lecture notes 2.pdf

2. Electric Potential Energy, Potential and Capacitance

EPF 0024 PHY II 1

OutlineOutline

2.1 Electric Potential Energy and the Electric Potential2.2 Energy Conservation2.3 The Electron Volt (eV)2.4 Electric Potential of a Point Charge2 5 E i t ti l Li d S f2.5 Equipotential Lines and Surfaces 2.6 Capacitor2 7 Dielectric2.7 Dielectric2.8 Storage of Electrical Energy2.9 Electric Energy and Power in Current Electricity

EPF 0024 PHY II By Dr. John Ojur Dennis 2

gy y

Topics for today's lecture:Topics for today s lecture:

Electric Potential Energy and the Electric Potential.

Energy Conservation.

The Electron Volt (eV).


Objectives of this lecture:Objectives of this lecture:

Explain the concepts of Electric Potential Energy,Electric Potential and Electric Potential Difference.

Discuss energy conservation in electric fields

Introduce a new unit of energy called the Electron Volt(eV) and its relation to the joule (J).

Solve Problems related to these concepts.


2.1 Electric Potential Energy and the Electric Potential

In Physics I the concepts of work ,W, & energy,U, have been introduced in a mechanical context.In this section these concepts are used in theIn this section these concepts are used in thecontext of electric fields.

A charged particle moves in an electric field asa result of the field doing work on it. A chargedparticle in an electric field therefore has electricparticle in an electric field therefore has electricpotential energy , U.


Fig. 2.1: Electric & Gravitational Potential Energy

Fi 2 1 h if A t t h i l dFig. 2.1 shows a uniform E. A +ve test charge q0 is placedat A and experiences a downward electric force ofmagnitude F = q0E which displaces it to point B.

Electric field

+ ++ +Gravitational field

mA

+ q0 A

yA

AA mgyU =AA EyqU 0=

y

x

BE B

yB

yA

BB mgyU =BB EyqU 0=g

F = mgF = q0E

yB


Ground level = zero potential

2.1.1 Electric Potential Energy

The work done WAB by the electric force F:

2.1.1 Electric Potential Energy

The work done WAB by the electric force F:

While the change in potential energy U is -ve:( )( ) ( )( ) yEqyFyyFW ABAB === 0 (2.1)

While the change in potential energy U is ve:

yEqEyqEyqUUU ABAB 000 === (2.2) Therefore

UWAB = (2.3)Where U is set to zero is arbitrary. The absolute

U of a test charge q0 is set to zero with respect toh h th i fi it l f t


a charge q when they are infinitely far apart.

2.1.2 Electric Potential2.1.2 Electric Potential

Electric potential energy U is associated withelectric potential that is known as voltage V inelectric circuits. The electric potential V isdefined asdefined as

UV =or

0q (2.4)VqU 0=

qVU =

V is a scalar quantity and has the SI unit of thej l / l b lt b l V

q0


joule/coulomb = volt, symbol V.

2.1.3 The Electron Volt (eV)2.1.3 The Electron Volt (eV)

W i ti (2 4) th t th l t iWe see in equation (2.4) that the electricpotential energy is given by .qVU =

This equation suggests, at the atomic level, aconvenient and commonly used unit of energyy gycalled the Electron Volt (eV) defined as

( )( ) ( )( ) J10601V1C10601V1e1eV1 1919 === (2 5)( )( ) ( )( ) J10601V1C10601V1e1eV 1 === .. (2.5)


2.2 Electric potential difference (p.d.)

The electric p.d. V between the

2.2 Electric potential difference (p.d.)

ptwo points yA and yB in Fig. 2.1 is:

EyEqWUV AB 0 (2 5)

+ ++ ++ q0 A AA EyqU 0=

From which we obtain the

yEq

yqqq

V AB 0

0

00

==== (2.5)E B

yA

BB EyqU 0=From which we obtain theimportant relation between E andV, that is

VF = q0E

yB

yVE = (2.6)

(The SI Unit for E can therefore


(The SI Unit for E can therefore also be volts/meter, V/m).

2.3 Energy Conservationgy

Gravitational potential energy U of a mass mdropped in gravitational field decreases whiledropped in gravitational field decreases whileits kinetic energy K increases. The total energy,however, remains the same (i.e. energy ishowever, remains the same (i.e. energy isconserved).

