Electricity

Electricity

Electric Charges: the basis of electricity is charge.

• The charge on an atom is determined by the subatomic particles that make it up.

Proton- has a positive charge and is located in the nucleus.

Neutron- has no charge (is neutral) and is also located in the nucleus as it fills in the spaces between the protons.

Electron- has a negative charge and is located outside of the nucleus in an electron cloud around the atom.

Particle charges:• Electrons and protons have the

same magnitude of charge (elementary charge, e).

• Electron (-e): -1.60 x 10-19 C

• Proton (+e): +1.60 x 10-19 C

• This is why electrons are forced to orbit around the nucleus.

• Electrostatic Forces hold atoms together.

• The Law of Charges- Like charges repel, and unlike charges attract.

Because atoms have the same amount of protons and electrons

they are electrically neutral. (Nitrogen has an atomic number

of 7- 7 protons orbited by 7 electrons).

Subatomic Particle Sizes

• A proton and neutron have about the same mass- 1.67 x 10-27 kg

• An electron has a much smaller mass- 9.11 x 10-31 kg

To put this into perspective it’s like comparing the sizes of a penny and large bowling ball. The proton is obviously the bowling ball and the electron is represented by the penny.

How do atoms become “charged?”

• Atoms become charged when they become more positive or more negative.

• How can this happen?

Remove or add a proton or an electron.

Protons and neutrons are bound together by the Strong Nuclear Force and it is very hard to separate them.

Electrons, however, can be more easily removed.

Ions

An atom with a deficiency of electrons is positively charged.

An atom with an excess of electrons is negatively charged.

ATOMS DO NOT GAIN OR LOSE PROTONS!!!!

Charge is a fundamental quality like mass.

• Charge is denoted as q.• Charge has a fundamental unit of a Coulomb (C).• Charges are usually really really small numbers

(10-).• So what is 1 C?

– An object would have to have 6.25 x 1018 extra electrons to amount to –1 C of charge.

– A lightning bolt is estimated to carry a charge of 10 C.

• Revisit the charges on an electron and proton.

Charges can ONLY be in multiples of e

• Remember:– -e = an electron = -1.60 x 10-19 C– +e = a proton = +1.60 x 10-19 C

• An object that has a net charge of 8.0 x 10-19 C has a net charge of what multiple of e? Hint: How many electrons would need to be removed to create this charge?

The net charge would be +5e, 5 electrons were removed

Multiples of Charges Chart

1e 1.6 x 10-19

2e 3.2 x 10-19

3e 4.8 x 10-19

4e 6.4 x 10-19

5e 8.0 x 10-19

Electrostatic Force• This is a non-contact force (like the gravitational force except

instead of two masses exerting force on each other the two objects charges exert a force of repulsion or attraction).

• ANY charged object can exert the electrostatic force upon other objects- both charged and uncharged objects.

Coulomb’s Law- formula for electrostatic force

Again this is similar to the gravitational force…

Fg = GmM r2

charge (q) is now responsible for the force

Fe = kq1q2

r2

Just like G was a constant so is k. k is the electrostatic constant

and = 8.99 x 109 N•m2/C2

2x 3x 4x

1/4 = 2.24

1/9 =1.09

1/16 =.61

1x

Remember this…the relationship between the gravitational force and the distance from the object…this is the inverse square law

9.81 m/s2

Fg

r2

Fg = GmM r2

Fe = kq1q2

r2

Fe

r2

Try this…The distance between a proton and an electron

in a hydrogen atom is 5.3 x 10-11 m. Find both the gravitational force and the electrostatic force between the two particles.

ANY charged object, whether positively charged or negatively charged, will have

an ATTRACTIVE interaction with a neutral object.

A balloon when rubbed on your head becomes charged by picking up extra electrons from your hair.

That same balloon, because it is charged, will attract a neutral object like pieces of paper.

-+

-

So we are able to predict the charge on objects based on their interaction with other objects.

They can either both be positive or both be negative.

They can have opposite charges or one object

is charged and the other is neutral.

Try this…• On two occasions, the following charge interactions between

balloons A, B and C are observed. In each case, it is known that balloon B is charged negatively. Based on these observations, what can you conclude about the charge on balloon A and C for each situation.

positive or neutral

negative

positive (if it was neutral it wouldn’t repel C)

positive

Why does the balloon stick to the wall?

• When a balloon is rubbed with a piece of cloth electrons are transferred between the two objects. Usually the balloon attracts extra electrons and then receives an overall negative charge.

balloons static

http://www.colorado.edu/physics/phet/simulations/balloon/webstart.jnlp

When the balloon is placed against the wall the excess electrons will repel the electrons in the wall and be attracted

to the positive charges.

What happens to your hair when you rub a balloon on your head?

• The balloon, after being rubbed and then pulled away, removes some of the electrons in your hair which give each strand a positive charge. Like charges want to repel and each strand is repelling from the others and “sticking up.”

http://sciencecheerleader.com/wp-content/teddycrazyhair.jpg

Getting Shocked

• As you walk across a carpet, electrons are transferred from the rug to you.

• Now you have extra electrons. • Touch a door knob (conductor) and ZAP! • The electrons move from you to the knob.

Lightning

• Lightning is a REALLY big shock.

• Positive charges tend to go up, negative charges tend to go down.

• When the attraction reaches a critical level you get a lightning bolt.

