Unit 2 Topic 4

Lokman awad 2013/2014 Section 1: Electric Charges There are two types of electric charge. 1) Positive charges 2) Negative charges Innormalconditions,anatomisuncharged becausethenumberofelectrons(negative charges)equaltothenumberof protons(positive charges) When a balloon is rubbed against your hair, some of yourhairselectronsaretransferredtotheballoon. Thus, the balloon gains a certain amount of negative chargeandbecomenegativewhileyourhairloses an equal amount of negative charge and hence is left with a positive charge. 1) Like charges repel ; Unlike charges attract 2) The force (f) between charges is called electrostatic force Properties of charged body Electric Current The current in the wire is like the water in the pipe . The amount of water that flows depends on the flow rate and the time. Its the same with electricity- current is the rate of flow of charges Q:charge measured in coulombs(C) I: current measured in Ampere(A) Onecoulomb(C)isdefinedastheamountofcharges that passes in 1 second when the current is 1ampere Remember: Convectional current flows from + to -, the opposite way from electron flow QItA=ANote Q: electric charge measured in C coulomb n : number of electrons e:charge of each electron C The charge of a single electron is 1.6x10-19 C Q = n x e Example 1) How much charge flows through the filament of an electriclampin1hourwhenthecurrentinitis 250mA? 2) Theelectronchargeis1.6x10-19C.Howmany electrons flow through the filament during this time Answer: t=1h = 3600s I= 250 m A = 250 x 10-3 A Q = I x tQ = 250 x 10-3x3600=900 C -191921 900 1.6 109001.6 105.6 10Q n ennn electrons= = == Current in series circuits The current can be measured by connecting an ammeter in series with the component. Ammeters have a low resistance so that they do not affect the current that they are measuring. Inseries the current is the same in each component(rate at which electrons leave any component must be the same) Potential Difference Potentialdifference(voltage)istheamountof electricalenergyconvertedintootherforms(thermal, light, mechanical) of energy when unit charge passes from one point to another int arg sin done differenceargpassingelectrical energy converted o other formsPotential differenceCh e pas gWorkPotentialCh eWVQ===The unit for (p.d) is joules per coulomb(J C-1) Thisunitiscalledvolt.Thus,ap.dof1Vexistsb/w twopointsinacircuitif1J ofenergyisconverted when 1C of charge passes b/w the points Example Theelementofanelectrickettlethattakesacurrent of12.5Aproduces540kJofthermalenergyinthree minutes a) Howmuchchargepassesthroughtheelementin these three minutes b) What is the potential difference across the ends of the elementAnswer: a) Q = I x t = 12.5 x 3 x 60 = 2250 C b) 3540 102402250WV VQ= = =Example A12Vpumpforafountaininagardenpondcan pumpwateruptoaheightof0.8matarateof 4.8liters per minute 1) Howmuchworkdoesthepumpdoperminute whenraisingthewatertoaheightof0.8m?(lliter of water has a mass of 1 kg) 2) Ifthepumpif75%efficient,howmuchcharge pass through the pump motor in one minute? 3) What current does the motor take when operating under these conditions Answer W= m g h= 4.8 x 9.81 x 0.8 =37.6J