The same applies to electric fields, i.e.:BBAA UKUK +=+

( )BBAA

BBAA

VVqUUvvm

UmvUmvUKUK

+=+++

)( 221

2212

21 (2.7)


( )BABAAB VVqUUvvm == )(21

Example 1Example 1

The work done by the electric force as a testcharge (q0 = + 2.0 106 C) moves from A to B isW + 5 0 10 5 J ( ) Fi d U U UWAB = + 5.0 105 J. (a) Find U = UB UAbetween these points. (b) Determine V = VB VA.


SolutionSolution

(a) WU(a)

J100.5

5=

= ABWU

(b) = UV

V25J100.5

50

==

qV

V25C102.0

6 ==


Example 2p

Find U as a charge of (a) +2.20 106 C or(b) 1.10 106 C moves from a point A to B,(b) 1.10 10 C moves from a point A to B,given that V = VB VA = +24.0 V (positiveimplies B at higher potential compared to A).


SolutionSolution

(a) ( )( )V 0.24102.20

6==

CVqU

J105.28 5=

(b)( )( )V02410101

6==

CVqU

( )( )J1064.2

V 0.241010.1 5=

= C


Example 3 yExample 3x

The work done by the electricforce as a test charge(q =+2 0 106 C) moves from A

++++

A +q0(q0=+2.0 10 C) moves from Ato B is WAB = +5.0 105 J. (a)Determine U = UBUA. (b) Find

( )

F = q0E

V = VB VA. (c) What ismagnitude of E if distancebetween A and B is 0 25 m?

EB

between A and B is 0.25 m? _ ___


Solution

(a) J100.5 5=== ABAB WUUU

Solution

(a) J100.5 ABAB WUUU

(b)

== UVVV AB( )

V 25C102 0J100.5 6

50

== qAB

C102.0 6

(c) ( )( ) V/m100V 25 ==== AB VVVE(c) ( ) V/m100m 25.0 AB yyyE

Magnitude of E is 100 V/m and the -ve


Magnitude of E is 100 V/m and the ve indicates that E points in ve y direction

Example 4Example 4

A uniform electric field with a magnitude of6250 N/C points in the positive x direction.6250 N/C points in the positive x direction.Find the change in electric potential energywhen a + 12.5 C charge is moved 5.50 cmin (a) the positive x direction, (b) the negativex direction, and (c) in the positive y direction.


SolutionSolution

xqEVqUVE

= xqEU(a)

xqEVqUx

E === ,

( )( )( ) mJ 30.4m 0550.0N/C 6250C105.12 6 ==

xqEU(a)

( )( )( ) mJ30.4m0550.0N/C6250C105.12 6 ===

xqEU(c)

( )( )( )(b) ( ) 00 === qExqEUEPF 0024 PHY II By Dr. John Ojur Dennis 19

(b) ( ) 00 === qExqEU

Example 5Example 5

When an ion accelerates through a potentialdifference of + 2850 V the change in itsdifference of + 2850 V the change in itselectric potential energy is 1.37 1015 J.What is the sign and magnitude of the chargeon the ion?


SolutionSolution

15

= VqU

C10814V 2850

J10371

19

15

== .

VUq

C10814 19= .

It is a negative charge with magnitude of 4.81 1019 C


Todays topics include:Today s topics include:

2.4 Electric Potential of a Point Charge

2.5 Equipotential Lines and Surfaces



To determine an expression for the potential dueto a point charge.p g

To explain the meaning of equipotential lines.p g q p

To solve problems related to these concepts.To solve problems related to these concepts.


2 4 Electric Potential of a Point Charge2.4 Electric Potential of a Point Charge

Fig 2.2 shows a point charge +q fixed at origin.g p g q gA +ve test charge +q0 is held at rest at A adistance rA from the origin.

