Lecture Notes 3

26
3.0 Electric Current EPF 0024: Physics II 1

description

Physic 2

Transcript of Lecture Notes 3

Page 1: Lecture Notes 3

3.0 Electric Current

EPF 0024: Physics II 1

Page 2: Lecture Notes 3

OutlineOutline

3.1 Production of direct electric current3 2 Oh ’ L3.2 Ohm’s Law3.3 Resistivity3 4 El t i E d P3.4 Electric Energy and Power

EPF 0024: Physics II 2

Page 3: Lecture Notes 3

Today's lecture Include:Today s lecture Include:

Production of direct electric current

Ohm’s Law

EPF 0024: Physics II 3

Page 4: Lecture Notes 3

ObjectivesObjectives

To explain the basic principles of a simplecell and define the electric current.

State and explain Ohm’s law.

Solve Problems.

EPF 0024: Physics II 4

Page 5: Lecture Notes 3

3.1 Principle of a Simple Cellp p

A simple cell (Fig. 3.1) + − Zinc Carbon

+ Terminal − Terminal

A simple cell (Fig. 3.1)consists of two rods(electrodes) of carbon andzinc immersed in solution

electrode (−)electrode (+)

zinc immersed in solution,e.g. diluted sulfuric acid(electrolyte).

The terminals of the cell isthe portion of the electrode

t id th l t l t A

Acid

Electrical symbol for a celloutside the electrolyte. ABattery are several cellsconnected together (in

i )

Electrical symbol for a cell

+ −

EPF 0024: Physics II 5

series). Fig. 3.1 A Simple Electric Cell

Page 6: Lecture Notes 3

3 1 1 Operation of a simple cell3.1.1 Operation of a simple cell

Electrolyte dissolves zinc and each zinc atomyleaves 2 electrons behind and enters theelectrolyte as positive ion making zinc electrode–vely charged and electrolyte +vely charged.y g y y gThe +ve electrolyte pulls off electrons from thecarbon electrode making it +vely charged and ap d now exists between the two terminalsp.d. now exists between the two terminals.

The p.d. (voltage) that exists between terminalsp ( g )is called the electromotive force (emf). Allowingcharges to flow externally results in more zincbeing dissolved to maintain constant voltage at

EPF 0024: Physics II 6

being dissolved to maintain constant voltage atterminals. Eventually, the zinc will be used up.

Page 7: Lecture Notes 3

3.1.2 The Electric Current

A simple electric circuit ispshown in (Fig. 3.2 (a)).Closing the switch S results ina net motion of electrons fromthe negative terminal to thepositive terminal.

Motion of charges isrepresented as flow ofconventional electric currentconventional electric currentfrom +ve terminal to negativeas shown in Fig. 3.2 (b). Fig. 3.2 (a) flashlight (simple

electric circuit) & (b) direction

EPF 0024: Physics II 7

electric circuit) & (b) direction of current & electron flow.

Page 8: Lecture Notes 3

The electric current I is defined as the netamount of charge that passes through a givencross-section of a conductor per unit time:

(3.1) tQI = Constant current

tQIΔΔ

= Variable current

The SI unit of I is coulomb per second (C/s)and is known as the ampere (symbol: A)

EPF 0024: Physics II 8

and is known as the ampere (symbol: A).

Page 9: Lecture Notes 3

ExampleExample

A steady current of 2.5 A flows in a wire for4 0 minutes (a) How much charge pass4.0 minutes. (a) How much charge passthrough any point in the circuit. (b) How manyelectrons would this beelectrons would this be.

EPF 0024: Physics II 9

Page 10: Lecture Notes 3

SolutionSolution

(a) ( )( ) C 600s 604C/s 5.2 =×== Itq

(b)( )C600

= neq( )

( ) electrons. 108.3C101.6

C600 2119 ×=

×== −e

qn

EPF 0024: Physics II 10

Page 11: Lecture Notes 3

3 5 Ohm’s Law3.5 Ohm s Law

States: The current flowingth h d t i di tl

I as a function of V is astraight line through theorigin (Fig 3 3)through a conductor is directly

proportional to the potentialdifference applied to its ends. RV

I 1slope =ΔΔ

=

I (A)

origin (Fig. 3.3).

IR VRVI == or (3.2)

I (A)

ΔI

Where the proportionalityconstant R is called resistance( it h (Ω)) V (V)

ΔV

(units = ohm (Ω)).

Fig. 3.3: I versus V for

V (V)

EPF 0024: Physics II 11

Circuit symbol for Rg

conductors

Page 12: Lecture Notes 3

Resistors are an indispensable part of allelectronic components (Fig. 3.4).

EPF 0024: Physics II 12

Fig. 3.4 Electronic components

Page 13: Lecture Notes 3

Table 3.1 shows the convention to determinethe value of a resistor using color codes

Table 3.1: Color code for resistorsColor 1st digit 2nd digit Multiplier Tolerance (%)

the value of a resistor using color codes.

Color 1 digit 2 digit Multiplier Tolerance (%)Black 0 0 1Brown 1 1 10Red 2 2 102Red 2 2 102

Orange 3 3 103

Yellow 4 4 104

G 5 5 105Green 5 5 105

Blue 6 6 106

Violet 7 7 107

8Grey 8 8 108

White 9 9 109

Gold 0.1 5

EPF 0024: Physics II 13

Silver 0.01 10No color 20

Page 14: Lecture Notes 3

Fig. 3.5 is an example indicating the decodingof the actual value of a resistor

Using the color code in Table 3.1 the resistor

of the actual value of a resistor.

