrons through the wire and the discs resulted. The device is called Volta's pile.

The unit for potential difference, the volt, was named for Volta.

These primitive devices evolved into the batteries that we use today – dry cells (your basic alkaline energy cell type thing) and rechargeable batteries (nickel cadmium and lead acid batteries).

Properly speaking a battery is a collection of cells. A single AA Duracell thingee is, to be correct about the thing, called a cell, not a battery.

A French physicist named George Leclanche (1839 - 1882) invented the dry cell over a hundred years ago. These cells were made up from a zinc can – the zinc acts as the cathode - and a carbon rod that serves as the anode. The electrolyte is a paste made up of manganese oxide (MnO4), ammonium chloride (NH4Cl), and powdered carbon. Modern dry cells include the alkaline type dry cells. These are constructed much the same as the older (and cheaper) zinc carbon types. The main difference is the electrolyte. Both use carbon rods and zinc cans. The electrolyte in the alkaline battery is basic rather than acidic. It turns out that the zinc can lasts longer in a basic solution. In this type, the NH4Cl is replaced with potassium hydroxide (KOH) or sodium hydroxide (NaOH).

Both of these cells produce approximately 1.5 V.

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The amount of charge that can be provided depends upon the size of the cell. A large D cell can produce a greater amount of charge than can a small AAA cell.

All of these types of cells or batteries are called primary cells. They produce electricity as long as the chemical reactions take place. When they cease, the battery is “dead”. At which point you go out and buy a new one. You use them and then throw them away. A second major type of cell or battery is also in use, these are called secondary cells or storage cells. These batteries, once they've been drained of current, can be recharged. Examples would be your old automobile lead acid battery and nickel cadmium batteries used in portable electric devices.

Batteries work because of a chemical reaction that takes place within the thing. These are known as oxidation/reduction reactions. You will learn all about them in advanced or AP chemistry. The name is often shortened to redox reactions.

In a redox reaction, electrons are transferred from one reactant to the other. The reactant that gains electrons is "reduced", and the reactant that loses electrons is "oxidized". This causes electrons to move from one chemical substance to another. Any movement of electrons is called an electric current.

Electric Current: Electric current is simply the flow of electrons through a conductor. Two things are needed for current to flow:

1. A voltage source (a dry cell will do nicely).2. A complete path through witch the electrons may move.

The carrier of charge in a solid conductor is the electron. Positive charge cannot move as it is tied up in the protons of the atoms, which are essentially fixed in position.

There are two different kinds of electric current; dc and ac. dc stands for direct current. This is where the electrons all flow in the same direction. ac stands for alternating current. In an ac circuit the direction of the current changes. We will deal mainly with dc current.

To maintain the flow of electrons - the electric current - a constant potential difference must be maintained. Such potential differences are produced by voltage sources. Common voltage sources are: dry cells, batteries, generators, magnetos, solar cells, fuel cells, and alternators – just to name a few.

The symbol for electric current is I. (This doesn’t make a whole bunch of sense. You have to understand, however, that this convention comes from France, so what did you expect?)

The unit for current is the ampere, which is abbreviated as A or amps.

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Electric current is defined as the rate at which the charge flows through the conductor. The equation for current is:

Equation for current.

I is the current in amperes, Q is the charge in Coulombs, and t is the time in seconds.

The equation given on the AP Physics Test is:

But we’ll use because it’s easier to write.

The ampere is a fairly large quantity, so it is very common to use the mA.

1 mA = 10-3 A

3.25 x 1017 e- flow past a point in 0.235 s. What is the current?

We first have to convert the number of electrons to the charge that they have in Coulombs.

Direction of Current Flow: This is an area of controversy. Doesn’t seem logical that people would argue about this, but they do. The convention that we will use is this one. The direction of current flow is the direction that a positive charge carrier would travel in the circuit.

Now this makes no sense at all, and wise people don’t use it. The U.S. Navy does not use this convention; that’s how bad it is.

Anyway for us:

electricity flows in the same direction as positive charge.

The reason this is so stupid is that there is no positive charge carrier. Only a negative charge carrier, the ubiquitous electron.

Actually, the only reason we use this stupid convention is that it is the one used on the AP Physics test.

So set your mind to using this stupid convention.

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Current Flow Basics: The path that the electrons use as they flow around and around is called a circuit. The simplest possible circuit would have a voltage source and a conductor that connects the positive and negative side of the voltage source.

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Refrigeration and Batteries, A Public Radio Commentary by Bill Hammack

I was walking through the hallways when a colleague motioned to me. In a very conspiratorial whisper she said "I need your help." Then added quickly "do you know what bizarre thing my husband does?" I looked away, not wishing to know, but she continued. "He stores, in the refrigerator, batteries!"

Now, I listened carefully, because one of the most alarming things to me about marriage was to find batteries taking over the refrigerator: D batteries guarding the luncheon meat, C cells surrounding the milk, and double As scattered throughout conducting, apparently, some kind of reconnaissance. My colleague asked me "Can you as an engineer shed any light on the need for putting batteries in the refrigerator?"

So, what does science say about batteries in the refrigerator?

First, why would it even seem sensible to do this? It's because a battery generates a current by a chemical reaction. When the chemicals exhaust themselves, the battery dies. This reaction is only supposed to take place when the battery's being used but - and here is the inroad for the refrigerator/battery enthusiast spouse - the chemical reaction does go on even when the batteries not being used. Over time the reaction will corrode the battery, covering the end with a brown film. So, how do you stop this reaction? The answer: Lower the temperature.