The test charge experiencesl i f fa repulsive force of

magnitude given byrA

20

ArqqkF =


The work done by the electric field in moving theThe work done by the electric field in moving thetest charge +q0 from position rA to rB using integralcalculus is

1( )1

1. 20r

r

r

r

r

r

rAB

qqqqqqqq

drr

qkqdrrFdW

B

B

A

B

A

B

A

=== rF

(2.9)

F th b i th h i l t i

1 00000BAABr rqqk

rqqk

rqqk

rqqk

rqkq

A

=

+=

=

From the above expression the change in electricpotential energy of test charge is:

ABABBA

AB UUrqqk

rqqk

rqqk

rqqkWU ==

== 0000 (2.10)


Therefore U at any point for a point charge isgiven by (dropping the subscripts):

A d V i i b(2.11)

rkqqU 0=

And V is given by:AB kqkqUUUV === (2 12)

Finally, V at any point is given by (dropping theAB rrqq

V ===00

(2.12)

y y p g y ( pp gsubscripts:

kqV = (2.13)


r

The net potential at a point due to a group ofThe net potential at a point due to a group ofpoint charges can be calculated by summingpotentials. For n point charges, net potential isp p g , p

==++++= n in qkVVVVVV (2 14) == ==++++= i ii in rkVVVVVV 11321 (2.14)


Example 1Example 1

What minimum work is required by an externalforce to bring a charge q1 = +3.0 106 C from aforce to bring a charge q1 3.0 10 C from agreat distance away (take r = infinity) to a point0.5 m from a charge q2 = +20 106 C?


SolutionSolution

The work W required isThe work, W, required is

( )== AA UUUW ( ) 212121 =

=

AA

AA

rqqk

rqq

rqqk

( )( )( )m0 5

C1020C100.3/CNm1099.8 66

229 =

AA rrr

J 08.1 m0.5

=


Example 2Example 2

Calculate the electric potential at points A andp pB due to charges q1 and q2 as shown Fig.

yy

60 cmA B

30 cm40 cm

40 cm

q = +50 C q = 50 Cx26 cm 26 cm


q2 = +50 C q1 = 50 C

SolutionSolution

The potential at point A isThe potential at point A is

1212 qqkqkq

( ) C1005C1005 55A1

1

A2

2

A1

1

A2

2A1A2

+=+=+=

rq

rqk

rkq

rkqVVVA

( )V1057V10750V1051

m 6.0C100.5

m 3.0C100.5/CN.m1099.8

566

229

==

+=

V105.7V1075.0V105.1 ==


Solution (continued)Solution (continued)

At point B

B1

1

B2

2

B1

1

B2

2B1B2B

+=+=+=

rq

rqk

rkq

rkqVVV

( )m40

C100.5m40

C100.5/CN.m1099.8 55

229

B1B2B1B2

+=

rrrr

V 0 m4.0m 4.0

=

the potential is zero everywhere on the planeequidistant between the two charges. This planeis called an equipotential surface with V = 0


is called an equipotential surface with V 0.

Example 3

A charge q = 4.11 109 C is placed at the origin,d d h 2 i l d th iand a second charge 2q is placed on the x-axis

at x = 1.00 m. (a) Find V midway between the twocharges (b) V vanishes at some point P betweencharges. (b) V vanishes at some point P betweenthe charges; that is, for a value of x between 0and 1.00 m. Find this value of x.

+q 2q

+y

P

0

1 00 m

q

x

q+ x


1.00 m

SolutionSolution

(a) 21

= kqV(a)

( )( ) 21C10114/CN m10998 9229 =

=

BA rrkqV

( )( )V 9.73

m 50.0m 50.0C1011.4/CN.m1099.8

==

(b) Setting V = 0 we obtain( ) 03m 00.12 + xqkkqV ( )( ) ( )

103001

0m 00.1m 00.1

==+= xxxxV


m3

,03m 00.1 == xx

2.5 Equipotential Lines and surfaces2.5 Equipotential Lines and surfaces

The electric potential can be represented byequipotential lines or surfaces as shown by thegreen lines in Fig. 2.3.

EPF 0024 PHY II By Dr. John Ojur Dennis 35Fig. 2.3: Equipotential lines (green lines)

By definition, an equipotential line/surface isone in which all points on it are at the samepotential. That is to say, the potential difference(V) between any two points on it is zero(V) between any two points on it is zero.

E i t ti l li / f lEquipotential lines/surfaces are alwaysperpendicular to the electric field. Note that aconductor must be entirely at the sameconductor must be entirely at the samepotential in static case. The surface of aconductor is therefore an equipotential surface.


q p

Todays lecture include:Today s lecture include:

Capacitor

Dielectric

Storage of Electrical Energy

Electric Energy and Power in Current Electricity



To provide expressions for capacitance of ap p pcapacitor.

To explain dielectric.

