engineering physics 11 electromagnetism

8/12/2019 Engineering Physics 11 Electromagnetism

1/57

Prof. Guna MagesanRoom No B301 Extn: 1010

[email protected]

Engineering Physics

BEN503

Week 12


2/57

Engineering PhysicsBEN503

ElectromagnetismSl no Topic

1 Electric Charge & Electric Field

2 Electric Charge and Electric Flux3 Current, Resistance & Electromotive Force: Ohms Law

4 Capacitance and Dielectrics

5 Direct Current CircuitsKirchoffsVoltage & Current Laws

6 Alternating Current: RLC Series Circuit, Reactance, Impedance, Phase Angle

(Lead and Lag), Resonance


3/57

Electromagnetism

In general, when a force acts on an object it is incontact with the object.

There are exceptions where the force can act on

an object without being in contact. Gravity

Electricity

Magnetism

Electric forces occur between two objectsbecause each have a special character called

charge.


4/57

Electric Charge

Electric chargeis an intrinsic characteristic of thefundamental particles making up those objects.

There were two types of charge: positive and negative.

How was the concept of electric charge discovered?

History: Earliest experiments on charge involved simplestatic attraction and repulsion effects with dielectric rods

rubbed on cloth or fur. For example, If you rub a glass rod with silk, a positive charge appears on

the rod.

Negative charge of equal magnitude appears on the silk.

Now both objects have become charged.


5/57

Electric Charge

Rubbing does not create charge but only transfers it fromone body to another upsetting the electrical neutrality of each body during the

process.

Electric charge is conserved: the net charge of any isolatedsystem cannot change.

The most basic charges are the protonand electron.

Proton have positive chargeand electrons have the same

amount of negative charge.

Chargeis measured in Coulombs, abbreviated C. The charge of a proton is 1.6 x 10-19C

The charge of an electron is -1.6 x 10-19C


6/57

Electric Charge

Smallest unit of charge, e=1.6 x 10-19C, is theamount of charge on a proton or electron.

It is a constant of nature and it is called the elementarycharge.

Electric charge is quantized: any charge can bewritten as ne, where nis a positive or negativeinteger and eis elementary charge.

Any two charges create an electric force on eachother. Two charges of same sign repeleach other.

Two charges of opposite sign attracteach other.


7/57

Electric Charge -

Applications The attraction and repulsion between charged

bodies have many industrial applications: Electrostatic paint spraying and powder coating

Fly-ash collection in chimneys

Nonimpact ink-jet printing, and Photocopying


8/57

Conductors & Insulators

Materials fall into different categories:

Conductors: materials in which charge can movefreely (metals, water with dissolved impurities).

Insulators: materials in which charge cannot movefreely (rubber, plastic, pure water).

Semiconductors: materials that are intermediatebetween conductors and insulators (silicon andgermanium in computer chips).

Superconductors: materials that areperfectconductors.


9/57

Coulombs Law If two charged particles are brought near each other,

they each exert a force on the other. This force ofrepulsion or attraction due to the charge propertiesof objects is called an electrostatic force.

Coulombs lawdescribes the electrostatic forcebetween small (point) electric charges q1and q2atrest (or nearly at rest) and separated by a distance r.

The constant k= 8.99x109N.m2/C2 = 1/4peo eo= 8.85x10-12C2 /N. m2is the permittivity constant

221

221

4

1

r

qq

r

qqkF

ope


10/57

Coulombs Law

The force of attraction or repulsionbetweenpoint charges at rest acts along the linejoiningthe two charges.

If more than two charges are present, CoulombsLaw equation holds for each pairof charges.

The net force on each charge is then found, using

the superposition principle, as the vector sum ofthe forces exerted on the charge by all theothers.

We will solve a problem on this topic in the next tutorial!


11/57

Electric Field

A single charge (or a collection of charges)creates an electric field in its surrounding space.

A point in space has an electric field, and a

charge at that point experiences a force from theelectric field.

