Current Electricity

Chapter 20

Potential Energy

If you do work against gravity, the gravitational field stores that energy as Gravitational Potential Energy, GPE

If you do work against an electrostatic force, the electric field stores that energy as electric Potential Energy, EPE

Low PE

High PE

Electric Potential Energy Electrostatic Force is conservative Electrical potential energy is

the energy contained in a configuration of charges.

Like all potential energies, when it goes up the configuration is less stable; when it goes down, the configuration is more stable.

The unit is the Joule.

Electric Potential EnergyElectrical potential energy increases when charges are brought into more unstable configurations.

+ + +

Lower PE Higher PE

dFe

+

Moving q1 closer to q2 requires work and that will increase the PE of the charge. Work against electric force increases electric PE

Stable Unstable

Electric Potential EnergyElectrical potential energy decreases when charges are brought into more stable configurations.

+ -

Higher PE Lower PE

d

Fe+ -

q1 will naturally move or fall towards q2 in the direction of E. No work is required and the PE of the charge will decrease. Work with the electric force decreases electric PE

StableUnstable

Electric Potential EnergyMoving a charge in an electric field requires work. This work stores potential energy.

Vq

U

q

W

UW

elel

elel

Electric Potential, V, is electric potential energy per charge. At every location, a charge has a position-dependent potential. Potential difference is simply the difference in electric potential at any 2 points

Electric PotentialUNITS: J/C or Volt

Electric energy provides the means to transfer large quantities of energy over great distances with little loss.

Producing Electric Energy

Because electric energy can so easily be changed into other forms, it has become indispensable in our daily lives.

Producing Electric Energy

Circuit

Low V

High V

Charges flow from high to low V through conducting wireThis flow of positive charge is called conventional current

The flow stops when the potential difference between A and B is zero.

A

B

Potential Difference and Current

Lower V

High V

To be a circuit, charges must flow continuously thru a loop, returning to their original position and cycling thru again. To do so requires energy input, a charge pump that raises the electric potential of the charge

Circuits require an electrical energy source – VOLTAGE SOURCE, DV

A circuit is simply a closed, conducting loop through which charges can continuously move

Flow of charge is CURRENT, I

Energy Pump +DV

1. Voltaic or galvanic cell converts chemical E to electric E.

A battery is made up of several galvanic cells connected together.

2. Photovoltaic cell, or solar cell—changes light energy into electric energy.

Producing Electric Current (Charge Pump)

Requirements of a Circuit1. Closed conducting loop

2. An energy source that maintains an electric potential difference, DV, across the ends of the circuit.

+-DV

Energy Source

Pumps charge from – to + terminal

maintains a DV across the circuit

Low VLow energy

High VHigh

energy

CurrentOnce the two requirements of a circuit are met, charge will flow.Rate of charge flow is called CURRENTElectric current is represented by I I = DQ/t

Unit of current:

Ampere (A)1 A = 1 C/s

Current A 2 mm long cross section of wire is isolated and 20 C of charge is determined to pass through it in 40 s.

I = _____________ A20C/40s = 0.5

Conventional Current DirectionCurrent is in the direction of + charge flow.

In reality, the particles that carry charge through a wire are mobile electrons which move in a direction opposite conventional current.

+-

I

DV

Vdrift ≈ 10-6 m/s or 1 m/hr (SLOW!!!)

DV

I = Q/t

The charge carriers are densely packed into the wire, so there does not have to be a high speed to have a high current. That is, the charge carriers do not have to travel a long distance in a second, there just has to be a lot of them passing through the cross section.

The amount of current in a circuit depends on BOTH the potential difference across the circuit, DV, AND the total resistance in the circuit, R.

Resistance

LOAD

Energy Source

II

An electron traveling through the wires and loads of a circuit encounters resistance, R. Resistance is a hindrance to the flow of charge.