To analyze energy stored in a capacitor andelectric energy in current electricity.

To solve problems related to these concepts.EPF 0024 PHY II By Dr. John Ojur Dennis 38

p p

2 6 Capacitor2.6 Capacitor

A capacitor consists of a pair of metal plates of area At d b ll di t h i Fi 2 5 fseparated by a small distance as shown in Fig. 2.5 for a

parallel- plate capacitor.

Circuit symbol


Fig. 2.5: A Parallel-Plate Capacitor

2 6 1 Charging a Parallel-Plate Capacitor2.6.1 Charging a Parallel-Plate Capacitor

When p d (voltage) V isWhen p.d. (voltage) V isapplied to a capacitor, it ischarged (Fig 2.6). Charge Q

i d b h l t iacquired by each plate isdirectly proportional to V

Q

where C is proportionalityVQ CCVQ == or (2.15)

where C is proportionalityconstant called capacitanceof the capacitor. SI unit C/V

f d (F)

Fig. 2.6: Charging a Parallel-Plate Capacitor


or farad (F).

2.6.2 Calculating the Capacitance2.6.2 Calculating the Capacitance

For a parallel-plate capacitorFor a parallel plate capacitorwith area A and separation d(Fig. 2.7), the capacitance is

l ti l t A dalso proportional to A andinversely proportional to d:

A

22120

0

/N.mC1085.8 ==

dAC (2.15)

The proportionality constantis found to have the value 0,th iti it f f

0

Fig. 2.7: A parallel-


the permitivity of free space. plate capacitor

2.6.3 Calculating the E in a capacitor

Using equ. 2.6 and taking thelt t l t t b Vvoltage at +ve plate to be V

and at ve plate to equalzero (Fig. 2.8), we obtain anzero (Fig. 2.8), we obtain anexpression for E inside acapacitor thus:

xVE

= (2.16)

dV

dV

xxVV =

==

00

12

12

Fig 2 8: A parallel-


Fig. 2.8: A parallel-plate capacitor

Example 1Example 1

(a) Calculate the capacitance of a parallel-platecapacitor whose plates are 20 cm by 3 cm andare separated by a 1.0 mm air gap. (b) What isthe charge on each plate if the capacitor isconnected to a 12 V battery? (c) What is theconnected to a 12 V battery? (c) What is theelectric field between the plates?


SolutionSolution

AC(a)

m1006 232212

0

=

dAC (a)

pF53m100.1m100.6)/N.mC1085.8( 3

2212 =

=

( )( ) 1012( )( ) C104.6V12F1053 1012 === CVQ(b)V/m102.1

m100.1V12 4

3 === dVE(c)


2 7 Dielectric2.7 Dielectric

In most capacitors, an insulating sheet called aIn most capacitors, an insulating sheet called adielectric is placed between the plates. Thisincreases the capacitance of the capacitor.

Consider a parallel-plate capacitor whose platest d b i If th it iare separated by an air gap. If the capacitor is

isolated (i.e. not connected to a battery) andcarries a charge Q and has voltage V across itscarries a charge Q and has voltage V0 across itsterminals, the capacitance will be given byC0= Q/V0. Let the electric field be E0.


0 Q 0 0

A dielectric is placed in between the platesbecomes polarized as shown in Fig. 2.8.


Fig. 2.9: Dielectric effect on E inside a capacitor.

There will be net -ve charge on outer edge of theg gdielectric facing +ve plate, and net +ve chargefacing the -ve plate. An electric field is set upinside the dielectric opposite to original field Einside the dielectric opposite to original field E0and reduces it by a factor say. Net E inside thedielectric is now E = E0/ .0

Hence the voltage must decreased by a factorHence the voltage must decreased by a factor also, i.e. V = V0/ , since V = Ed.


The charge Q on plates remain constantThe charge Q on plates remain constant(remember capacitor is isolated). Therefore from

VVQC 0d VVQC 0

find we

and ==(2.17)

dA

CVQC 00

0

===Implies decrease in E by a factor will increase Cby the same factor .


Capacitance for parallel plate capacitor withdielectric between plates is therefore given by

AAC (2 18)

where = is called the permittivity of thedd

C == 0 (2.18)where = is called the permittivity of the

material.0

is known as dielectric constant, a factorindicating by how many fold capacitance isincreased Table 2 1 shows the dielectric


increased. Table 2.1 shows the dielectricconstants of some substances.