Objects that tend to give up electrons and become positive:

• Glass

• Nylon

• Fur

• Hair

• Wool

Objects that tend to attract electrons and become negative:

• Rubber

• Polyester

• Styrofoam

• Saran Wrap

• PVC

Insulators and Conductors

• Different materials hold electrons differently.

• Insulators don’t let electrons move around within the material freely.– Ex. Cloth, Plastic, Glass, Dry Air, Wood, Rubber

• Conductors do let electrons move around within the material freely.– Ex. Metals- Silver, Copper, Aluminum

Try this…

• A charged plastic rod is brought close a neutral metal sphere. How would the distribution of charges be in the metal sphere?

Try this…

• Which of the diagrams below best represents the charge distribution on a metal sphere when a positively charged plastic tube is placed nearby?

Law of Conservation of Charge

• Charges within a closed system may be transferred from one object to another,

but charge is neither created nor destroyed.

The diagram below shows the initial charges and positions of three metal spheres, R, S, and T, on insulating stands.

0e -8e +6e

R S T

Sphere R is brought into contact with sphere S and then removed. Then sphere S is brought into contact with sphere T and removed. What is the charge on sphere T after this procedure is completed?

Note that the net charge of the system is -2e.

0e -8e

R S

When the spheres come in contact the charge will be distributed evenly between both spheres.

-4e -4e +6e

T

0e -8e

S T

-4e +6e-4e

R

-4e + 6e = 2

+2e = 2

+e

+e +e

Note that the charge of the system is conserved- the initial charge is the same as the

final charge.

-4e +e +e

R S T

-4e + e + e = -2e

Electroscopes- instruments used to detect charge

The yellow arms or leaves on both instruments will move to show the charge.

Charged Electroscope

Excess of the same charge in both leaves causes them to diverge/repel,

+-

++-+

+-+

+-+

Uncharged Electroscope

Leaves are neutral so they are not diverging/repelling or converging/attracting.

+-

+- +-

+-+- +-

Steps in charging an electroscope by INDUCTION:

1. Uncharged electroscope:

+-+-+-

+-+-+-

+-+-+-

Leaves are just hanging straight down.

Net charge is zero.

2. A negatively charged rod is brought near the electroscope:

++++

+-+-+-

+-+-+-

-- -- --- -- ----- ---- -- --- -- ----- - -

---

Net charge is zero

-- -- --- -- ----- ---- -- --- -- ----- -

3. Electrons move to the leaves:

++++

+- +- +-

-+-+-+

Net charge is still zero

----

-- -- --- -- ----- ---- -- --- -- ----- -

4. Leaves diverge:

++++

+--+--+--+-+-

-+-

Net charge of zero

+--+--+-

-+-+--+-

-- -- --- -- ----- ---- -- --- -- ----- -

5. The electroscope is grounded:

++++

+--+--+--+-+-

-+-

Grounding is the process of removing the excess charge on an object by means of the transfer of electrons between it and another object of substantial size.

-- -- --- -- ----- ---- -- --- -- ----- -

6. Electrons go to the ground:

++++

+-+-+--+-+-+

-- -- Net charge is now positive

-- -- --- -- ----- ---- -- --- -- ----- -

7. Leaves converge and ground is removed:

++++

+-+-+-

-+-+-+

+-+-+--+-+-+

-- -- --- -- ----- ---- -- --- -- ----- - +

+++

+-+-+-

-+-+-+

8. Negatively charged rod is removed:

++++

++-+-

+-+-+

9. Electrons redistribute throughout the electroscope and move up toward the top:

--

++++

++-+-

+-+-+

--

10. Leaves now have the same charge and diverge:

++-+- +-+-+

Electric Fields

• Electric fields are similar to gravitational fields.

• The only difference is that two objects with mass will always attract each other.

• Charges can either repel or attract when held some distance apart.

Here are the gravitational field lines around the Earth. Notice how they

point toward the center and are perpendicular at the surface of the

Earth.

Charges produce similar fields…

• A charged object creates an electric field - an alteration of the space in the region which surrounds it.

• Other charges in that field would feel the unusual alteration of the space.

-

+

Electric field between two opposite charges…

Electric field between two like charges…

5 Rules for Electric Fields:1. Electric field lines always go from positive to negative.

This is the path that a positive “test charge”

would follow.

5 Rules for Electric Fields:2. Electric field lines always enter and leave the charge perpendicularly.

5 Rules for Electric Fields:3. Electric field lines never cross.

This is just like contour lines.

5 Rules for Electric Fields:4. Where the electric field lines are closer the electric field is stronger.

Again remember when the contour lines were closer together the slope was

steeper.The field is also stronger the closer you are to a charge.

5 Rules for Electric Fields:5. Make sure the electric field lines are in contact with charged objects.

+++++++

-------

Charged Parallel Plates

What would the electric field look like around these charges?

- -

What would the electric field look like around these charges?

+ +

-

Analyzing the path of a test charge:

A test charge is a positively charged object that is used to test the electric field around other charged objects.

-+

+

+

+

Electric Field Strength (E)

• This is also similar to the gravitational field strength.

• The gravitational field strength of the Earth is 10 m/s2.

g = w m

= Fg

m= GmM r2

m

g = GM r2

E = FE

q= kqQ r2

q

E = kQ r2

What is the magnitude of the electric field strength at a point in a field

where a positive test charge of 8.00 x 10-2 C experiences a 2 N force?

