Electric Potential. Let’s review... 10 kg 12 m KE + PE g = E

Electric PotentialElectric Potential

Let’s review...

10 kg

12 m

KE + PEg = E

Force by Hand?

10 kg12 m

KE + PEg = E

WE,b = 100N

NR,b = 100N

a = 0

Workext = Fext x= 100N m

=J

Workext?

Which buckets were filled?

10 kg

12 m

KE + PEg = E

KE + PEg = E

WORKPEg = mgh

Energy sloshes

10 kg

12 m

KE + PEg = E

KE + PEg = E10 kg

B AX

qtest=+1 C

Like Gravitational potential energy a charged object can have potential energy due to it’s location in an electric field. Just like work is required to lift an object, work is required to push a charged particle against the electric field of a charged body.

Electrical potential = electric potential energy divided by charge

V = PE/q1 volt = 1 Joule/Coulomb PE = q V

Current, voltage Current, voltage and resistanceand resistance

Consider a flashlight …..

battery battery

Wiring Diagram or a Schematic

Battery

Battery

Resistor

Switch

Predict the relative brightness

A B C

D

E

F G

H

I

1

32

4

5

6

7

CURRENT:

• Measure of the rate of flow of charge.

•I = q/t• Measured in coulombs/second,

which we define as the Ampere (A).

How do we measure current?

• use an ammeter• How do we insert it?

I = ?A

What drives the current?

Electromotive Force (EMF)

• The battery acts like a (charge) pump.• EMF is the energy that one coulomb of

charge gains in passing through a battery.

• Measured in J/C or volts.• We’ll just call this the “battery voltage”

• Also given the symbol batt

EMF and Voltage

Loses 12 J/C

batt = 12 volts

A

B

VAB= J/C

D

C

VCD= J/C

Loses 12 J/C

= 12 volts

Loses 12 J/C

= 12 volts

Loses 12 J/C

= 12 volts

Loses 12 J/C

= 12 volts

Loses 12 J/C

= 12 volts

Loses 12 J/C

= 12 volts

Loses 12 J/C

= 12 volts

Loses 12 J/C

= 12 volts

Loses 12 J/C

= 12 volts

Loses 12 J/C

= 12 volts

How do we measure voltage?

• Voltage is a measure of the electric potential. • use a voltmeter• How do we insert it?

V

10.000 A

.001 A9.999 A

VA,B

A

B

Consider a simple circuit

• What if we increase the push?• more push means more flow

• What if we change the bulb?• not all identical pushes produce

identical flows

Resistance...Resistance...

……is futileis futile

QuickTime™ and aGIF decompressor

are needed to see this picture.

Resistance

• Represents the ability of a circuit element to impede the flow of current.

•R = V/I• volt/ampere = ohm • Georg S. Ohm (1789-1854)

V = I R

Which has more resistance?

steel

wood

Which has more resistance?

steel

steel

Which has more resistance?

steel

steel

Which has more resistance?

Steel at 0oC

Steel at 25oC

More flow more glow!

AC Delco- +

adjustable battery

RR

Vbatt = IbattRcircuit Vbatt = IbattRcircuit

A B

indicator bulb

Which box contains the greater resistance?

A B

Which is box A and which is box B?

R R

R R

B A

A B

The bigger the R

the smaller the current

R R

R R

B Athrough the battery.

of the circuit,

Junction

R R

Vbatt = IbattRcircuitVbatt = IbattRcircuit

A B

Which is box A and which is box B?

R R

A B

Adding in Series• What happens to Ibatt if a bulb or network of bulbs is inserted between these two

bulbs?

• In all cases, it goes down. Why?• Clog up an existing path,• increase the resistance of the circuit, and• decrease the current through the battery.

A

B

Summary• When a bulb or network of bulbs is added in series, the resistance of the

circuit increases.

• The important thing is not what is added but how it is added.

• If you have to break the circuit, you are adding something in series.

A

B

Adding in Parallel• What happens to Ibatt when a bulb (or network) is added in parallel to bulb B?

