We define resistance, R as

And we have for ohmicmaterials Ohm’s Law: RIV =

ALR ρ=

Resistance R

Units of R: Ω

Temperature dependence (ohmic materials):

)](1[)( 00 TTRTR −+= α

Example: A 80.0 m Cu wire 1.0 mm in diameter is joined to a49.0 m iron wire of the same diameter. The current in each is 2.0 A.a.) Find V in each wire.

Cu

Fe

VIRVVAIRV

mmm

ALR

mmm

ALR

FeFe

CuCu

FeFeFe

CuCuCu

5.1246.3)73.1)(0.2(

24.6)105.0(

)0.49)(1010(

73.1)105.0(

)0.80)(107.1(

23

8

23

8

===Ω==

Ω=×Ω×

==

Ω=×Ω×

==

−

−

−

−

πρ

πρ

b.) Find E in each wire.

mV

LVE

mV

LVE

ELV

Fe

FeFe

Cu

CuCu

255.0

43.0

==

==

=

I

VCu

VFe

Two cylindrical resistors are made from the same material, and they are equal in length. The first resistor has diameter d, and the second resistor has diameter 2d.

2) Compare the resistance of the two cylinders.

a) R1 > R2 b) R1 = R2 c) R1 < R2

3) If the same current flows through both resistors, compare theaverage velocities of the electrons in the two resistors:

a) v1 > v2 b) v1 = v2 c) v1 < v2

Preflight 15:

Lecture 15, ACT 1Two cylindrical resistors, R1 and R2, are made of identical material. R2 has twice the length of R1 but half the radius of R1. – These resistors are then connected to a battery V as shown:

VI1 I2

– What is the relation between I1, the current flowing in R1 , and I2 , the current flowing in R2?

(a) I1 < I2 (b) I1 = I2 (c) I1 > I2

1A

Lecture 15, ACT 1

• The resistivity of both resistors is the same (ρ).• Therefore the resistances are related as:

11

1

1

1

2

22 88

)4/(2 R

AL

AL

ALR ==== ρρρ

• The resistors have the same voltage across them; therefore

112

2 81

8I

RV

RVI ===

• Two cylindrical resistors, R1 and R2, are made of identical material. R2 has twice the length of R1 but half the radius of R1. – These resistors are then connected to a battery V as shown:

VI1 I2

– What is the relation between I1, the current flowing in R1 , and I2 , the current flowing in R2?

(a) I1 < I2 (b) I1 = I2 (c) I1 > I2

Electromotive Force and CircuitsHere we discuss “sources of electromotive force”, such as batteries, electric generators and solar cells, and we consider how they behave in a closed circuit. Below is an analogy between basketball and current in a closed circuit. The player does work to move the ball up and the ball loses energy in the viscous oil giving off heat. The battery does work tomove q from − to + terminals, gaining energy qV. The charge q movesthrough wire to resistor and loses energy in the resistor and goes to the – terminal.

Bat

tery

V

vol

ts

resi

stor

+

−

Vis

cous

oil

Ener

gy g

oes i

nto

Hea

ting

resi

stor

i

i

Electromotive Force (EMF)

A battery is a device that keeps a & b terminals at a fixed potentialdifference and will move a positive charge or current from the a to the b terminals by some process such as electrolysis (storage battery) or the photoelectric effect (solar cells). This “force” is called theelectromotive force or EMF or script ɛ.

The battery circuit diagram is given by perpendicular lines, with + terminal with the bigger line.

+

+−

−

Idealized Battery

+−

The ideal battery with potential ɛ has no internal resistance and is represented by

The real battery with potential ɛ has a small internal resistance r and is represented by

“Real Battery”

rɛ

+−

ɛ

Voltage, between a & b, is slightly less than ɛ

a b

RI

IRVV RB

εε

=

==+

−

I

ɛV across battery is ε no matterwhat current is drawn! Not very physical.

Electrical Meters

Voltmeter, measures voltage and has large internal resistance.An ideal voltmeter has infiniteresistance and does not conduct any current.

V

Ammeter, measures current and has very small internalresistance. An ideal ammeter has zero resistance and behaves like a conducting wire.

A

Real Battery with 12volts and an internal resistance of 2 ohms in a circuit with a resistor of 4 ohms

Voltage drop across 2Ohms is 4 volts

Voltage drop across 4Ohms is 8 voltsVab= Va′b′ = 8 volts

AVrR

I

rRIIRIr

VV BAAB

26

12)(

''

=Ω

=+

=

=+=−

=

εε

ε

+-

+ -

-

Terminal potential difference VAB = 12 – 4 = 8 V

Real life example of previous circuit is a car battery and a headlamp

R

r+ −

R = resistance in headlightr = internal resistance in battery

Real Battery with 12volts and an internal resistance of 2 ohms in a circuit with a resistor of 4 ohms

0=−− IRIrε+-

+ -

-

IRIr =−εRearranging:

Can interpret this asThe sum of all potentialdifferences around a closed loop must add tozero.

Kirchoff’s voltage law or loop rule.

Can define a voltage rise as positive OR a voltage fall as positive.

Electric Potential/Voltage Diagram of the circuit

Voltage drop across 2Ohms is 4 volts

Voltage drop across 4Ohms is 8 volts

NOTE 4Ω resistor hasonly 8 of the 12 volts.

+ + +- - -

The development of the battery has a curious history. Henry Cavendishstudied sting rays (called Torpedoes) to investigate how they createdelectricity around 1776. Around ~1790, Luigi Galvani applied currentto frog’s legs and observed them twitching. He believed there was an“animal electricity” stored in the frog’s brain. About 1800, AllesandroVolta, invented the “pile”, a series of silver and zinc disks in salt water.Volta tested his electricity by attaching electrodes to his tongue andeventually demonstrated his pile before Napoleon.

In 1803, George Forster, a convicted murdered was hanged at Newgateprison, and Galvani’s nephew, Giovanni Aldini, applied current from aVolta pile to the corpse causing the jaws, eyes, hands and legs to move. This process of resuscitation was eventually applied to drowning victims. This created philosophical and religious controversy about life, death and electricity. This gave Mary Shelley the idea for the 1818 novel, Frankenstein, about a corpse brought to life by electricity.

Early History of Current and Batteries

lightning rod attached to froggy!!

Drawing of one of Luigi Galvani’s experiments

LeydenJar

Manuscript on the invention of the battery sent to the Royal Society of London. Cardboard soaked in salt water was placed in between the zinc and silver disks.

Drawing of Volta’s Pile or Battery

Demonstrating before Napoleon

Y&F 25.36The circuit shown in the figure contains two batteries, each with an emf and an internal resistance, and two resistors. A) Find the direction and

magnitude of the current in the circuit

B) Find the terminal voltage Vab of the 16.0-V battery.

C) Find the potential difference Vbc of point b with respect to point c .

I

+

+

+

+-

-

-

-

-

Y&F 25.36The circuit shown in the figure contains two batteries, each with an emf and an internal resistance, and two resistors. A) Find the direction and

magnitude of the current in the circuit

B) Find the terminal voltage Vab of the 16.0-V battery.

C) Find the potential difference Vbc of point b with respect to point c .

AAI

IVIVIIIV

471.0178

017800.984.10.56.116

==

=Ω−=Ω−−Ω−Ω−Ω−A.) Use Kirchoff’s

loop law:

(+) so assumed direction correct

B.) Vab = 16V – I(1.6 Ω) = 16V – (0.471 A)(1.6 Ω) = 15.2 V

C.) Vbc = – I(9.0 Ω) = – (0.471 A)(9.0 Ω) = -4.24 V