So, this means batteries stored in the refrigerator will, in theory, last longer. So far, all seems to be in favor of the refrigerator/battery enthusiast’s spouse. But, the key question is, how much longer will they last?

Consumers Report magazine took exactly 432 double A, C and D batteries. They stored some in the refrigerator, and some at room temperatures. At the end of five years they found that indeed the refrigerated batteries had more charge, but not by much. The room temperature batteries still had 96 percent of the charge of the refrigerated ones. So, is this enough to merit filling a refrigerator with batteries?

I suppose rational spouses could disagree, but to me it seems the answer is "no." Particularly when you consider the inconvenience of having to wait for the battery to warm up. Also, as the batteries come up to room temperature water condenses on them, which could damage electronic equipment.

Should you present your refrigerator/battery enthusiast spouse with these cold, hard scientific facts? No, I suggest you instead follow the advice of a poet. Ogden Nash once wrote: "To keep your marriage brimming, with love in the wedding cup, whenever you're wrong, admit it; whenever you're right, shut up." So, even though science is on our side, we will forever have batteries in our refrigerators. I'd suggest that as non-refrigerator battery storing spouses, we, instead, form a support group.

As the electrons move through the conductor, their motion is blocked somewhat by the atoms in the conductor, so the current encounters a kind of opposition to its flow. We define this as resistance. The unit for resistance is the Ohm. The symbol you ask? Okay it is this; .

The thing that uses up the electricity and provides the resistance is called the load. A load could be a light bulb, an electric motor, a solenoid, a heating coil, etc.

The resistance of a conductor is very small, so electrons travel through it very easily. If it had a lot of resistance it would be called an insulator. Many circuits have components placed in them to deliberately add resistance. These are called resistors.

For current to flow through a resistor there must be a potential difference across it – one side is at one potential and the other side is at another potential, the electrons flow through the thing attempting to equalize the potential. We call this voltage difference the voltage drop.

Circuit Symbols: There are a number of standard symbols that are used when a drawing is made of a circuit. These drawings are called schematics. In old timey days electronic devices would come with a schematic showing how the circuit worked. You don’t see this much these days, mainly because the circuitry today is usually shrunk down and put on a microchip. Can’t do much to repair a broken microchip.

The symbol for a resistor is:

The symbol for a battery is:

The large line is the positive side of the battery and the small line is the negative side of the battery. Where you have only a large line and a small one, you have a cell.

Conductors are simply straight lines.

The symbol for a switch is: It is shown in an “open” position. Which means that it has created a gap in the circuit so that current cannot flow.

Let’s look at a simple circuit.

The circuit has a battery, a switch, and a resistor. The resistor represents the load on the circuit. The load is the thing in a circuit that uses up the electricity. The load for a power plant is all the various places on the network that use the current that is produced. The “load” in a flashlight for example is the light bulb.

The direction of current is the direction of a positive charge carrier. This would be from the positive side of the battery to the negative side of the battery. So the current goes up (when the

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or+

switch is closed, making a complete circuit), makes a right turn, goes through the switch, then turns right, goes down, then turns left, travels right to left through the resistor, then turns right, then goes up and back into the battery.

Ohm’s Law: We now have three quantities we can measure in a simple circuit: voltage (potential difference), current, and resistance.

These three quantities are related to each other by a simple formula known as Ohm’s Law.

A potential difference across a conductor causes current to flow. The flow of the current is opposed by the resistance of the circuit. The resistance “uses up” some of the energy in the current and converts it to heat.

A waffle iron draws 7.1 A when plugged into a 120 V circuit. What is its resistance?

Resistivity: We consider short lengths of conductors to have essentially zero resistance, but this is not true for long lengths. The resistance of a conductor is given by this equation:

R is the resistance, is the resistivity of the conductor, l is the length of the conductor, and A is the cross sectional area of the conductor.

This equation is only good for wires that have a constant cross-section.

The resistivity depends on the metal, each metal has it own value. Generally you look the thing up when you want to do a problem.

The longer the wire, the greater its resistance will be. What about the size of the wire; its diameter? You can see from the equation that as the cross sectional area of the wire increases, the resistance will decrease. So a thick fat wire will have less resistance than a very thin one will.

Interesting. Think of it this way, in a big fat wire there are lots of paths for the electrons to go through. In a skinny wire, there aren’t as many paths and the

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Resistivities of Common Materials

Material

SilverCopperGoldAluminumTungstenIronPlatinumLeadNichromeCarbonGermaniumSiliconGlassHard RubberSulfurQuartz

Resistivity(m)

1.59 x 10 - 8

1.70 x 10 - 8

2.44 x 10 – 8

2.82 x 10 - 8

5.60 x 10 - 8

10.0 x 10 - 8

11.0 x 10 - 8

22.0 x 10 - 8

150 x 10 - 8

3.50 x 10 5

0.466401010 - 1014

1013

1015

75 x 1016

electrons are sort of scrunched together trying to get through, so they get held up more. This means more resistance.

If you look at the table of resistivity values, you see that the best conductor is silver with copper coming in a close second. Silver, great conductor that it is, is quite expensive, so copper is the conductor of choice when wiring things together.

Gold is sometimes used, not because it is the best conductor, which, from the table, you can see, but rather because it is pretty inert stuff and doesn’t react with its surroundings or corrode. Copper is quite happy to corrode. So electrical connections with copper wire that is exposed to the elements can corrode. That wouldn’t happen with gold.

Here is a lovely model of electricity flowing through a circuit.

We have a coal mine which puts 100 tons of coal into each car that goes through. The cars come in empty, they leave with coal. The mine represents the voltage source. Each car receives 100 tons of “potential energy”. This is equivalent of the energy provided by a battery.