A

B

A

• It goes up. Why?• Add a new path (opportunity),• decrease the resistance of the circuit, and• increase the current through the battery.

Summary• When a bulb or network of bulbs is added in parallel, the resistance of the

circuit decreases.

• Again, the important thing is not what is added but how it is added.

• If you don’t have to break the circuit, you are adding something in parallel.

A

B

A

A B

Which is box A and which is box B?

R R

A B

A B

Which is box A and which is box B?

R R

B A

• Rank all the bulbs below for brightness. Explain your reasoning.

A

CB

A>B=C

A gets all of the current while B and C each get half of total the current.

B and C share equally since each branch has equal resistance.

Be the river

Rank the bulbs• Rank all the bulbs below for brightness. Explain your reasoning. Batteries

and bulbs are all the same.

A

CBF

ED

A=F>B=C=D=E Both circuits have the same current. Which comes first does not matter.

• Current favors the branch of least resistance.

C<D Less current will take the left branch since it has more resistance than the right branch.

• Which is brighter, C or D? Explain.

A

C

BD

Ranking: A>D>B=C

Be the river

A

C

B

D

E

E>A=B>C=D

Be the river

Short Circuits

A

B

What happens to the brightness of each bulb when the switch is closed?

SP 117

Rank the Resistances

A C

R R

Rank the Resistances

B CA

RR R

The need for more

A

CB

Closed: A>B=C

Open: A=B, C out

What happens to B as switch is opened?

Review of Current Model

More flow more glow. The current through the battery

depends on the resistance of the circuit. It’s not what you add but how you add

it. Current favors the path of least

resistance. Branches connected directly to a

battery are independent.

Add a bulb in series

What about the currents now?

12 J/C

6 J/C

6 J/C

What do we mean by sharing?

= 12 volts

= 12 volts

= 12 volts

= 12 volts

= 12 volts

= 12 volts

= 12 volts

= 12 volts

= 12 volts

= 12 volts

= 12 volts

= 12 volts

-1

= 12 volts

Start Here

Start Here

Start Here

PE/person = 75 kg x 10 N/kg x 1000m

PE/person = 750,000 J

PE/person = ?

Start Here

PE/person = 75 kg x 10 N/kg x 1000m

PE/person = 750,000 J

PE/person = ?

PE/person = ?

R

R

I

I

What if there is more than one path?

How bright are the bulbs?

12 J/C

What if there is more than one path?

12 J/C

What if there is more than one path?

6 J/C

What about the currents?

Voltage Model

• Voltage: how much energy each coulomb of charge gains (battery) or loses (bulb) in going through an element.

• More voltage, more glow.• More volt, more jolt.

• What goes up, must come down!• voltage rises and drops must be equal

around any loop.

Applying The Voltage Model

=12 V

Rank the bulbs.

Applying The Voltage Model

=12 V

Rank the bulbs.

Va

>6 volts=6 volts<6 volts

Vb Vc

Va + Vb + Vc = 12 volts -or- Va + Vb = 12 volts

Applying The Voltage Model

=12 V

Which meter reads more?

Vb Vc

Va

Applying The Voltage Model

=12 V

Which meter reads more?

Vb Vc

Va V

V

Which network has the greater resistance?

R

R

Applying The Voltage Model

=12 V

For series networks, the voltage drop is greater across the larger R.

Vb Vc

Va V

V

R

R

Voltage Model

• Voltage: how much energy each coulomb of charge gains (battery) or loses (bulb) in going through an element.

• More voltage, more glow.• More volt, more jolt.

• What goes up, must come down!• voltage rises and drops must be equal

around any loop.

Voltage Model

The voltage for series elements DIVIDES.

The voltage divides equally for identical Rs.

Otherwise, the voltage drop is greater across the greater/lesser R.

A

B

R.

Be careful with your words...

2

4

12 volts

200

200 4 volts

8 volts

The greater the resistance,

...the greater the voltage.

the share of

the share of

6 volts

6 volts

Parallel Series

I

I/2 I/2

I

I

parallel

series

current voltage

splits same

same divides