Table 2 1: at 20C for different substancesTable 2.1: at 20 C for different substances

Material Material Vacuum 1.0000Air (1 atm) 1.0006P ffi 2 2Paraffin 2.2Rubber, hard 2.8Vinyl (plastic) 2.8 4.5Paper 3 7Quartz 4.3Glass 4 7Porcelain 6 8Mica 7Ethyl alcohol 24


Ethyl alcohol 24Water 80

ExampleExample

A parallel-plate capacitor has plates of area0.0280 m2 and separation 0.550 mm. The spaceb t th l t i fill d ith di l t i fbetween the plates is filled with a dielectric ofdielectric constant . When the capacitor isconnected to a 12 0-V battery each of the platesconnected to a 12.0-V battery, each of the plateshas a charge of magnitude 3.62 108C. What isthe value of the dielectric constant ?


SolutionSolution

First we determine C thus:First we determine C thus:

( ) F10023C1062.3 98 === QCN t fi d

F1002.3V 0.12

===V

C

Next we find : 0= d

AC

( )( )( )( ) 70.602800/NC10858 m10550.0F1002.3 22212319

0

===

ACd

d


( )( )m0280.0/N.mC1085.8 222120 A

2 8 Storage of Electric Energy2.8 Storage of Electric Energy

A Charged capacitor stores electric energy equal to workd t h it Th lt V th it idone to charge it. The voltage V across the capacitor isdirectly proportional to the charge q already accumulatedand increases from 0 to V0 linearly .(Fig. 2.10).

V (V)V0

021 V

021 Q Q (C)0 Q0


Fig. 2.10: Voltage across a capacitor vs. charge accumulated

V (V)V0

The average voltage02

1 V

The average voltage(from Fig. 2.10)

0V +02

1 Q Q (C)0 Q0

0210

20 VVVav =+= (2.19)

The average energy stored in a capacitor

002

1000 2

VQVQVQU avav =

== (2.20)


Dropping the subscript 0 in eq.2.20 and usingpp g p q g

f

and 21 CQVQVU ==

we obtain alternative expressions for energystored in a capacitor as

2

or(2.21)C

QQVU2

21

21 ==

or

221

21 CVQVU ==


Energy stored in a capacitor is stored in theelectric field between the plates. Using V=Edand C=o (A/d)

A

Ad is volume Therefore we may define the( ) AdEEd

dACVU 202

1202

1221 =

== (2.21)

Ad is volume. Therefore we may define theenergy density u as energy per unit volume:

21 EUu (2 22)and for a Capacitor with dielectric

021 E

Adu == (2.22)


202

1221 EEu == (2.23)

Example 1Example 1

What is the potential difference between theplates of a 3.3-F capacitor that stores

ffi i t t t 75 W li ht b lbsufficient energy to operate a 75-W light bulbfor one minute.


SolutionSolution

21 2= CVU

22 ===

CPt

CUV

PtU

( )( )( ) V 52F3 3

s 60 W752 ==CC

F3.3


Example 2Stopped here

Example 2

A parallel-plate capacitor has plates with area ofA parallel plate capacitor has plates with area of1.2 cm2 and separation 0.88 mm. The spacebetween the plates is filled with dielectric ofdielectric constant 2.0. (a) What is the potentialdifference between the plates when the charge

l t i 4 7 C (b) Will t ton plates is 4.7C (b) Will your answer to part(a) increase, decrease, or stay the same if thedielectric constant is increased? Explain (c)dielectric constant is increased? Explain. (c)Calculate the potential difference for the casewhere the dielectric constant is 4.0.


SolutionSolution

QdQ(a)

( )( ) V1091m1088.0C107.4 6360

==

==

AQd

CQV

(a)

( )( )( )( ) V109.1m 10.21N.mC108.852.0 242212 == (b) ill d b i i i l(b) V will decrease because it is inversely

proportional to .

(c)( )( )( )( ) V107.9m1021/N mC108 854 0 m1088.0C107.4 4242212

36

==

V


( )( )m10.21/N.mC108.854.0

Example 3Example 3

Find the electric energy density between theplates of a 225-F parallel-plate capacitor.plates of a 225 F parallel plate capacitor.The potential difference between the plates is315 V, and the plate separation is 0.200 mm.