The train travels along the track, which represents the conductor in a circuit.

As the cars travel along they pass through a car rate meter which measures the number of cars that pass in a minute. This represents the current, which is the number of Coulombs of charge per second. Here we have a rate of 5 cars/minute.

The next thing the train must travel through is a coal detector. This is the gizmo represented by This device measures the amount of coal in each car going in and out and measures the difference. For the first one, the tons of coal in each car is the same going in as coming out. This meter represents the volt meter which measures potential difference across a component. There is no potential difference in a short piece of conductor just as there is no difference in the amount of coal in each car traveling along the straight track.

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Pow er P lant5 0 0 tons/ min

C oal M in eadd s 1 0 0 tonsto each car

each car begins w i th 1 0 0 ton s

VM ass D ifference is 1 0 0 tons

V V

V

AA

5 cars/ minute

each car retu rns w i th no coa l

M ass D ifference is 1 0 0 tons

N o mass D ifference

Car rate M eter

The coal arrives at a power plant. This, the power plant, represents the load. Each car dumps the 100 tons of coal it began with. There is a coal mass difference across the plant of 100 tons per car. This represents the voltage drop of the load.

After the power plant, once again there is no mass difference between the cars. Also the car rate is still the same, 5 cars/minute.

Back at the mine, there is a mass difference as the cars go into the mine and leave the mine of 100 tons per car. This represents the voltage drop across the battery.

What a lovely little model. Is everything clear? What if there were 2 power plants?

What is the resistance of a copper wire, which has a diameter of 1.00 mm and a length of 52 m?

The resistivity of copper is 1.7 x 10-8 m. All we have to do is plug and chug.

Electrical Energy & Power: A battery contains potential energy which, when it is discharged, is converted into the kinetic energy of the electrons. The resistance of the circuit converts the kinetic energy of the electrons to thermal energy.

Ever notice how warm a light bulb gets after its been switched on for a few minutes? Toy ovens that little girls play with actually use the heat produced by a light bulb to bake cookies.

The circuit uses up the energy provided by the voltage source.

We know that electrical potential energy is:

The rate at which the potential energy is provided is the power:

becomes

But is the current, I. so

This gives us a simple equation for power:

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P is power, I is current, and V is the potential difference.

This equation will be provided to you on the AP Physics Test.

If we apply Ohm’s law to the power equation, we can develop a few other power equations as well. But you won’t have these for the test, although you can easily develop them yourself. You will have Ohm’s law to play around with, so this is a useful thing.

Here is the power equation:

Using Ohm’s law, we can plug IR into the power equation for V.

Also, from Ohm’s law

So

This gives us 3 equations for Power:

When electrons flow through a conductor, the power loss due to the resistance of the conductor is equal to .

This is called the or joule heating. It can be very significant in long conductors. For

example, power lines can have very large losses of energy to the loss.

When you buy electricity from the power company, you are not buying power. You are actually buying work, i.e. they sell you energy which you do work with. The unit that the friendly power company sells their energy in is not the joule. They sell it by the kilowatt hour. But what is a kilowatt hour?

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Power is the rate at which work is done: so

We can take a kilowatt hour and convert it to joules. This should convince you that it is in fact a unit of energy.

An electric heater provides 50.0 V to a nichrome wire of resistance 8.0 . Find the (a) current through the wire and (b) power of the device.

(a) We can easily find the current in the wire using Ohm’s law:

(b) Find the power:

You have a 75 W light bulb. It goes into a 120 V circuit. So what is (a) the resistance of the bulb and (b) the current that flows through it?

(a) Resistance:

(b) Current:

You watch TV for 2.8 hours. The TV is rated at 3.8 kW. If electricity costs 9.5 cents per kWh, how much did your little TV program cost you to watch?

All we need figure out is the number of kilowatt hours used up, then aply the 9.5 cents per kilowatt hour conversion factor.

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We didn’t round it off to two significant figures. This is because you just know that the power company would never round anything off (except to round it up), the greedy blighters!

An oven operates drawing 24.0 A on a 120 V household circuit. If you run it for 5.0 hours, how much do it cost (at 8.0 cents a kWh)?

Cost:

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Dear Cecil:How long does it take for electricity to travel from the generator where it is produced to the light bulb at my house? -- Roberto Chevres

Cecil replies:As is my practice with technical questions, I convened the Straight Dope Science Advisory Board. One honorable member proposed that we figure in the inductance, capacitance, resistance, phase velocity, and for good measure the wind direction, Planck's constant, and Avogadro's number. The motion was voted down and its author shot. Another member ventured that "the signal velocity would be the speed of light divided by the square root of the dielectric constant of the insulation." Jeers from the back benches, author left in tears. Finally a junior member chirped, "Let's call it one-third the speed of light and adjourn to the refreshments." Approved by acclamation followed by a rush to the bar. So there you have it. Incidentally, while the juice as a whole moves pretty fast, the individual electrons don't. This being alternating current, they dance frantically to and fro and never get anywhere. Just like thee and me.

AP Physics - Electric Circuits, DCCircuits are classified by the type of path that the electricity follows as it goes around the circuit. There are two types of circuits - series circuits and parallel circuits. In a series circuit there is only one path. All the circuit components are in line, connected by the conductor, so that all the electrons flow through each component. A parallel circuit offers different paths – some of the electrons can go this way and some go a different way.

Series Circuits: Here’s a simple series circuit with three resistors; R1, R2, and R3. When the electrons leave the battery (opposite the direction of the current), they all go through the first resistor they encounter. Then all of them go through the next one and the next one. Then they all go back into the battery.