DV

Flow rate depends on- pump pressure- resistance of pipe which depends on pipe length and diameter

Water Flow through pipe

Current depends on

- Voltage difference, DV- Resistance of circuit elements

Current through Circuit

ResistanceMaterials and elements in a circuit offer

resistance to the flow of charge.

A

LR

pHet Sim

A

L

Resistivity ( W m)Conductors: r ~10-8 WmInsulators: r ~1011-1016 Wm

d

L

d

2L

2d

L

2d

2L

ResistanceRank the following circuit elements from highest resistance to lowest resistance.

A.

B.

C.

D.

A > C > B > D

R

VI

Current flow does NOT depend only on voltage. Charge traveling through the wires and loads of a circuit encounters resistance, R. Resistance is a hindrance to the current. The higher the resistance, the smaller the current.

VI

To produce electric current, I, a potential difference, DV, is required. Simon Ohm established experimentally that the current in a metal wire is proportional to the potential difference applied to its ends.

Ohm’s Law

IRV ResistanceUnits: Ohms ( )W

CurrentUnits: Amperes ( )A

Electric potentialUnits: Volts (v)

Every element in a circuit obeys Ohm’s Law

Circuit Components

Voltage Source

+ -

+ -Battery

Wire

Light bulb

Switch

Resistor

In which circuit does the light bulb have highest resistance? What are the resistances of each bulb?

RA = ΔVA / I = 6/1 = 6 Ω RB = ΔVB / I = 6/2 = 3 Ω

A. B.

1A

+ -6 V

2A

+ -6 V

6V

6V

0V

0V

ΔVA = -6VΔVB = -6V

A

Which of the following will cause the current through an electrical circuit to decrease? Choose all that apply.

a. decrease the voltageb. decrease the resistancec. increase the voltaged. increase the resistance

Kirchoff’s Rules1. Junction Rule At any junction point

in a circuit, the sum of all the currents entering the junction must equal the sum of all currents leaving the junctionCurrent into junction = Current out of junction(Conservation of

charge)

I = I1 + I2

I

I2

I1I

R1

R2

outin II

4 A

10 A What is the current in the unknown wire and what are the directions of the currents?6 A

4 A

5 A

11 A

2 A

14 A AI

I

II outin

14

11524

?

Kirchoff’s Rules2. Loop Rule The sum of the potential

differences across all elements around any closed loop of a circuit must be zero. (Conservation of

energy) loop

V 0

+ -

R1

R2

R3

DVbat

DV is negative (voltage drop) across elements in the circuit.

DV is positive (voltage gain) across the battery.

+ -+6 V VV

VV

B

Bbat

6

0

VV

V

VVV

B

B

BBbat

3

026

0

Outer Loop

What is the voltage drop across each of the bulbs? The bulbs have the same resistance.

-6 V

-3 V -3 V

Inner Loop

Two Types of ConnectionsWhen there are 2 or more electrical devices in a circuit with an energy source, there are a couple of ways to connect them.

Series ConnectionWhen devices are in series, all of the current going through one device goes through the other 1 2 I

Device 1 and 2 are in series. To go from A to B, all the charge passing through 1 must go through 2.

The SAME CURRENT goes through resistors in series. The total R is greater than each individual resistor

A B

Parallel Connection When devices are in parallel, the current splits; some current goes through one device, some through the other. In going from A to B, a charge goes through one device only. 1

2

I

Device 1 and 2 are in parallel. More charge (current) goes through path of least resistance

Resistors in parallel have the SAME VOLTAGE DROP. The total R is less than each individual resistor

A BI1I2

IKJR

I = I1+I2

Circuit Connections

Series Circuit–How do the brightnesses compare?

Which circuit has the greater current flow?

Does the charge get used up?

+ - + -

1 2 3brighterdimmer

More I (less resistance)

less I(more resistance)

the circuit with less load or less resistance

In a closed loop, current flow is the same through every element (light bulbs 1, 2, and 3 are equally bright)

I + - + -

Parallel ConnectionsAs the number of light bulbs (resistance) increases,

• what happens to the current through the circuit?