The Electric fieldis defined as the force per unit

charge i.e. the electric field at a point in space is equal to

the force that a 1 Coulomb charge (a unit charge)would feel at that point.


12/57

Electric Field Chargescreate electric field, and electric field creates

forces on charges.

Q E F

Each of these steps produces a vector, either field or

force. That means, an electric fieldhas a magnitude

and a direction at every point.

The electric field vector points awayfrom a positive

charge and towardsa negativecharge.

The electric field at a point due to a collection ofcharges is the vector sumof the electric field created by

every charge.


13/57


14/57

Field Line representation Maxwell came up with the field line representationas a

way to visualizethe electric field.

The field linesshow the directionof the electric field andthe density of linesis related to the strengthof the field.

http://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html
http://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.htmlhttp://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.htmlhttp://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.htmlhttp://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.htmlhttp://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.htmlhttp://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.htmlhttp://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.htmlhttp://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.htmlhttp://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.htmlhttp://phet.colorado.edu/sims/charges-and-fields/charges-and-fields_en.html


15/57

Field Line representation Electric field linesprovides a means for visualizing

the directionand magnitudeof electric fields.

The electric field vector at any point is tangentto afield line through that point.

The density of field linesin any region is proportionalto the magnitudeof the electric field in that region.

Electric field lines start at positivecharges and end at

negativecharges. Field lines cannot cross!

The field line model is a useful representation rather

than a quantitative method.


16/57


17/57

Electric Flux

We define Ato be a vector having amagnitude equal to the area of thesurface, in a direction normal to thesurface.

The amount of surfaceperpendicular to the electric field isA cos

Therefore, the amount of surfacearea effectively cut through by the

electric field is A cos

AEffective= A cos so E= EAEffective= EA cos . EE A

E

A


18/57

Gauss Law Gauss lawrelates the net flux of an electric fieldthrough a

closed surface (a Gaussian surface) to the net chargeqencthatis enclosed by that surface.

The net charge qencis the algebraic sum

of all the enclosed positive and negativecharges, and it can be positive, negative,or zero.

If qencis positive, the net flux is outward;

if qencis negative, the net flux is inward


19/57

Capacitance

A capacitoris a device that stores charge. A capacitor consists of two conductorsseparated by an

insulator.

The simplest capacitor consists of a pair of parallel metal

plates.

When a capacitor is charged, its plates have charges ofequal magnitudes but opposite signs: +qand -q.

However, we refer to the charge of a capacitor as being q,the absolute value of these charges on the plates.

When charge is moved from one plate to the other a

voltage developsbetween the plates.


20/57


21/57

Capacitance The charge qand the potential differenceVfor a

capacitor are proportional to each other.

q = CV

The proportionality constant Cis called thecapacitance

of the capacitor. Its value depends only on the geometryof the conductors (e.g.

area of plates and their separation) and noton their charge orpotential difference.

The capacitance is a measure of how much charge must be put

on the plates to produce a certain potential difference betweenthem.

The greater the capacitance, the more charge is required.

The units of capacitance are coulomb per voltwhich is

named the farad(1 F = 1 C/V).


22/57


23/57

Charging a Capacitor

Plate h, losing electrons, becomes positively charged. Plate l, gaining electrons, becomes negatively charged.

As the plates become oppositely charged, that potentialdifference increases until it equals the potential difference V

between the terminals of the battery.

With the electric field zero, there is no further drive ofelectrons.

The capacitor is then said to be fully charged, with a potentialdifference Vand charge q.


24/57

Capacitors in Parallel When a potential difference Vis

applied across several capacitorsconnected in parallel, thatpotential difference Vis appliedacross each capacitor.

The total charge qstored on thecapacitors is the sum of thecharges stored on all thecapacitors.

Capacitors connected in parallelcan be replaced with anequivalent capacitorthat has thesame totalcharge qand the samepotential difference Vas the

actual capacitors.


25/57

Capacitors in Parallel To derive an expression for Ceq, we first use q= CV

to find the charge on each actual capacitor.


26/57

Capacitors in Series When a potential difference Vis

applied across several capacitorsconnected inseries, the capacitorshave identical charge q.