The current is the same at every point in the circuit.

Each resistor has a potential difference across it.

The voltage, current, and resistance in a series circuit behave according to the following rules.

Rules For Resistance in Series:

1. The current in every part of the circuit is the same.

2. The total resistance in the circuit is equal to the sum of all the resistances.

3. The voltage provided by the voltage source is equal to the sum of all the voltage drops across each of the resistors

Mathematically:

I = I1 = I2 = = In

On the AP Physic Test, the equation for series resistance is given as:

This simply says that the total resistance is the sum of the individual resistances.333

V

R1 R2

R3

I I

II

No equations for voltage and current are given. You’ll have to remember that the current is the same at every point in the circuit and the voltage drops add up to the total voltage provided by the voltage source.

Time to do the odd problem.

Find (a) the total resistance and (b) the current in the circuit to the right.

(a) To find the total resistance, we simply add up the three resistances.

(b) Find I: V = IR,

That was a phun problem, may I have another? You bet.

Figure out (a) the current and (b) the total voltage in this circuit.

(a) Find current through R1 :

This is the total current.

(b) We can use Ohm’s law to find the voltage. First we find the total resistance.

R = 45.0 + 75.0 = 120.0

V = IR V = 0.267 A (120.0 A)

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2 5 3 5

3 2 18 V

4 5 7 5

12 V

Parallel Circuits: Parallel circuits offer the current more than one path.

The current, I, comes to the junction between R1 and R2 and splits. Some of the current goes through R1 and some goes through R2. If you add up the current in each leg, I1 and I2, their sum would equal I.

Each leg of the parallel circuit sees the same electric potential, V.

Imagine a 12 V car battery. You drive two vertical rods into the soft lead electrodes of the battery. The potential difference across the electrodes and now the rods is 12 V. You clip the leads of a light bulb onto the vertical rods. The voltage drop across the bulb (which acts like a resistor) is 12 V. The metal rod and the wire leads for the bulb have essentially no resistance and

there is no potential difference from one spot to the other in a short length of conductor – we discussed this before. Only the bulb has a potential difference. Next, we add a second bulb. But what have we done to change the voltage drop? Nothing! Each of the bulbs sees a voltage difference of 12 V. Each side of the two bulbs are connected together and have to be a the same potential. We could even add a third bulb. In fact we could add as many bulb as we like. Each would see a voltage drop of 12 V.

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R

R

1

2

V

I I1

I2

I1I

Power

Pow er

P ower

1 2 V1 2 V1 2 V

12 V12 V12 V

Rules for Parallel Circuits:

1. The voltage drop is the same across each leg of parallel circuit.

2. The total current is the sum of the currents through each leg.

3. The total resistance is less than the resistance of any one branch. The reciprocal of the total resistance is equal to the sum of the reciprocals of the resistance of each branch.

The equation to find the total resistance:

Which basically means:

To find the total current:

I = I1 + I2 + + In

And, of course, the voltage is the same for all legs in the parallel circuit.

Ohm's law applies separately to each branch.

Look at this circuit: find (a) the total resistance, (b) total current, and (c) the current through each leg.

(a)

(b)

(c) To find the current through each leg, we use Ohm’s law for each branch:

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3 5

4 5 V 8 5

R

R

1

2

Combination Circuits: Sometimes we have a circuit that has components in series with one another and components that are in parallel. We call these combination circuits. To solve problems, we merely simplify things by finding the equivalent circuit. Basically you take all the resistances and, by adding the series ones and solving the parallel ones, you end up with one equivalent resistor. You basically are finding the total resistance for the entire circuit. This is what you do when you find the total resistance of a parallel circuit, isn’t it?Equivalent Circuits: The idea here is to take a complicated circuit that has series and parallel components and work through the thing finding and adding various resistances until you end up with a single equivalent resistance that has the same value of resistance as the entire circuit.

Here’s a simple example - this circuit has two resistors in parallel.

Using the parallel resistance equation, we can calculate the total resistance, which is the equivalent resistor.

Here are some other examples:

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What is (a) the total resistance and (b) the total current?

(a) The first resistor is in series with the other two resistors that are in parallel. So we have to add the first resistor’s resistance to the resistance for the parallel part of the circuit.

(b) Use Ohm’s law to find the current since we know the voltage and the total resistance.

Here’s another lovely circuit. Find the total voltage.

We know the current that goes through R3 – we’ll call it I3. We also know the values of all the resistors. We can calculate the total resistance. But how do we find the current?

We can find the voltage drop across R3 using Ohm’s law.

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5 5

2 4 V8 5

R R1 2

R33 5

3 5

11

R

R

1

2R315

R4

17

We also know that the same voltage drop will exist on the other leg of the parallel segment of the circuit. We know that the equivalent resistor must drop this same voltage and we know the value for this resistor, good old R123. So using Ohm’s law we can find the current through the parallel segment. Since it is in series with R4, this current must be the total current.

Now we can use Ohm’s law to find the voltage supplied by the battery. We have found the total resistance and the total current.

You have three 75 W bulbs that you will use in a household circuit – so figure 117 V. What type of circuit (series or parallel) would give you the greatest brightness (assume that the bulbs that operate with the greatest power are the brightest).

Let’s look at a series circuit first.

We know that one bulb will operate at 75 W. Using this, we can figure out the resistance of the bulb.

Power is: Ohm’s law is:

We can solve Ohm’s law for I:

We plug this into the equation for power and solve for R, this will give us the bulb’s resistance.

The resistance doesn’t change – it’s pretty much constant, no matter how much potential difference we apply. Now we can find the power developed by each bulb in the series circuit.