+

E = FE

qE = 2 N 8.0 x 10-2 C

E = 2 N 8.0 x 10-2 C

E = 25 N/C

Electric Potential Difference- VOLTAGE

Let’s first revisit gravitational potential energy…

PE = mgh

What is gravitational potential energy dependent on?

2h

h

m

2mWork (energy) is required to lift these rocks against the force of gravity.

Work is also required to move charges in an electric field. If the direction of an electric field is such that it

opposes the motion of a charged particle, work must be done to move the particle in that direction.

+

+++

+

+

+

++

W = Fd+

Doing work on a system transfers energy. This influences the electric potential

(think potential energy) of the charge. The electric potential of this positive test charge changes as it is moved closer to the positive metal sphere.

Work must be done to keep 2 like charges

together or two opposite charges apart.

For each situation below determine if work is done on the test charge to move it from point A

to point B.+ + + + + + +

- - - - - - -

+ + + + + + +

- - - - - - -

+

+

B

B

A

A

NO work is done. The + charge is moving with nature; work is not required when it moves

with the electric field.

YES work is done. The + charge is moving against nature; work is required when it moves against the electric field.

ElectricField Lines

Now let’s look at the electric potential of the charges. Are the charges losing or gaining potential energy when

they are moved from A to B?+ + + + + + +

- - - - - - -

+ + + + + + +

- - - - - - -

+

+

B

B

A

A

The positive test charge will naturally move in the direction of the

electric field. No work is done in moving it. The potential energy will decrease (just like an object

falling with gravity).

The positive test charge does not want to naturally move against the

electric field (same charges repel) so work needs to be done to move it to point B.

The potential energy will increase (just like raising an object up against gravity).

ElectricField Lines

We can now conclude that the high energy location for a positive test charge is a location

nearest the positive source charge; and the low energy location is furthest away.

++

+

Higher PE

Lower PEElectricFieldLines

Electric Potential Difference (V)

The electric potential difference between two points in an electric field is the work done per unit charge.

V = W q

1 J/C = 1 volt

If the electric potential difference between two locations is 1 volt, then 1 C of charge will gain 1 J of potential energy when moved between those two locations.

++

+ElectricFieldLines

If 8 Joules of work are required to move 2 Coulombs of charge through a 3-ohm

resistor, what is the potential difference across the resistor?

V = W q

V = 8J 2C

V = 4V

Units- Joules vs. eVHow much work is done to move an elementary charge (+/- e) against an electric field through a potential difference of 1 volt?

V = W q

qV = Wq q

W = Vq

W = (1V)(1.6 x 10-19 C)

= 1.6 x 10-19 J

1.6 x 10-19 J = 1 eV (electronvolt)

Electronvolts are like inches and Joules are like miles.

How much energy in eV is needed to move one electron through a potential

difference of 1.0 x 102V?

W = energy

W = Vq

W = (1.0 x 102V)(-1.6 x 10-19C)

W = 1.6 x 10-17J

1 eV = 1.6 x 10-19 J

= 100 eV

How many electronvolts are in 320 x 10-19J of energy?

1 eV = 1.6 x 10-19 J

200 eV

eVand

Joulesare bothunits

of ENERGY!

Static Electricity vs. Electric Current

Static electricity is a one time event.

Electric current is a constant flow.

Electric current (I) is the rate at which charge passes a given point in a

electric circuit.

An electric circuit is a closed path along which charged particles move.

Current

I = Δq t

I = amps (A)

An ammeteris a device

used to measurecurrent.

Determine the current for these situations:

1. A cross section of wire is isolated and 20 C of charge

are determined to pass through it in 40 s.

20 C

40 s

I = Δq t

= 20 C 40 s

= .5 A

2. A cross section of wire is isolated and 2 C of charge

are determined to pass through it in .5 s.

2 C

.5 s

I = Δq t

= 2 C .5 s

= 4 A

Requirements necessary for an electric current:

1. There must be a closed conducting path which extends from the positive terminal to the negative terminal of a cell or battery

(combination of cells).

Requirements necessary for an electric current:

2. There must be a difference in electric potential between the two end points in the circuit. The potential difference may be supplied by a cell or battery. In other words you need an energy source.

The potential difference or voltage of a circuit can be measured using a voltmeter.

Batteries provide the voltage (electric pressure) necessary to move current

through a circuit.

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animationsindex.htm

Light Bulbs

• Using only a light bulb, a wire, and a battery come up with 4 ways to create a complete circuit so that the light bulb lights up.

You need to create a closed circuit.

Conventional flow vs. THE TRUTH

Ben Franklin envisioned positive charges as the carriers of charge. Because of this an early convention for the direction of an electric current was established to be in the direction which positive charges would move. The direction of an electric current is by convention the direction in which a positive charge would move. However, we now know that the positive charges (protons) are NOT the charged particles that are moving. IT’S THE ELECTRONS. And the negatively charged electrons would be moving in the opposite direction.

Alternating Current Direct Current

AC - The electrons in alternating current flow in one direction, then in the opposite direction—over and over again. Electricity from a power plant is alternating current.

DC - The electrons in direct current flow in one direction. The current produced by a battery is direct current.