The sum of the potential differencesacross all the capacitors is equal to theapplied potential difference V.

Capacitors that are connected in series

can be replaced with an equivalentcapacitorthat has the same charge qand the same total potentialdifference Vas the actual series

capacitors.


27/57

Capacitors in Series To derive an expression for Ceq, we first use V = q/Cto

find the potential difference of each actual capacitor.

h l


28/57

Capacitor with a Dielectric

A dielectric, is an insulating material such asmineral oil or plastic, and is characterized bya numerical factor k, called the dielectricconstant of the material.

In a region completely filled by a dielectricmaterial, all electrostatic equationscontaining the permittivity constant oareto be modified by replacing o with ko.

The introduction of a dielectric To increase the capacitance of a capacitor

To limit the potential difference that can beapplied between the plates to a certain valueVmax, calledthe breakdown potential.


29/57

Current

An electric current iin a conductor is defined by:

where

dQis the amount of (positive) charge that passes in time dt

through a hypothetical surface that cuts across a conductor.

SI unitof current is the Ampere(= Coulomb/second) = 1 C/s

dt

dQi

Electric Currentis the rate of flow of chargepast a point in

space (e.g. in a wire or a beam of charged particles).

Typically given the symbol ior I.


30/57

Current

Heres a really simple circuit where

a current arrow is drawn:

The current is in the direction of flow of positive charge

opposite to the flow of electrons, which are usually thecharge carriers.

+-

current

+ -

current electrons

An electronflowing fromto +gives rise to the same

conventional current as a protonflowing from + to.

Currentis a scalarquantity, and it

has a sign associated with it.

i


31/57

Resistance

A perfect conductor lets charge flow freely whereasan insulator will not let charge flow at all.

In practice, materials and their geometry (e.g. thinwire versus thick wire), offer some impediment to theflow of charge. This property is known as resistance.

In other words, the resistanceof a material is ameasure of how easily a charge flows through it.

A perfect conductorwould have zero resistance.

A perfect insulatorwould have infinite resistance.

R i & Oh L


32/57

Resistance & Ohms Law

The resistanceRof a conductor is defined as

where Vis the potential difference across the

conductor and iis the current.

This relation is known as Ohms law.

The difference in voltage (or potential)between twosides of a resistor is equal to the currentthrough theresistor timesthe resistanceof the resistor.

The SI unitof resistance is the Ohm ()(1 volt perampere).

i

VR

iRV

R i & R i i i


33/57

If we write the equation as

we see that, for a given V, the greater the resistance, thesmaller the current.

Resistor & Resistivity A conductor whose function in a circuit is to provide a

specified resistanceis called a resistor.

In a circuit diagram, we represent a resistor and aresistance with the symbol

R

Vi

The resistance Rof a conducting wire of length Landuniform cross sectional areaAis

where is the resistivityof the wire.

A

LR

P


34/57

Power & Resistive Dissipation

Power, or the rate of energy used in the resistor, is

The units of power is volts times amps = watts

(1 V) (1 A) = (1 J/C) (1 C/s) = 1 J/s = 1 W

iVP

Resistive Dissipation:

For a resistor or some other device with resistance

R, we can combine R= V/iand P= iVto obtain, forthe rate of electrical energy dissipationdue to aresistance, either

RiP 2R

VP

2

El i F


35/57

Electromotive Force

Any device which transforms a form of energyinto electric energy is called a source of emf.

emf is an abbreviation for electromotiveforce, but emfis not a force!

That term no longer popular.

The emfdevice of initial interest is the battery.a dc-

power supply.

El t ti F


36/57

Electromotive Force

An emfdevice does work on charges to maintaina potential energy difference between itsterminals.

the emf(defined as the work done per unit charge)

The SI unit of emfis volt, (as for potentialdifference)

emf= work/charge (Joule/Coulomb = Volt).

dQ

dWmfe

Di t C t Ci it


37/57

Direct Current Circuits

Circuit Symbols The resistor

The battery (positive terminal long line,negative short line). Current always flowsfrom positive to negative.