We’ll find the total resistance and then the total current. This is the same current for each bulb, so we can use it to find the power consumed by a bulb.

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1 2 0 V

The voltage drop for a bulb is:

The power is:

So in a series circuit with three other bulbs, the 75 W bulb is really an 8.8 W bulb. So it’s not so bright.

Now let’s look at the same three bulbs in a parallel circuit.The voltage drop for each bulb, since they are in series, is 117 V. Therefore each bulb produces 75 W and is as bright as it was designed to be.

So the bulbs are brightest in the parallel circuit.

House are wired so that all the lights and the things you plug in will be in parallel. This way each item has a voltage drop of 120 V.

What happens to the current when you add a resistor R to (a) a series circuit and (b) a parallel circuit? The series circuit has two identical resistors R. The parallel circuit also has two of the same resistors R in parallel.

Series Circuit: In a series circuit, the total resistance is the sum of the resistances. When you add a third resistor you end up changing the resistance from 2R to 3R. The current is equal to:

When the resistance is increased, the current gets smaller. It goes from:

Parallel Circuit: Let’s look at the total resistance for a parallel circuit.

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1 2 0 V

V

V

The current is:

When we add a third resistor:

So the current becomes: The current gets bigger with each extra resistor.

Measuring Current, Voltage, and Resistance: Measuring these values is done with an appropriate meter. Specifically, the voltmeter, ammeter (current meter), and Ohmmeter (resistance meter).These are simple to use. Usually they are combined together into a thing called a multimeter.

Measuring Voltage: Let’s start with the voltmeter. The voltmeter has a huge internal resistance. To use the thing, you place it in parallel with the thing you want to read the voltage of. (Ugh, what a wretched sentence.)

Let’s look at a simple circuit with a single load (in this case, a resistor).

The voltmeter is placed in parallel with the load – our single resistor.

The voltmeter, because it’s in parallel with the resistor has the same potential difference across it as does the resistor, correct? It has a huge resistance – millions of Ohms worth, so the current that goes through it is very very small and the equivalent resistance of the parallel circuit we’ve formed (which is what we are measuring) is essentially the resistance of the single resistor. Don’t believe the old Physics Kahuna (good,

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RV

RV

V

RV

1

R m

you should doubt everything!)? Let’s attach some representative numbers to the thing and see what happens.

Let R1 be 35.0 and Rm (the resistance of the meter) be 5.50 M. The voltage is 12.0 V. Let us calculate the equivalent resistance of the parallel circuit and the current through each of the legs (this would be I1 and Im).

Okay, let’s figure out the equivalent resistance:

So the equivalent resistance is simply the resistance of the lone resistor!

We know the voltage drop is the same for each resistor, so we can find the current.

The current before the meter was placed into the circuit was:

You can see that the current is still the same after adding the meter (since the equivalent resistant hasn’t essentially changed).

The current through the 35.0 resistor is 0.343 A. The current through the meter is:

This value is so small that we can ignore it – it is insignificant.

So when we add a voltmeter, we don’t change anything in the circuit. The voltage is the same, the current is the same, and the resistance is the same. The reading for the voltage drop is a correct one.

Measuring Current: To measure current we put the ammeter in series with the circuit. We know that the current in a series circuit is the same at all points in the circuit. So we can accurately

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measure the current. The resistance of the ammeter is very very small so that the voltage drop it causes is so small that it too is insignificant.

Measuring Resistance: To measure the resistance of a component or of a circuit you have to remove the items of interest from the circuit. Then you hook in your Ohmmeter. You place it in parallel with the components of interest. The Ohmmeter makes its own little circuit. It provides a small current at a given volume, measures the current, and gives you the resistance for the circuit.

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Dear Doctor Science,

Which will do less damage, plugging a 110 volt appliance into a 220 volt circuit, or plugging a 220 volt appliance into a 110 volt circuit?-- Bryan Osborn from London, England

Dr. Science responds:Either is capable of ending life as we know it. That's why they make the plugs different, so they won't fit. But there's always somebody who will go to great length to overcome the safeguards built into the system. Tell me, did you stuff matchheads into electrical conduit as a youngster, and then crimp both ends? I bet you have a secret gripe against the Underwriters Laboratory, and are trying to prove them wrong so you can cop some cheap self-esteem at somebody else's expense. Well sir, you're the one who's going to be held responsible when your hometown is just a memory.

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AP Physics - Internal Resist/CapacitanceA nine volt battery produces, well, 9 volts would be good, right? Actually it produces a bit more voltage than advertised and uses some of it up before it even gets a chance to leave the battery. This is because batteries have internal resistance.

The voltage drop across the battery is the potential difference that the battery produces. This is called the terminal voltage. The voltage the battery actually produces is called the emf (electromotive force) of the battery.

Voltage drop of battery terminal voltage

Voltage produced emf

The symbol for emf is .

The terminal voltage of the battery is given by the following equation:

r internal battery resistance

source of emf

The terminal voltage is often called the open-circuit voltage

= V when the current is zero

But the terminal voltage is equal to IR.

If the external load, R, is much greater than the internal resistance, r, then we can safely ignore r.

We can think of battery as a actually having inside of it a voltage source and an internal resistor r:

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r

A battery has an emf of 12.0 V delivers 150.0 mA when connected to a 75.0 load. What is the internal resistance of the battery?

Capacitor Combinations: We don’t have to do much with capacitors, but we do have to know how capacitance adds up in parallel or in series. Basically it’s just the opposite of the way resistors work. So:

Capacitors in parallel:

The capacitance of the capacitors add up like resistances that are in series.

Capacitors in Series:

They add up like resistors in parallel.