In the United States, the current flow alternates 120 times per second. (In Europe it alternates 100 times per second.) The current supplied to your home by the local utility is alternating current.

War of CurrentsDC vs. ACEdison carried out a campaign to discourage the use of alternating current, including spreading disinformation on fatal AC accidents, publicly killing animals, and lobbying against the use of AC in state legislatures. Edison directed his technicians to preside over several AC-driven killings of animals, primarily stray cats and dogs but also unwanted cattle and horses. Acting on these directives, they were to demonstrate to the press that alternating current was more dangerous than Edison's system of direct current.

EDISON

WESTINGHOUSE

TESLA

http://upload.wikimedia.org/wikipedia/commons/b/ba/Edison.jpg

http://upload.wikimedia.org/wikipedia/commons/5/55/George_Westinghouse.jpg

http://upload.wikimedia.org/wikipedia/commons/d/d4/N.Tesla.JPG

Warmup:On the circuit drawing below label

conventional current flow and electron flow. Why is conventional current flow

technically the “wrong” way?

How many electrons flow through a battery that delivers a current of 3 A for 12 s?

(HINT: 1 C = 6.25 x 1018 e)

I = q t

tI = qt t

q = It

q = It

q = (3A)(12s)

q = 36 C

#e = (36C)(6.25 x 1018e)

#e = 2.25 x 1020

Resistance

(I) Current is the flow of electrons through a circuit.

(R) Resistance discourages or controls the flow. It is caused by “things” that get in the way of a direct path.

(V) Electric potential difference (voltage- like a battery) is necessary for current to flow.

(Electric Pressure).

The resistance of a conductor (like a wire) is the ratio of the potential difference applied to the circuit and the current that

flows through it…

This is Ohm’s Law…

R = V IR = Volts Amps

R = ohm (Ω)

Resistors are “things” in a circuit that limit current flow. They can be controlled resistors or electrical

devices like light bulbs or lamps.

Circuit Analogy• The pipe is the counterpart of the wire in the electric circuit. • The pump is the mechanical counterpart of the battery. • The pressure generated by the pump, that drives the water through the pipe, is

like the voltage generated by the battery to drive the electrons through the circuit.

• The seashells plug up the pipe and constrict the flow of the water creating a pressure difference from one end to the other. In a similar manner the resistance in the electric circuit resists the flow of electricity and creates a voltage drop from one end to the other. Energy is lost across the resistor and shows up as heat.

R = V I

Try this…

A student measures a current of .10 A flowing through a light bulb connected by short wires to a 12 V battery.

What is the resistance of the light bulb?

R = V I

R = 12 V .10 A

R = 120 Ω

Try this…

A lamp with a resistance of 20 Ω is in a circuit that has a current of .05 A flowing through it. What is the potential

difference across the lamp?

R = V I

I R = V I I

V = I R

V = I R

V = (.05 A)(20 Ω)

V = 1 V

Recall slope…What are the slopes for the following graphs?

d

t

v

t

F

x

F

a

v = d t

velocit

y

accelera

tion

a = v t

sprin

g consta

nt

k = F x

m = F a

mass

Guess what the slope is for this

graph:

V

I

Resistance

(R)

R = V I

Wires made of a certain types of metals are used in circuits because they are

good conductors and allow the electrons

within them to move relatively freely.

Although wires allow current to flow through a circuit they also control the current or have some resistance.

Factors that affect the resistance of a wire:

1. LENGTH: Increasing the length (L) of the wire will increase the resistance of the wire.

This is because the current (electrons) will now have further to travel and will encounter and collide with an increasing

number of atoms.


2. AREA: Increasing the cross-sectional area (A) of a wire will decrease the wires resistance.

This occurs because in making the wire thicker there is now more spaces between atoms through

which the electrons can travel and thus flow easier. The wire isn’t resisting the flow as much.

A = πr2


3. RESISTIVITY (ρ) – this is a characteristic of a materialthat depends on its electronic structure and temperature. If a wire is made of a material that has a high resistivity then it will have a high resistance.

Combining all these factors gives us the following equation for the

resistance of a wire:

R = ρL A

Short/thick/cold wires = low resistance (easy for electrons to flow)

Long/thin/hot wires = high resistance (hard for electrons to flow)

http://phet.colorado.edu/sims/resistance-in-a-wire/resistance-in-a-wire_en.html

http://phet.colorado.edu/sims/resistance-in-a-wire/resistance-in-a-wire_en.html

Try this…

Determine the resistance of a 4.00 m length of copper wire having a diameter of 2 mm. Assume the temperature of 20°C.

4 m

d= 2 mm

ρ copper = 1.72 x 10-8 Ω•m

R = ρL A

= (1.72 x 10-8 Ω•m)(4m) π(.001m)2

= .0219 Ω

Determine the length of a copper wire that has a resistance of .172 Ω and cross-sectional area of 1 x 10-4 m2. The resistivity of copper is

1.72 x 10-8 Ω•m.

AR = ρLA A

AR = ρL ρ ρ

L = AR ρ

L = AR ρ

L = (.001m2)(.172Ω) (1.72 x 10-8 Ω•m)

L = 10,000 m

• Which one of the five wires has the largest resistance?