The ground connection

Resistances in Series & Parallel


38/57


A B

Put your finger on the wire at A. If you can move along the

wires to B without ever having a choice of which wire to follow,

the circuit components are connected in series.

A B

Put your finger on the wire at A. If in moving along the wires to B

you ever have a choice of which wire to follow, the circuit

components are connected in parallel.



39/57


+ -

V

parallel

+- V

series

It matters where you put the source of emf.

V

I

If resistors see the same potential difference, they are in parallel.

If resistors see the same current, they are in series.



40/57


V

R1

R2

VR1

R2

If possible see if the resistors can be combined using theparallel and series combination rules.

Simplifying the circuit by replacing parallel and seriesresistors by their equivalent values can allow the circuit

parameters to be determined.

21 RRReq 21

111

RRReq

parallelseries

R i t i S i


41/57

Resistance in Series

Current flows

in the steady state, the same current flows through all resistors

there is a potential difference (voltage drop) across each resistor.

An electric charge qis given a potential energy qVby the battery.

As it moves through the circuit, the charge loses potential energy qV1asit passes through R1, etc.

The charge ends up where it started, so the total energy lostmust equal

the initial potential energy input: qV= qV1+ qV2+ qV3.

R3R2R1

+ -

VI

III

V1 V3V2

Heres a circuit with

three resistors and a

battery:


42/57

R i t i P ll l


43/57

Resistance in Parallel

Current flows different currentsflows through

different resistors

but the voltage dropacross eachresistor is the same.

In the steady state, the current Isplits into I1, I2, and I3at point A.

I1, I2, and I3recombine to make acurrent I at point B.

The net current flowing out of A andinto B is I = I1+ I2+ I3.

V

V

V

R3

R2

R1

+ -

VI

I3

I1

I2

A B

I

R i t i P ll l


44/57

Resistance in Parallel

Now imagine replacing the threeresistors by a single resistor, having aresistance R such that it draws thesame current as the three resistors inparallel.

From above, I = I1+ I2+ I3, and

Because the voltage drop across eachresistor is V:

1 2 3

1 2 3

V V VI = I = I =R R R

V

Req

+ -

VI

I

A B

I

So thateq 1 2 3

V V V V= + + .

R R R R

Ki h ff C t L


45/57

KirchoffsCurrent Law(KEERKOFFsJunction Rule)

At any junction point, the sum of all currentsentering the junction must equal the sum of allcurrents leaving the junction.

Also called Kirchhoffs First Rule.

This is just conservation of charge: charge in =charge out.

at any junctionI=0

Kirchoffs Voltage Law


46/57

KirchoffsVoltage Law(KEERKOFFs Loop Rule)

The sum of the changes of potential around any closedpath of a circuit must be zero.

Also called Kirchhoffs Second Rule.

This is just conservation of energy: a charge ending up whereit started out neither gains nor loses energy (Ei= Ef ).

around any closed loopV=0

For a resistor, the sign of the potential difference is negativeif yourchosen loop directionis the sameas the chosen current directionthrough that resistor; positive if opposite.

For a battery, the sign of the potential difference is positiveif yourchosen loop directionmoves from the negative terminal towards the

positive; negative if opposite.


47/57

Solving Circuits


48/57

Solving Circuits

Apply the junction rule.

Apply the loop potential rule around all the loops ofthe circuit.

Each loop will give one equation relating thepotentials and the loop currents.

Use simultaneous equations to solve for the

quantity of interest.

Resistor: I

loop

V is -

+

-Battery:

loop

V is +

Solving Circuits


49/57

Solving Circuits

We have 3 unknowns (I1, I2, and I3), so we will need 3 equations. We

begin with the junctions.

Junction a: I3I1I2= 0 --eq. 1

Junction d: -I3+ I1+ I2= 0

Junction d gave no new information, so we still need two more equations.

1 We1= 85 V

1 W e2= 45 V

20 W

40 W

ab c

d

eg f

h

I3

I2

I1

30 W

da

Solving Circuits


50/57

Solving Circuits

There are three loops.