The charge on each capacitor will be the same and can be found by using the total capacitance.

Four capacitors in series as shown: Find equivalent capacitance & charge on the 12 F capacitor.

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C1 C2 V

C1 C2V C3 C4

6 .0 F3 .0 F

24 F12 F

18

C1 C2

V

Now we can find the charge:

The voltage provided by the battery is 18.0 V in this circuit. Find equivalent capacitance & stored charge:

The circuit really looks like this:

This gives us an equivalent circuit for the parallel part of two capacitors in parallel. So they add up to be:

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V1 8

1 .0 F

1 .5 F

2 .0 F 1 .8 F

2 .2 F

C1

C2

C3

C4C5

V1 8

1 .0 F1 .5 F

2 .0 F

1 .8 F

2 .2 F

C1C2

C3 C4C5

This gives us an equivalent circuit that has three capacitors in series; C1, C234 and C5.

We can add these up to get the equivalent capacitance of the circuit.

Find Charge:

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AP Physics - Electric Circuits, DCKirchhoff’s Rules:

1. Sum of current entering a junction = sum of current leaving the junction.

2. Sum of the potential difference across all the elements around any closed-circuit loop must be zero.

Page 589 Problem Solving Strategy using Kirchhoff’s Rules:

Find: the current through each resistor.

1. Assign currents and directions:

2. Apply the junction rule:

3. Apply the Loop Rule:

a to b - across battery:V

b to c through R1: - I1R1

c to d, through R2: - I2R2

They add up to zero:

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I out

I outI in

12 5

5 5 .0

7 5 .0 12 .0 V

I1 I2I3

a

bc d

Moving clockwise through small loop:

c to d through R3: - I3R3

d to c through R2: I2R2

They must add up to Zero:

Now have 3 equations and 3 unknowns:

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Do Another way:

Find current through each resistor in this circuit:Find total Resistance:

Find total current, I:

Find V in parallel loop:

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8 9 .0

12 5

17 5 2 4 .0 V

Find I through R1 and R2:

Look at this circuit:

Find: the total resistance and the voltage supplied by the battery.

1. First step: find the equivalent resistance for the one leg of the parallel circuitLet R3 = 15.0 , R4 = 10.0 , and R5 = 35.0

Thus, the equivalent resistance, Re, for the one leg is:

Re = R3 + R4 = 15.0 + 10.0 = 25.0

parallel resistance is:

Rp = 14.6

This resistance is in series with all the other resistors, so we just add them up:

R = R1 + R2 + Rp +R6

R = 15.0 + 10.0 + 14.6 + 25.0

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R = 64.6

2. Find the voltage supplied by the battery:

V = IR V = 0.1858 A (64.6 )

V = 12.0 V

Capacitors:

Symbol for capacitor

Parallel Combination:

Total charge is:

V is same for all capacitors in parallel

Find the equivalent capacitance for this circuit, the charge on the 1.85 F capacitor, and the total charge in all the capacitors.

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q1

q2

c1

c2

V

V = 1 2 .0 V

2 .3 5 F

1 .8 5 F

3 .1 5 F

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Series Combination:

V across capacitors may be different

Q for each is the same.

Find the equivalent capacitance for this circuit, the total charge in all the capacitors, and the charge on the 2.85 F capacitor.

.

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V = 12 .0 V

3 .3 5 F

2 .8 5 F

5 .2 5 F

AP Physics – Electric Current Wrapup“What equations will I have to play with that deal with electricity and magnetism?” Asks the concerned AP Physics student.

“Why, you will have these.” Responds the kindly Physics Kahuna.

This is the definition of capacitance as a function of charge and potential difference.

This is capacitance as a function of plate area and plate separation.

The potential energy stored in a capacitor.

The equation for current showing that it is the rate at which charge is flowing.

The equation to determine the resistance of a long conductor. Short lengths of conductors are considered to have zero resistance, but long power lines and things like that have a significant amount of resistance. This equation will figure it all out.

This is Ohm’s law. A very important equation in the world of electricity.

Power.

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Capacitance of capacitors add algebraically when they are in parallel – you just add them up.

1 over C

Capacitors in series add up like resistors in parallel.

Resistors in series add up algebraically. Sum them up.

This is the equation for finding the equivalent resistance for a set of parallel resistors.

Next the Physics Kahuna will run the standards – stuff that a student has to be able to do. There’s a bunch of things here.

Capacitors and Dielectrics:

1. You should know the definition of capacitance so you can relate stored charge and voltage for a capacitor.

Use the capacitor equation. Pie.

2. You should understand energy storage in capacitors so you can:

a. Relate voltage, charge, and stored energy for a capacitor.

You use the capacitor equation and the potential energy of a capacitor equation. Pretty simple stuff. Definitely pie.

b. Recognize situations in which energy stored in a capacitor is converted to other forms.

The energy in the capacitor is released when it is hooked up so that there is a path for the electrons on the negative plate to flow to the positive plate. Thus do current flow, at least until the capacitor is discharged. Then nothing happens.

3. You should understand the physics of the parallel-plate capacitor so you can:

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a. Describe the electric field inside the capacitor, and relate the strength of this field to the potential difference between the plates and the plate separation.

The electric field inside the capacitor looks like this:

The lines of force between the charged plates are parallel to each other and equally spaced, indicating a uniform electric field.

The field strength depends on the voltage and the separation between the charges:

So as the voltage increases, the field increases. As the distance between the plates increases, the field decreases.

b. Determine how changes in dimension will affect the value of the capacitance.