3I

More I

less I

Increases, There are more pathways so less

resistance

Series ConnectionsIf one resistor is turned off (a light bulb goes out), what happens to the other resistors in the circuit?

If one resistor goes out, there is no longer a closed loop for current flow and all other devices in series will go out. There is an OPEN CIRCUIT

Parallel ConnectionsIf on resistor is turned off (a light bulb goes out), what happens to the other resistors in the circuit?

If one resistor goes out, there is still a closed loop for current flow and so the other devices in series will stay on

Energy Transfer and Power When a LOAD is put on the circuit

(light bulb, beeper, motor…), electrical energy is transformed to other, useful forms of energy.

An electrical circuit is simply an energy transformation tool. Rate of energy transformation/transfer is POWER

LOAD

Energy Source

Energy Transfer and PowerPOWER, P, is the rate

that energy is supplied to the load or the rate of work done on the charge.

LOAD

Energy Source

RVRIIVP

t

qV

t

W

t

E

timeP

dtransformeEnergy

/22

Unit of Power: Watt (W)

1 W = 1 J/s

Energy transfer and PowerPOWER, P, is the rate that energy is

supplied to the load or the rate of work done on the charge.

LOAD

Energy Source

60 Watt light bulb means 60 J of

energy delivered to bulb every second

OR 60 J of energy used

by the bulb per second

Electric heater. An electric heater draws 15.0 A on a 120 V line. How much power does it use and how much does it cost per month (30 days) if it operates 3.0 h per day and the electric company charges 10.5 cents per kW-h?

kWWIVP 8.11800)120)(15( To operate it for 30 days, 3 hr/day would total 90hrs and would use

hrkWhrskW 162)90(8.1

17$)105.0($162 kWhrCost

Will a fuse blow? Determine the total current drawn by all the devices used at once. Will they blow a 20-A fuse?

AVWVPILB 8.0120/100/

AVWVPIH 15120/1800/

AVWVPIS 9.2120/350/

AVWVPID 10120/1200/

A7.28YES

Equivalent Resistance – Resistors in Series

0321 VVVVb

IRV Each element obeys

Ohms Law

)(

0

321

321

RRRIIR

IRIRIRIR

Total

Total

321 RRRReq

(Kirchoff’s Loop Rule)

I

R1 R2

+ -

R3

-DV1 -DV2 -DV3

DVbat

I

REq

+ -

DVbat

Equivalent Resistance – Resistors in Series

This circuit can be replaced by

...

...

...

...

321

321

321

321

PPPP

IIII

VVVV

RRRReq

+ -R1

R2

R3I

+ -

Req

I

this one whereV

V

I I2

I3

I1

321 IIII

311

311

3

3

2

2

1

1

1111

)111

(

RRRR

RRRV

R

V

R

V

R

V

R

VI

total

total

+ -

+DV

Equivalent Resistance – Resistors in Parallel

R1

R2

R3

Kirchoff’s Junction Rule:

Equivalent Resistance – Resistors in Parallel

This circuit can be replaced by

...

...

...

...1111

321

321

321

331

PPPP

IIII

VVVV

RRRReq

this one where

+- R1 R2 R3

I I1 I2 I3

I

V

+- Re

q

I

V

Series or Parallel? The light bulbs are identical and have identical resistance, R. Which configuration produces more light? Which way do you think the headlights of a car are wired?

More pathways,

Less resistance,

More current

More light

DO NOW What is the equivalent resistance of the circuit? What is the current through the circuit?

+ -4 W

I+ -

Req = 18W

I

12 V

6 W

8 W

12 V

AIIIRV eqReq67.01812

IRV Ohms Law

Rank, from highest to lowest, the voltage between A and B, C and D, D and EE and F

4W

12 V

6W

8W

AI 67.0

IRV

+ -

I

AB

C D E F1 (highest)

23

4

-5.36V

-4.02V

-2.68V

If the resistors were light bulbs, which would be brightest?The 8 W one

+ -

Example - What is the equivalent resistance of the circuit?