Any two loops will produce independent equations. Using

the third loop will provide no new information.

Loop 1. Loop 2. Loop 3.

1 We1= 85 V

1 W e2= 45 V

20 W

40 W

ab c

d

eg f

h

I3

I2

I1

30 W

Solving Circuits


51/57

Solving Circuits

Three equations, three unknowns; the rest is algebra.

Make sure to use voltages in V and resistances in W. Then currents will be in A.

5

I

loop

V is - + -

loop

V is +

The green loop (a-h-d-c-b-a):

(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0

- 30 I1+ 45 - 41 I3= 0 --eq. 2

The blue loop (a-b-c-d-e-f-g):

(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0

41 I3 -130 + 21 I2= 0 --eq. 3

Solving Circuits


52/57

Solving Circuits Collect our three equations:

I3I1I2= 0 - 30 I1+ 4541 I3= 0

41 I3 -130 + 21 I2= 0

Rearrange to get variables in right order:

I1I2+ I3= 0 - 30 I1- 41 I3+ 45 = 0

21 I2+ 41 I3130 = 0

Use the middle equation to eliminate I1:

I1= (41 I345)/(-30)

There are many valid sets of steps to solving a system of equations.Any that works is acceptable.

Final answers

I1= - 0.94 AI2= 2.72 A

I3= 1.78 A

Verification

1.78(-0.94)2.72 = 0

- 30 (-0.94) + 45 - 41 (1.78) = 0.22

- 41 (1.78) -130 + 21 (2.72) = 0.10

Solving Circuits


53/57

Solving Circuits

38

8

10

6

1 9 V

38

8

10

6

1 9 V

83

8

10

6

1 9 V

8 3

8

10

6

1 9 V

Solving Circuits


54/57

Solving Circuits

38

8

10

6

1 9 V

3

10

6

1 9 V

4

3

10

1 9 V

10

3

1 9 V

5

Example:Two 100 Wlight bulbs are connected (a) in series and (b) in


55/57

parallel to a 24 V battery. (i) What is the current through each bulb and

(ii) what is the equivalent resistance of each circuit? (iii) For which circuit

will the bulbs be brighter?

(a) Series combination.

Req= R1+ R2

V = I Req

V = I (R1+ R2)

I = V / (R1+ R2) = 24 V / (100 W+ 100 W) = 0.12 A

The same current of 0.12 A flows through each bulb.

The equivalent resistance is Req= R1+ R2

Req= 100 W+ 100 W= 200 .

R2R1

+ -

IV = 24 V

Example:Two 100 Wlight bulbs are connected (a) in series and (b) in

( )


56/57

parallel to a 24 V battery. (i) What is the current through each bulb and

(ii) what is the equivalent resistance of each circuit? (iii) For which circuit

will the bulbs be brighter?

(b) Parallel combination.V

V

R2

R1

+ -

V = 24 VI

I2

I1

I

eq 1 2

1 1 1 = +

R R R

1 2

IV =

1 1+R R

1 2

1 1I= V +R R

1 1I= 24V +

100 100

200I= 24 = 0.48 A10000

V = I Req

2

eq

1 1 1 200 = + =

R 100 100 10000 eqR = 50

(iii) Which one is brighter?


57/57

(iii) Which one is brighter? To answer the question, we must calculate the power dissipated in the

bulbs for each circuit. The more power consumed, the brighter thebulb.

In both circuits, the bulbs are identical and have identical currentspassing through them (for a given circuit). We pick either bulb for thecalculation.

Series circuit:

We know the resistance andcurrent through each bulb, sowe use:

P = I2R

P = (0.12 A)2 (100 W)

P = 1.44 W

Parallel circuit:

We know the resistance andvoltage drop across each bulb,so we use:

P = V2/ R

P = (24 V)2 / ( 100 W)

P = 5.76 W

Compare:

Pseries= 1.44 W; Pparallel= 5.76 W

engineering physics 11 electromagnetism

Documents