To do this, you use the equation: This equation tells us that as the area A of

the plates increases, the capacitance C also increases. Capacitance and plate area are directly proportional. Also it tells us that as the distance between the plates increases the capacitance decreases. If the separation decreases the capacitance increases. Capacitance and plate separation are inversely proportional.

Current, Resistance, Power:

1. You should understand the definition of electric current so you can relate the magnitude and direction of the current in a wire or ionized medium to the rate of flow of positive and negative charge.

Just use the equation for current, you know, charge divided by time.

2. You should understand the conductivity, resistivity, and resistance so you can:

a. Relate current and voltage for a resistor.

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This is just your basic old Ohm’s law.

b. Describe how the resistance of a resistor depends upon its length and cross-sectional area.

Use the resistivity equation.

c. Apply the relationships for the rate of heat production in a resistor.

This sounds weird – did we study any stuff about this? Well, we did. Remember that the resistor basically turns electricity into heat. So figure the amount of work done by the resistor and that is equal to the heat produced by the resistor. Power is equal to work

divided by time so the heat developed is simply:

Steady-State Direct Current Circuits with Batteries and Resistors Only:

1. You should understand the behavior series and parallel combinations of resistors so you can:

a. Identify on a circuit diagram resistors that are in series or in parallel.

Please don’t try to tell the Physics Kahuna that you can’t do this!

b. Determine the ratio of the voltages across resistors connected in series or the ratio of the currents through resistors connected in parallel.

This is using Ohm’s law for different sorts of circuits. Recall how much phun we had doing this sort of problem.

c. Calculate the equivalent resistance of two or more resistors connected in series or parallel, or of a network of resistors that can be broken down into series and parallel components.

This is pretty simple, just use the equations for equivalent resistance.

d. Calculate the voltage, current, and power dissipation for any resistor in such a network of resistors connected to a single battery.

You use the power equation, Ohm’s law, the rules for parallel and series circuits, etc. We did us several of these.

e. Design a simple series-parallel circuit that produces a given current and terminal voltage for one specified component, and draw a diagram for the circuit using conventional symbols.

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Use Ohm’s law, the rules for parallel and series circuits, etc. to figure out the needed circuit.

2. You should understand the properties of ideal and real batteries so you can:

a. Calculate the terminal voltage of a battery of specified emf and internal resistance form which a known current is flowing.

The Physics Kahuna was proud to show you how to do this. Treat the battery like a small circuit. The emf is the voltage and the internal resistance is the resistor. You can find the voltage drop of the resistor. Subtract this voltage drop from the emf and you get the voltage the battery actually produces. Simple.

3. You should be able to apply Ohm’s law and Kirchoff’s rules to direct-current circuits in order to: a. Determine a single unknown current, voltage or resistance

Okay, what the heck is Kirchoff’s rules? Well, we didn’t study them. Mainly because the rules are a very confusing way to solve circuit problems. The Physics Kahuna showed you a better way. What you do is analyze the circuit using the rules for parallel and series circuits and apply Ohm’s law to the thing to find the unknown value.

4. You should understand the properties of voltmeters and ammeters so you can:

a. State whether the resistance of each is high or low.

Ammeter – low resistance. Voltmeter – high resistance.

b. Identify or show correct methods of connecting meters into circuits in order to measure voltage or current.

Ammeters are placed in series with the circuit to measure current. Voltmeters are placed in parallel to measure the voltage of a circuit or component of a circuit.

The voltmeter has very high resistance. When placed in parallel with a component, the voltage drop is the same across both the meter and the component, but because the resistance is very large in the meter, the current through it is extremely small.

The ammeter has a very low resistance. It is placed in series with the component. It measures the current, but because its resistance is so small, it doesn’t add anything significant to the total resistance so that the current it measures is essentially the current in the circuit without the meter.

Capacitors in Circuits:

1. You should understand the behavior of capacitors connected in series or in parallel so you can:

a. Calculate the equivalent capacitance of a series or parallel combination.

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Use the equations. We did several of these.

b. Describe how stored charge is divided between two capacitors connected in parallel.

Let’s look at a parallel capacitor circuit:The top plate of each capacitor has the same potential difference as the other. The equivalent capacitance of the circuit is simply the sum of the capacitance of the two capacitors. This is also true for the charge. The total charge stored in such a parallel circuit is equal to the sum of the two charges:

2. You should be able to calculate the voltage or stored charge, under steady-state conditions, for a capacitor connected to a circuit consisting of a battery and resistors.

a. You should develop skill in analyzing the behavior of circuits containing several capacitors and resistors so you can:

(1) Determine voltages and currents immediately after a switch has been closed and also after steady-state conditions have been established.

Immediately after the switch is thrown, current flows as if the capacitors did not exist. As the capacitors gain charge, the current decreases until, when the capacitors are fully charged, no current flows at all. So when the switch is initially thrown, solve for voltage, current, or resistance as if the capacitors did not exist.

Once steady state conditions are established, no current flows in the circuit and the capacitor(s) is/are fully charged.

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C1 V C2

Sample AP Physics Test Problems Involving DC Electricity:

The year 2000:

Three identical resistors, each with resistance R, and a capacitor of 1.0 x 10-9

F are connected to a 30 V battery with negligible internal resistance, as shown in the circuit diagram above. Switches S1

and S2 are initially closed, and switch S3 is initially open. A voltmeter is connected as shown.

(a) Determine the reading on the voltmeter.

V through R1:

(b) Switches S1 and S2 are now opened, and then switch S3 is closed. Determine the charge Q on the capacitor after S3 has been closed for a very long time.