Wire Material Length Diameter

A iron 2 m 6.4 x 10-4 m

B copper 2 m 6.4 x 10-4 m

C copper 2 m 1.2 x 10-3 m

D copper 1 m 1.2 x 10-3 m

E Iron 2 m 1.2 x 10-3 m

ρ iron = 9.7 x 10-8 Ω•m ρ copper = 1.7 x 10-8 Ω•m

High resistance = long/thin/hot wires

Wire A


A iron 2 m 6.4 x 10-4 m




E Iron 2 m 1.2 x 10-3 m


• Of the five wires, which one has the smallest resistance?

Small resistance = short/thick/cold

Wire D


A iron 2 m 6.4 x 10-4 m




E Iron 2 m 1.2 x 10-3 m


Which of the wires carries the smallest current when they are connected to identical batteries?

How does current (I) relate to resistance (R)?

R = V I

When current is low, resistance is high.

High resistance = long/thin/hot wiresWire A

Resistance Variables LabCoil # Metal Length

(meters)

Cross

Sectional

Area

_____

Gauge

V

Volts

(V)

I

Current

(A)

R

Resistance

(Ω)

1 Copper 10 m .325 mm2

22


28


22


22

5 Copper-Nickel

10 m .325 mm2

22

First use R = ρL/A to predict the resistance of each circuit.

ρ cop-nick = 4.9 x 10-8 Ω•m ρ copper = 1.7 x 10-8 Ω•m

Schematic Circuit Diagrams

There are many different ways to represent circuits. Below is an artist’s drawing and a schematic drawing. We will be using schematic drawings with symbols.

Types of Circuits: Series

A series circuit can be constructed by connecting light bulbs in such a manner that there is ONE PATH for charge flow (CURRENT); the bulbs

are added to the SAME LINE with no branching point- charge passes through every light bulb.

Since there is only one current path in a series circuit, the current is the same through each resistor.

I = I1 = I2 = I3


I = I1 = I2 = I3

If the current is the same everywhere within a series circuit then what is changing?

V = I R

http://phet.colorado.edu/sims/ohms-law/ohms-law_en.html



Let’s look at resistance:

There are 3 resistors in this circuit. The total resistance of the entire circuit is called the equivalent resistance (Req). If we could

replace all 3 resistors with just 1 resistor it would have the equivalent resistance of the entire circuit.

Req = R1 + R2 + R3

Types of Circuits: SeriesLet’s look at voltage: The cell or battery of a circuit provides the maximum voltage allowed for the circuit.

It you have a 12 V battery hooked up to your circuit then there is 12 V of “electrical pressure”

pushing the current through the circuit. Current moves from an area of high pressure (12 V at the positive terminal of the battery)

to an area of low pressure (0 V at the negative terminal of the battery). This is why you may often hear the term “voltage drop”

across resistors. The voltage is decreasing as the current passes through resistors in the circuit because everything is together on one path.

12 V

V = V1 + V2 + V3

Series Summary:

I = I1 = I2 = I3

Req = R1 + R2 + R3

V = V1 + V2 + V3

3V

12

6

a b

cd

30 Ω

10 Ω

In this animation you should notice the following things:•The battery or source is represented by an escalator which raises charges to a higher level of energy. •As the charges move through the resistors (represented by the paddle wheels) they do work on the resistor and as a result, they lose electrical energy. •The charges do more work (give up more electrical energy) as they pass through the larger resistor. •By the time each charge makes it back to the battery, it has lost all the energy given to it by the battery. •The total of the potential drops ( - potential difference) across the resistors is the same as the potential rise ( + potential difference) across the battery. This demonstrates that a charge can only do as much work as was done on it by the battery. •The charges are positive so this is a representation of Conventional Current (the apparent flow of positive charges) •The charges are only flowing in one direction so this would be considered direct current ( D.C. ).

Try this…

36 V

4 Ω

8 Ω

6 Ω

Determine the (a) equivalent resistance of the entire circuit, (b) the current through each resistor, and (c) the potential drop across each resistor.

I = I1 = I2 = I3

Req = R1 + R2 + R3

V = V1 + V2 + V3

a. Req = R1 + R2 + R3

Req = 4 Ω + 6 Ω + 8 Ω

Req = 18 Ω

b. I = I1 = I2 = I3

I = V Req

I = 36 V 18 Ω

I = 2 A

36 V

4 Ω

8 Ω

6 Ω

c. I = 2 A

V = I R

V1 = I R1

V1 = (2A)(4 Ω)

V1 = 8 V

V2 = I R2

V2 = (2A)(6 Ω)

V2 = 12 V

V3 = I R3

V3 = (2A)(8 Ω)

V3 = 16 V

V = V1 + V2 + V3

V =8V +12V+ 16V

V = 36 V

Try this…

24 V

4 Ω

6 Ω

A2 A

The diagram shows a circuit with three resistors. What is the resistance of resistor R3?

I = I1 = I2 = I3

Req = R1 + R2 + R3

V = V1 + V2 + V3

Req = V I

Req = 24 V 2 A

Req = 12 Ω

R3 = Req – R1 – R2

Req = R1 + R2 + R3

R3 = 12 Ω – 4 Ω– 6 Ω

R3 = 2 Ω

Light bulbs in Series

Remember how a light bulb works?

What will happen to LB 2 if LB 1 blows out?

LB 1LB 2

The circuit is no longer closed- LB 2 will also go out.

Christmas lights are connected in series. If one goes out they

should all go out. But small Christmas light bulbs are designed so this doesn’t

happen.