1st digit (red)

gvalue is determined to be 25 kΩ ± 10%.

1st digit (red)

2nd digit (green)

Tolerance (silver)

Multiplier (orange)

( )

EPF 0024: Physics II 14Fig. 3.5: Decoding of actual resistor value

Page 15: Lecture Notes 3

Example 1p

A small flashlight bulb draws 300 mA from its1 5 V b tt ( ) Wh t i th i t f th1.5 V battery. (a) What is the resistance of thebulb? (b) If the voltage is dropped to 1.2 V,how would the current change?how would the current change?

EPF 0024: Physics II 15

Page 16: Lecture Notes 3

SolutionSolution

(a) Applying Ohm’s law we find:(a) Applying Ohm s law we find:

( ) Ω05V 5.1===

VR

(b) If the voltage drops to 1 2 V assuming the

( ) Ω0.5A 0.3

===I

R

(b) If the voltage drops to 1.2 V, assuming theresistance stayed constant, then

( )( )( ) mA. 60 of drop aor A 24.0

Ω 5.0V2.1

===RVI

EPF 0024: Physics II 16

Page 17: Lecture Notes 3

3 6 Resistivity3.6 Resistivity

For conductors:

(3 3)LRLR (3.3)

Wh i i ti it L l th d A

AρR

AR =∝ or

Where ρ is resistivity, L length and A cross-sectional area. From equation (3.3) wededuce the SI unit of resistivity to be Ω.m.deduce the SI unit of resistivity to be Ω.m.

EPF 0024: Physics II 17

Page 18: Lecture Notes 3

ExampleExample

Given that the resistivity of a copper wire is1 7 × 10−8 Ω m Find (a) the diameter of a 20 m1.7 × 10 8 Ω.m. Find (a) the diameter of a 20-mcircular wire if the resistance of the wire is0.10 Ω. (b) What is the voltage drop across( ) g p

the wire if the current flowing through the wireis 12 A.

EPF 0024: Physics II 18

Page 19: Lecture Notes 3

Solution

(a) 4==

ρρ LLR( )

4

2

=

==

ρπ

RLd

dAR

( )( )( )( ) mm 2.1m101.2

Ω0 10m 20Ω.m107.14 3

8

=×=×

= −−

πR

( )Ω0.10π

(b) Using Ohm’s

( )( )Ω 10.0A 12 == IRV

EPF 0024: Physics II 19

V 1.2 =

Page 20: Lecture Notes 3

Today's lecture Include:Today s lecture Include:

Temperature Effect on Resistance.

SuperconductivitySuperconductivity.

EPF 0024: Physics II 20

Page 21: Lecture Notes 3

Objectivesj

E plain the effect of increasing temperat re onExplain the effect of increasing temperature onresistance.

Explain superconducting effect.

Consider some applications related to theseconceptsconcepts.

EPF 0024: Physics II 21

Page 22: Lecture Notes 3

3 7 Temperature Effect on Resistance3.7 Temperature Effect on Resistance

Resistivity of metals increases linearly withResistivity of metals increases linearly withincreasing temperature (for moderatetemperatures of up to 300oC) according to:

)1( ToT Δ+= αρρ (3.4)

where ρo is the resistivity at 0oC and α istemperature coefficient of resistivity. The rapidvibration of atoms at higher temperaturesvibration of atoms at higher temperaturescauses a conductor to have higher resistance.

EPF 0024: Physics II 22

Page 23: Lecture Notes 3

3.7.1 Resistance Thermometer

Variation of R with temperature is used forpprecise temperature measurement. If at 0°C Rfor Pt is 164.2 Ω. When placed in a solution, Rincreases to 187.4 Ω. What is the temperature ofthe solution if α for copper is 3.927 × 10−3 (oC)−1?

Since R0 = ρ0L/A , we can write equation 3.4 as( )

( ) C35.9Ω 2.164Ω 4.187C0 ,1

o1

0

0

=−

=−

=

=−=ΔΔ+=RRT

TTTTRR oα

EPF 0024: Physics II 23

( ) ( )C35.9

Ω 2.164C103.927 1o30 ×

−−RT

α

Page 24: Lecture Notes 3

3 7 2 Superconductivity3.7.2 Superconductivity

At low temperatures the resistance of certainpmetals and their alloys drop to zero. The effectis termed superconductivity and materialsexhibiting the phenomenon are calledsuperconductors.

It was first observed by Onnes in 1911, whenmercury was cooled down to below 4 2 K Inmercury was cooled down to below 4.2 K. Ingeneral materials become superconductingwithin a few degrees of absolute zero

EPF 0024: Physics II 24

within a few degrees of absolute zero.

Page 25: Lecture Notes 3

Resistivity of superconductors is practicallyy p p yzero. Current in a ring-shaped superconducting coil has been observed to flow for

f ffyears in the absence of a potential difference.

Earlier, the highest temperature at whichsuperconductivity is achieved was 23 K andso requires liquid hydrogen cooling Currentlyso requires liquid hydrogen cooling. Currentlysome alloys have been developed that canbe superconducting at 90 K requiring cooling

EPF 0024: Physics II 25

be superconducting at 90 K requiring coolingin boiling liquid nitrogen.

Page 26: Lecture Notes 3

Advantages of superconductivity :

(i) Smaller electric motors and generators.Electric cars will be practical.

(ii) Less power lost on transmission lines anduse of thinner wires feasible (cost saving).

(iii) Faster computer and more efficient highspeed train levitation.

EPF 0024: Physics II 26