+ -

Req = 1.8W

12 V

4 W6 W

8 W

12 V

The equivalent resistance of resistors in

parallel is less than the smallest single resistance

+ -

12 V

4 W

8 W

Rank the currents at points A-F from greatest to least

AB

CD

EF

A=B > F=E > D=C

If the resistors were light bulbs, which would be brighter?

The 4 W one

+ -

Series and Parallel Resistors – Two 100 W resistors are connected a) in series and b) in parallel to a 24.0 V battery. What is the current through each resistor? What is Req of each circuit?

+ -

Req = 33W

30 V

50 W

100 W

30 V

+ -I

+ -

Req = 150 W

I

30 V

100 W

30 V

50 W

I1

This is the current through each resistor

I3 I2

I1

I = 0.20 A

I1

I1 = 0.91 AI2 = 0.30AI3 = 0.61 A

+ -

Series and Parallel Resistors – Two 100 W resistors are connected a) in series and b) in parallel to a 24.0 V battery. What is the current through each resistor? What is Req of each circuit?

+ -

Req = 33W

30 V

50 W

100 W

30 V

AIII 90.060.030.032

I

I3I2

ARVI 30.0100/30/22 ARVI 60.050/30/3

12 V

I

I2

12W

2W 4W

I3

Find the current through each resistor and the current drawn from the battery.

1. Draw and label the current flow in the circuit2. Apply Kirchoff’s Rules

AIIIRIRIVV

204212012012

2

22

4222

42

Loop1

AIIRIV

101212012012

3

3

123

12

Loop2

Kirchoff’s Loop Rule

Kirchoff’s Junction Rule A

III312

32

Circuit Components

Voltmeter

Ammeter

V

A

W Ohmmeter

Measuring Current

Current to be measured must pass through the ammeter, so it must be placed in SERIES mode in the circuit.

Ideally ammeters have ZERO resistance so that they do not affect the energy of the circuit

Ammeter – measures current

Measuring VoltageVoltmeter – measures

voltage

Does NOT require the current to pass through it. It must be placed in parallel to the circuit element.

Ideally voltmeters have INFINITE resistance so that they do not draw current away from circuit.

For the circuit below, all resistors have the same value, R (10 )Wa) Draw and label the current flow

b) Calculate the equivalent resistance of the circuit

c) Calculate the total current provided by the battery.

d) Calculate the current in resistor 3

e) Calculate the current in resistor 5

R282V

I I3

I2

I3

I4I5

R1 R3

R4

R5

R6

R7R8

R2

I I3

I2

I3

I4I5

R1 R3

R4

R5

R6

R7R8

R2

I I3

I2

I3

I4

I5

R1 R3

R4 R5

6

R7R8

R2

II3

I2

R1 R3

R456

R7R8

R2

II3

I2

R1

R34567

R8

R2

II3

I2

R1

R34567

R8

I

R1

R234567

R8

IReq82V

Req = 2 8/11 R = 27.3W

For the circuit below, all resistors have the same value, R (10 )Wa) Draw and label the current flow

b) Calculate the equivalent resistance of the circuit

c) Calculate the total current provided by the battery.

d) Calculate the current in resistor 3

e) Calculate the current in resistor 5

I = 3A

R282V

I I3

I2

I3

I4I5

R1 R3

R4

R5

R6

R7R8

I3 = 33/40 A = 0.825AI5 = 0.275A

DV

bat

R1 R2

R3

R4R5

R6

I=2

I

I

I1I2=1.5

I3 = 1.0

I4

DV(v)

I(A)

R (W)

Bat 2.0

R1 5.0

R2 3.5

R3 1.5

R4 4.0

R5 1.0

R6 2.0

Example– Calculate the missing information in the table for the following series-parallel network. Find the equivalent resistance of the circuit.

Req = 7 W

14

2.5 0.5

0.5 7.0

6.0 4.0

1.0 4.0

4.0 4.0

4.0 2.0