Each bulb has a "shunt" of several turns of tiny wire inside

the bulb near the bead. The shunt is intended to conduct

current when the filament fails.

Electric Power (Watts): 3 different ways

P = V I P = V I

From Ohm’s Law V = I R

P = I R I

P = I2 R

P = V I

I = V R

P = V V R

P = V2

R

Power is the rate at which electrical energy is supplied to a circuit or consumed by a load.

Remember:1 W = 1 J/s

Try this…

A potential difference of 60 V is applied across a 15 Ω resistor. What is the power dissipated in the resistor?

P = 240 W

P = (60 V)2

15 Ω

P = 3600 V2

15 Ω

P = V2

R

Try these…Determine the ...a. ... current in a 60-watt bulb plugged into a 120-volt outlet.

b. ... current in a 120-watt bulb plugged into a 120-volt outlet.

c. ... power of a saw that draws 12 amps of current when plugged into a 120-volt outlet.

d. ... power of a toaster that draws 6 amps of current when plugged into a 120-volt outlet.

e. ... current in a 1000-watt microwave when plugged into a 120-volt outlet

I = P / V = (60 W) / (120 V) = 0.5 A

I = P / V = (120 W) / (120 V) = 1.0 A

P = V • I = (120 V) • (12 A) = 1440 W

P = V • I = (120 V) • (6 A) = 720 W

I = P / V = (1000 W) / (120 V) = 8 A

Electrical Energy (J)

W = Pt

P = I2R

W = I2Rt

W = V2t R

or P = V2

R

Remember:1 J = 1 N•m = 1 kg •m2/s2

Try this…

A current of.4 A is measured in a 150 Ω resistor. How much energy is expended by the resistor in 30 seconds?

W = I2Rt

W = (.4)2(150 Ω)(30s)

W = (.16)(150 Ω)(30s)

W = 720 J

Try this…

You are given 10 Ω, 20 Ω, and 30 Ω resistors, and a 3V battery.

a. Draw a schematic diagram of these components in a series circuit.

b. Determine the equivalent resistance of the circuit.c. Determine the current through each resistor.d. Determine the potential difference across each

resistor.e. Determine the power dissipated by the 20 resistor.

Try this…

Consider the following series circuit:a. What is the potential difference between: a

and b, b and c, c and d, d and a.b. What is the current in the circuit?c. What power is dissipated by each resistor?

3V

12

6

a b

cd

Types of Circuits: Parallel

A parallel circuit can be constructed by connecting light bulbs in such a manner that there are SEVERAL PATHS for charge flow (CURRENT);

the light bulbs are placed within a separate branch line, and a charge moving through the circuit will pass through only

one of the branches during its path back to the low potential

(negative) terminal of the battery. Since there is now several current paths, the total current of the

circuit equals the sum of the current in each branch.

I = I1 + I2 + I3

Types of Circuits: ParallelLet’s take a closer look at how the current is flowing through each branch.

12 A

3 A

9 A 9 A

3 A

6 A 6 A

6 A

6 A6 A

3 A

9 A9 A

3 A

12 A

Does this hold true?I = I1 + I2 + I3


Let’s look at voltage:

Each branch is hooked up to the same battery. Each branch has the same voltage (electric pressure).

V = V1 = V2 = V3

Outlets in a house are connected in parallel so you can use one appliance without having to turn them all on.


Let’s look at RESISTANCE:

The actual amount of current always varies inversely with the amount of overall resistance.

I = V Req

1 = 1 + 1 + 1Req R1 R2 R3

1 = 1 + 1 + 1Req R1 R2 R3

The total resistance of a parallel circuit is a fraction of all of the resistors added together!!!!



I = V Req

So the more resistors you add in parallel the lower the equivalent resistance will be. And because Req the I sometimes to a potentially dangerous level.


This is why we have circuit breakers or fuses inserted into the main line of our homes. When we are trying to use too

many electrical devices (resistors) in parallel the current gets to be too much and a fuse in blown.


Inside the fuse is a small piece of metal, across which current must pass. During normal flow of current, the fuse allows the current to pass unobstructed. But during an unsafe overload, the small piece of

metal melts, stopping the flow of current.

Circuit breakers are switches that are tripped when the current flow passes a unsafe limit. The excess of current typically triggers an electromagnet, which trips the circuit breaker when an unsafe limit is reached. Once tripped, the switches simply turn off. That stops the flow of electricity, which will remain

off until the switch is reset.

Parallel Summary:

I = I1 + I2 + I3

V = V1 = V2 = V3

1 = 1 + 1 + 1Req R1 R2 R3

In this animation you should notice the following things:•More current flows through the smaller resistance. (More charges take the easiest path.) •The battery or source is represented by an escalator which raises charges to a higher level of energy. •As the charges move through the resistors (represented by the paddle wheels) they do work on the resistor and as a result, they lose electrical energy. •By the time each charge makes it back to the battery, it has lost all the electrical energy given to it by the battery. •The total of the potential drops ( - potential difference) of each "branch" or path is the same as the potential rise ( + potential difference) across the battery. This demonstrates that a charge can only do as much work as was done on it by the battery. •The charges are positive so this is a representation of conventional current (the apparent flow of positive charges) •The charges are only flowing in one direction so this would be considered direct current ( D.C. ).