After the capacitor is fully charged, switches S1 and S2 remain open, switch S3 remains closed, the plates are held fixed, and a conducting copper block is inserted midway between the plates, as shown below. The plates of the capacitor are separated by a distance of 1.0 mm, and the copper block has a thickness of 0.5 mm.

(c) What is the potential difference between the plates? 30 V

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(d) What is the electric field inside the copper block? Zero. The copper plate is a conductor, there is not potential difference across it as every point within the copper plate is at the same potential.

(e) On the diagram above, draw arrows to clearly indicate the direction of the electric field between the plates. The copper plate has no effect. The lines of force point towards the negative plate, the direction of a force that would be exerted on a positive charge.

(f) Determine the magnitude of the electric field in each of the spaces between the plates and the copper block.

The next question is from the 1998 test:

In the circuit shown below, A, B, C, and D are identical light bulbs. Assume that the battery maintains a constant potential difference between its terminals (i.e., the internal resistance of the battery is assumed to be negligible) and the resistance of each light bulb remains constant.

A. Draw a diagram of the circuit in the box below, using the following symbols to represent the components in your diagram. Label the resistors A, B, C, and D to refer to the corresponding light bulbs.

B. List the bulbs in order of their brightness, from brightest to least bright. If any two or more bulbs have the same brightness, state which ones. Justify your answer.

A, D, and then B & C equally.

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+

A

B

CD

All the current from the battery goes thru A, so it has the most current, greatest power, and is brightest.

The current splits between D and B & C which are parallel – they receive less current and aren’t as bright.

D is by itself and receives the next most current.

B & C in series have higher resistance than D, so they get the least. They each receive the same amount of current since they are in series so they will have the same level of brightness.

Bulb D is then removed from its socket.

C. Describe the change in the brightness, if any, of bulb A when bulb D is removed form its socket. Justify your answer.

Bulb A becomes dimmer.

We now have a series circuit. The resistance is greater (the three bulb’s resistances add

up algebraically) so the current decreases. Light A receives less current and is less bright.

All bulbs have the same brightness – the current is same through all bulbs, so the power

and therefore the brightness is the same.

D. Describe the change in the brightness, if any, of bulb B when bulb D is removed from its socket. Justify you answer.

Bulb B becomes brighter.

It was receiving the least amount of current since D was an alternate pathway for the current. With D gone all the current must go through B, which is in series now. More current means B will be brighter.

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From 1996:

A student is provided with a 12.0-V battery of negligible internal resistance and four resistors with the following resistances: 100 , 30 , 20 , and 10 . The student also has plenty of wire of negligible resistance available to make connections as desired.

a. Using all of these components, draw a circuit diagram in which each resistor has nonzero current flowing through it, but in which the current from the battery is as small as possible.

Draw a circuit where all the resistors are in series. This maximizes the resistance, which will make sure that the current is the smallest amount possible.

b. Using all of these components, draw a circuit diagram in which each resistor has nonzero current flowing through it, but in which the current from the battery is as large as possible (without short circuiting the battery).

Draw a circuit where all the resistors are in parallel. This will minimize the resistance which will max out the current.

The battery and resistors are now connected in the circuit shown.

c. Determine the following for this circuit.i. The current in the 10 resistor.

This is the current in the 10 resistor since it is in series with everything else.

ii. The total power consumption of the circuit.

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1 2 V 1 2 V

1

2 3

4

iii. The current in the 10- resistor

d. Assuming that the current remains constant, how long will it take to provide a total of 10 kJ of electrical energy to the circuit?

From 1995:

A certain light bulb is designed to dissipate 6.0 watts when it is connected to a 12 volt source.a. Calculate the resistance of the light bulb.

b. If the light bulb functions as designed and is lit continuously for 30. days, how much energy is used? Be sure to indicate the units in your answer.

The 6.0 watt, 12 volt bulb is connected in a circuit with a 1,500 watt, 120 volt toaster; an adjustable resistor; and a 120 volt power supply. The circuit is designed such that the bulb and the toaster operate at the given values and, if the light bulb fails, the toaster will still function at these values.

c. On the diagram below, draw in wires connecting the components shown to make a complete circuit that will function as described above.

Okay let’s draw in some wires. The basic idea is that the bulb must have 12 V to produce the 6.0 watts. The toaster requires 120 volts to operate properly. So the bulb must be in series with the adjustable resistor – the resistor needs to have a voltage drop of 120 V – 12 V. Then the bulb will

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have a voltage drop of 12 V. The toaster needs to be in parallel with the bulb/resistor deal. Let’s draw the circuit (add the lines).

d. Determine the value of the adjustable resistor that must be used in order for the circuit to work as designed.

The bulb has a resistance of 24 (which we figured out in the first part of this question). We can calculate the current that flows through the bulb. This will be same current as the resistor gets.

Now we know the current and the voltage through the parallel leg of the bulb/resistor leg. The total voltage is the sum of the two voltages dropped by each component, the bulb and the resistor. The resistor drops 12 V, the total voltage is 120 V, so we can find the voltage drop for the resistor.

Now we can use Ohm’s law to find the resistance for the adjustable resistor.

e. If the resistance of the adjustable resistor is increased, what will happen to the following?

i. The brightness of the bulb. Briefly explain your reasoning.

If resistance goes up then current drops in the resistor. Current stays the same in

series, so it also drops in the light. If the light receives less current, then it will be less

bright. Recall that since the current flowing into the bulb is less, the power

is less, and the bulb is not as bright.

ii. The power dissipated by the toaster. Briefly explain your reasoning.

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Voltage is the same in parallel. The toaster still receives 120 V. It will draw the

current it needs from the power supply. The resistance change in the other leg of the

circuit has not effect on the toaster.

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