SolutionSolution

202

1= Eu

( )2

20

2 =

dV

( )( )( )23

2NmC12

m10200.02

V 3151085.8 2

2

=

3J/m 11.0 =


2.9 Energy and Power in Current Electricitygy y

Electric Energy can be easily transformed intoElectric Energy can be easily transformed intoother forms of energy.

E.g. Electric motors transform electric energyinto mechanical energy, Heaters transforminto mechanical energy, Heaters transformelectric energy into heat energy, while lightbulbs transform electric energy into light energy.


For a small amount of charge Q moved acrossa p. d. V, U changes by the amount

( )VQU = (2 24)Power P is the time rate of change of energy,

and therefore

( )Q (2.24)and therefore

( ) IVtVQ

tUP =

== (2.25)

SI unit of power is the watt (W).tt


Eq. 2.25 is power transformed by a deviceEq. 2.25 is power transformed by a devicewhere I is the current passing through it and Vis the potential difference across it

and also gives the power delivered by ag p ysource such as a cell or battery.


2 9 1 Alternative expressions for power

2.9.1 Alternative expressions for power

For conductors, using V = IR, alternativeexpressions for power are obtained as follows:

( ) RIIRIIVP 2=== (2.26)

orVVVIVP

2

(2 27)R

VR

IVP === (2.27)


2 9 2 Fire Hazard from Electrical Appliances2.9.2 Fire Hazard from Electrical Appliances

Electrical appliances have small resistance ifElectrical appliances have small resistance. ifthe current through electric wires is large, thewire heats up and produce thermal energy at ap p gyrate which equals to I2R and may start a fire.

To prevent overload, a fuse or circuit breaker isused. These are switches that open the circuitwhen current exceeds some particular value(Fig 2.10).


Fig. 2.10: (a) fuse (b) circuit breakerFig. 2.10: (a) fuse (b) circuit breaker

(a) Fuses


(b) Circuit breaker

Fig. 2.11: Connection of appliances to fuseg pp


ExampleExample

Calculate the resistance of 40 W automobilehead-light designed for 12 V.


SolutionSolution

( )V12 22V ( ) 6.3

W40V12 22 ===

PVR


2 9 3 Paying Electrical Bills

2.9.3 Paying Electrical Bills

Electric bills are paid based on total electricenergy U used. From U = P t we find analternative unit for energy that isalternative unit for energy, that is,1 joule (J) = 1 watt second (Ws).

Larger units such a kilowatt-hour (kWh) maybe used for energy 1kWh = 1000 W 3600 sbe used for energy. 1kWh 1000 W 3600 s= 3.6 106 Ws = 3.6 106 J.


Example 1Example 1

An electric heater draws 15.0 A on a 240 Vline How much energy does it use and howline. How much energy does it use and howmuch does it cost per month (30 days) if itoperates 3 hours per day and the rate is 10.5p p ycents per kWh. Assume the current flowssteadily in one direction.


SolutionSolution

= IVP

( )( ) kWh 324 Wh10324d 30 h/d 3V 240A 15 3 ===== IVtPtU

34.02 $ $/kWh 0.105 kWh 324Cost ==


Example 2Example 2

A 75-W light bulb operates on a p. d. of 95 V.Find (a) the current in the bulb and (b) theresistance of the bulb. (c) If this bulb is replacedwith one whose resistance is half the valuef d i t (b) i it ti tfound in part (b), is its power rating greater orless than 75-W? By what factor?


SolutionSolution

75 W 0.79 A95 V

PIV

= = =(a)

2 2(95 V) 120VR = = = (b) 120 75 W

RP

= = = (b)

(c) greater by a factor of 2


Example 3Example 3

Find the power dissipated in a 25- electrich t t d t 120 V tl theater connected to a 120-V outlet.


SolutionSolution

2

=R

VP

( ) W58025V 120

2

==R

25


Example 4Example 4

The current in a 120-V reading lamp is 2.3 A.If the cost of electrical energy is $ 0.075 perIf the cost of electrical energy is $ 0.075 perkilowatt-hour, how much does it cost tooperate the light for an hour?


SolutionSolution

( )( )( )hr 1V 120A 3.2 === IVtPtU

kWh 276.0 Wh276 ==

( )( )021.0$

kWh$ 075.0kWh 276.0Cost==


lecture notes 2.pdf

Documents