30 Ω 10 Ω

Series vs. ParallelSeries Parallel

I

V

R

I = I1 + I2 + I3

V = V1 = V2 = V3

1 = 1 + 1 + 1Req R1 R2 R3

I = I1 = I2 = I3

Req = R1 + R2 + R3

V = V1 + V2 + V3

The more resistors you have the less the equivalent resistance is, and current increases. Voltage never changes.

The more resistors you have the more the equivalent resistance is, and current never changes. The voltage acrosseach resistor adds up to the total voltage of the source (battery).

Hooking up Devices Properly

Ammeters, because of the way they are built, have very little resistance. So they can be placed in series with other devices in a circuit and not disrupt the current.

A

Voltmeters on the other hand have a very high resistance and when placed in series would disrupt the current so they are placed in parallel within the circuit- separate branch.

V

Try this…Calculate (a) the equivalent resistance, (b) the potential difference across each resistor, and (c) the current through each resistor.

12 V 4 Ω 6 Ω 12 Ω

I = I1 + I2 + I3

V = V1 = V2 = V3

1 = 1 + 1 + 1Req R1 R2 R3

a. 1 = 1 + 1 + 1 Req R1 R2 R3

1 = 3 + 2 + 1Req 12 Ω 12 Ω 12 Ω

Need a common denominator

1 = 1 + 1 + 1Req 4 Ω 6 Ω 12 Ω

1 = 6 Req 12 Ω

Req = 12 Ω 1 6

= 2 Ω

NOTE: Notice how the Req is less than any of the resistors in the circuit.

12 V 4 Ω 6 Ω 12 Ω

b. V = V1 = V2 = V3

V = 12 V

c. I = V R

I1 = V R1

I1 = 12 V 4 ΩI1 = 3 A

I2 = V R2

I2 = 12 V 6 ΩI2 = 2 A

I3 = V R3

I3 = 12 V 4 ΩI3 = 3 A

Try this…

Find the magnitude of current that is flowing through all 3 ammeters.

I = I1 + I2 + I3

V = V1 = V2 = V3

1 = 1 + 1 + 1Req R1 R2 R3

A1 A2

A3

12 V 10 Ω 15 Ω

A1 = I1 = V R1

I1 = 12 V 10 Ω

I1 = 1.2 A

A2 = I2 = V R2

I2 = 12 V 15 Ω

I2 = .8 A

A3 = I1+ I2

A3 =1.2A + .8A

A3 = 2 A

Which two of the resistor arrangements below have the same equivalent resistance?

A

B

D

C

1 Ω 1 Ω

8 Ω

8 Ω

2 Ω2 Ω

2 Ω

2 Ω

Which circuit below would have the lowest voltmeter reading?

6 V

20 Ω 40 Ω

V

6 V

20 Ω 40 Ω

V

6 V

20 Ω

40 Ω

V

6 V

20 Ω

40 Ω

V

A

B D

C

Arrange the schematic diagrams below in order of increasing equivalent resistance.

1 2 3 4

Find the resistance of R3.

R1 = 6 Ω R2 = 6 Ω R3 = ? Req = 2 Ω

1 = 1 + 1 + 1Req R1 R2 R3

Simplifying Resistors in Combination Circuits

3Ω 11Ω

4 Ω 18 Ω

4 Ω18 Ω

12

62

3

4

12V

4

2

3

4

12V

6

3

4

12V

ComboCircuitsQuiz

Now we can analyze other aspects of the circuit…in order to do this we must first simplify.

R1 = 1 ΩR2 = 8 Ω

R3 = 8 Ω

1 = 1 + 1 R2,3 R2 R3

1 = 1 + 1 = 2 = 4 ΩR2,3 8 Ω 8 Ω 8 Ω

R1 = 1 Ω R2,3 = 4 Ω

100 V 0 V

100 V 0 V

Next simplified the circuit down to one resistor.

R1 = 1 Ω R2,3 = 4 Ω100 V 0 V

RT = R1 + R2,3

RT = 1 Ω + 4 Ω

RT = 5 Ω

RT = 5 Ω100 V 0 V

We went from

R1 = 1 ΩR2 = 8 Ω

R3 = 8 Ω

100 V 0 V

toR1 = 1 Ω R2,3 = 4 Ω

100 V 0 V

toRT = 5 Ω

100 V 0 V

Next find the total current flowing through the simplified circuit.

RT = 5 Ω100 V 0 V

IT = VT RT

IT = 100 V 5 Ω

IT = 20 A

Now we can go back and un-simplify the circuit and find the current and voltage through specific resistors…

Find the current and voltage through R1. Hint: Always find current first, then voltage.

R1 = 1 Ω R2,3 = 4 Ω100 V 0 V

RT = 5 Ω100 V 0 V

We already found the current through RT and since this is a series circuit IT = I1 = I2,3 so I1 = 20 A.

Next find voltage…

R1 = 1 Ω R2,3 = 4 Ω100 V 0 V

I1 = 20 A I2,3 = 20 A

V1 = I1R1

In a series circuit VT = V1 + V2,3 so the voltages across both of these resistors will add up to 100 V.

V1 = (20A)(1 Ω)

V1 = 20 V

V2,3 = I2,3R2,3

V2,3 = (20A)(4 Ω)

V2,3 = 80 V

Let’s check to see if the voltages make sense: VT = V1 + V2,3 100 V = 20 V + 80 V

Now we can find the separate currents and voltages through R2 and R3.

R1 = 1 Ω R2,3 = 4 Ω100 V 0 V

I1 = 20 A I2,3 = 20 A

R1 = 1 ΩR2 = 8 Ω

R3 = 8 Ω

100 V 0 V

I1 = 20 A

I2,3 = 20 A

R1 = 1 ΩR2 = 8 Ω

R3 = 8 Ω

100 V 0 V

I1 = 20 A

I2,3 = 20 A

R2 and R3 are in parallel so we know that V2,3 = V2 = V3 therefore

V2 and V3 are 80 V.Now we can find I2 and I3.

I2 = V2,3

R2

I2 = 80 V 8 Ω

I2 = 10 A

I3 = V2,3

R3

I3 = 80 V 8 Ω

I3 = 10 A

V1 = 20 V

V2,3 = 80 V

Let’s try another one…

R3 = 6 Ω

R1 = 4 Ω R2 = 2 Ω

Resistors R1 and R2 are in series so R1,2 = R1 + R2

R1,2 = R1 + R2

R1,2 = 4 Ω + 2 Ω = 6 Ω

R3 = 6 Ω

R1,2 = 6 Ω

100 V 0 V

100 V 0 V

R3 = 6 Ω

R1,2 = 6 Ω

100 V 0 V

1 = 1 + 1 RT R1,2 R3

1 = 1 + 1 = 2 = 3 ΩRT 6 Ω 6 Ω 6 Ω

RT = 3 Ω100 V 0 V

R3 = 6 Ω

R1 = 4 Ω R2 = 2 Ω

100 V 0 V

R3 = 6 Ω

R1,2 = 6 Ω

100 V 0 V

RT = 3 Ω100 V 0 V

What do we do next?

RT = 3 Ω100 V 0 V

IT = VT

RT

IT = 100 V 3 Ω

IT = 33.33 A

Next we un-simplify the circuit and find the rest…

R3 = 6 Ω

R1,2 = 6 Ω

100 V 0 VIT = 33.33 A

Since R1,2 and R3 are in parallel they have voltage in common: VT = V1,2 = V3 therefore

V1,2 and V3 are both 100 V. We can now find each resistors individual current flow, IT = I1,2 + I3.

I1,2 = V1,2

R1,2

I1,2 = 100 V 6 Ω

I1,2 = 16.67 A

I3 = V3

R3

I3 = 100 V 6 Ω

I3 = 16.67 A

Now let’s separate R1 and R2…

R3 = 6 Ω

R1,2 = 6 Ω

100 V 0 V

R3 = 6 Ω

R1 = 4 Ω R2 = 2 Ω

100 V 0 V

I3 = 16.67 A

I1,2 = 16.67 A

R3 = 6 Ω

R1 = 4 Ω R2 = 2 Ω

100 V 0 V

I3 = 16.67 A

I1,2 = 16.67 A

R1 and R2 are in series with each other so they have the same amount of current flowing through them: I1,2 = I1 = I2

therefore the current flowing through both of them is 16.67 A which will help us find the voltage drop across each resistor.

V1 = I1,2R1

V1 = (16.67A)(4 Ω)

V1 = 66.68 V

V2 = I1,2R2

V2 = (16.67A)(2 Ω)

V2 = 33.34 V

More practice…

a)Which letter shows the graph of voltage vs. current for the smallest resistance?

b) Which letter shows the graph of voltage vs. current for the largest resistance?

a) What is the total resistance of this circuit?

b) What is the total current of this circuit?

c) What is the amount of current running through each resistor?

a) What is the total resistance of this circuit?b) What is the total current of this circuit?

a) Find the equivalent resistance. b) Find the current (IT) going through this circuit. c) Find potential drop across R1 & R2

a) Find combined resistance (RT).b) Find the current in R1.

c) Find I3.

d) Find R2.

e) Find value of the second resistor.

a) If the voltage drop across the 3 ohm resistor is 4 volts, then what would the voltage drop be across the 6 ohm resistor?b) Find the total voltage in this series circuit.c) Find combined resistance in this circuit. d) Find the total current in this circuit.

a) Current in this circuit?b) Potential difference in 20 ohm resistor?c) Equivalent resistance in the circuit?

Find R2.

What happens to the brightness of the light bulbs as we add more and

more bulbs into the circuit?


http://phet.colorado.edu/simulations/sims.php?sim=Circuit_Construction_Kit_DC_Only


http://phet.colorado.edu/simulations/sims.php?sim=Circuit_Construction_Kit_DC_Only

Charging an object by INDUCTION-

Starting with a neutral sphere

Step 1:

+A negatively charged object is held up to the sphere. Electrons are repelled from the balloon.

++

----

Step 2:

A “ground” comesin and allows an easy path for the electrons to travel through and the electrons leave the sphere.

-- +++

--

- -

A ground is simply a large object which serves as an almost infinite source of electrons or sink for electrons (sink in this case). A ground contains such vast space that it is the ideal object to either receive electrons or supply electrons to whatever object needs to get rid of them or receive them.

---

Step 3:

The balloon is removed and the positive charges are redistributed uniformly throughout the sphere now giving it an overall positive charge.

+ ++

This is charging an object by induction- it started off neutral and then became positive.

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Electricity