37.63.13 (100%)122.34WVQWQ CIf efficientVQ C== = ==(2.34)39.160QI mAt= = =Using Voltmeter Thep.db/wtwopointsismeasuredbyconnectinga voltmeter in parallel with the component Thecurrentthroughthelampis(I-i)where(i)isthe current through the voltmeter. In order to keep (i) as small as possible, voltmeter should have very high resistance Electromotive force A cell does work on charges just as pump on water. Chemical energy is converted to electrical energy in acell,sothecellissaidtoproducean electromotive force Electromotiveforce(emf)ofanelectricalsourceis definedastheenergy(chemical,mechanical) converted to electrical energy when unit charge passes through the source. int. .arg sin doneargpassingenergy converted o electrical energye m fCh e pas gWorkCh eWQcc=== The unit of e.m.f is the same as p.d (J C-1) or(V) Ohms Law Experiments showed that: Formetalsatconstanttemp.,thecurrentinthe metalisdirectlyproportionaltothepotential difference across it Anyelectriccomponentforwhichthecurrentis proportional to the voltage is said to be Ohmic Resistance The resistance (R) of an electric component is a measure of its opposition to an electric current flowing in it. Resistance is given by the equation: The unit of resistance is ohmVRI=Example What is the potential difference across a 22k resistor when the current in it is 50A? Answer: V=RI = 22x 103 x 50x10-6 = 1.1VNote R=Gradient 1RGradient=Power Power is defined as the rate of doing work The unit for electrical power is watts(W) WPtQVPtQP V I VtA=A== = Electrical energy WPtW P tW I VtA=AA = AA = APower dissipated in a resistor Electric power in electric component is: P= IV Ohms law: Rearrange the equation ; V=RI P = I2 R VRI=2VPR=Example ShowthattheelectricpowerasdefinedbyP=IVare consistentwiththoseformechanicalpower,defined as the rate of doing work -1;QI Cst=-1;JWV sQ=Example An electric filament lamp is rated at 240V, 60W When its operating under these conditions, what is: a) The current in the filament b) The resistance of the filament Answer: 0.25A ;960 Conservation of energy in circuit V1V2V3 Energy converted by battery = energy dissipated in the 3 resistors 1 2 31 2 31 2 3Q QV QV QVV V VutIR IR IRIRcccc= + += + += + += EBExample The fig shown in previous slide is left on for 2min. a) Calculate how much electrical energy is converted in the cell b) How much energy is dissipated in each resistor Answer: Charge flowing: Q= I x t = 2 x 120 = 240 C Energy converted by cell 240 24 5.76 W Q kJ c = = =2 212 222 23.(2) (2) 120 0.96.(2) (4) 120 1.92.(2) (6) 120 2.88Energy diss in R I Rt kJEnergy diss in R I Rt kJEnergy diss in R I Rt kJ= = == = == = =Internal energy Unfortunately,notallofthechemicalenergy converted to electrical energy. Someoftheenergyisusedtopushthecharges though the cell inotherword,itsusedtoovercometheinternal resistanceofthecell,whichisusuallygiventhe symbol r Rate of energyconverted in the cell = Rate of work done against internal resistance + Rate of work done lightening lamp 2 2I I r I RIr IRIr VV Ircccc= += += += An alternative rearrangement of the energy equation to find the current IR rc=+Example Atorchbatteryofe.m.f4.5Vandinternalresistance of 0.435 is connected across a lamp of resistance of 6.4 a) What is the current in the lamp b) How much power is : i. Dissipated in the lamp ii.Wasted in the cellAnswer: 4.50.666.4 0.4IR rI Ac=+= =+For power dissipated in lamp: P = I2 R=(0.66)2 x 6.4 = 2.8W For power wasted in cell: P = I2 r=(0.66)2 x 0.4 = 0.2W Example Inthecircuitbelow,thehighresistancevoltmeter reads1.55Vwhentheswitchisopenand1.49when the switch is closed 1)Explain why: a) Thee.m.fofthecellcanbeconsideredtobe 1.55V b) The voltmeter drops when the switch is closed 2) Calculate the internal resistance of the battery Answer: a) Asthevoltmeterhasaveryhighresistance,it takes virtually no current. Therefore with the switch isopen,thereisnegligiblecurrentacrossthecell, so 1.55V can be taken as the e.m.f (open switch I = 0) b) Whentheswitchisclosedthe10isbroughtinto thecircuit.Thiscausesa current , I, inthecircuit. So the p.d across the cell drops to 2) r= 0.4 V Ir c = Finding the e.m.f and internal resistance of the cell Figbelowshowsastandardcircuittofindinternal resistance and emf of a cell. Startingwiththevariableresistor(rheostat)atits highest value(to minimize any heating effect) Recordthecurrentinthecellandthepotential difference across the terminals Rearrange the equation: V IrV Ircc= = + Ifagraphisplotted,wewouldexpecttogeta straightlineofgradient randintercepteonthe y-axis Inpractice,thelinemaynotbestraightline becausetheinternalresistancemaynotbe constant( for large current) Example A typical set of observation is recorded inthe table below a) Plot a graph of this data b) Calculate the e.m.f and the internal resistance Answer: e.m.f = 1.56V r= 0.6I/A0.200.400.600.801.001.201.401.60 v/V1.441.321.201.091.950.840.730.59 Note Allpowersupplieshaveinternalresistance,for examplethecarbatterymusthavesmall resistance(0.01)becausethestartermotortakes a very large current about 200A Ontheotherhand(E.H.T)supply,forsafety reasonshasaveryhighinternal resistance(50M).thislimitsthecurrenttofraction of mill-amp. Example A 12V car battery has an internal resistance of 0.01 a) Whatisthep.dacrosstheterminalswhenthe engine is started if the started motor takes 200A b) Explainwhy,ifthedriverhastheheadlightson, theyarelikelytogetdimwhenhestartsupthe engine c) Calculatehowmuchpowerthebatterydeliversto the starter motor Answer: a) 10 V b) Theheadlightsoperateatfullbrightnesswhenthe p.dis12v.ifthisisreducedto10Vwhenthe engine is started, the headlight will go dim c) 2000W Resistors connected in Series I I I R1 R2R3 +++ + + - --- - 1) Thecurrent,I,passinginalltheresistorsisthe same( the reading of both ammeters are equal) 2) TheP.dacrosseachresistordependsonthevalue ofeachresistor,thusV1,V2&V3havedifferent values.ButthesumofallP.dsmustequalthetotal P.d across the battery 3) Theeffectiveresistanceofallresistorsequalsto their sum, Note:theseriesconnectionincreasetheeffective resistor thus reduce the current in the circuit Vt = V1 + V2 + V3

Rt = R1 + R2 + R3

Example 1 a) Mark with letter (I) the current drawn from the cell, and with letter (e) the path of electrons b) Connect ammeter to the circuit above c) Calculate the effective resistance d) Calculate the current in the circuit e) Calculate the voltage(p.d) across each lamp Solution c) Rt = R1 + R2 = 30 + 60 = 90 d) Using ; f) Across lamp 1 : V1 = I x R1 = 0.1 x 30 = 3 V Across lamp 2 : V2 = I x R2 = 0.1 x 60 = 6 V 90.190VI AR= = =Example A7.0resistorisconnectedinserieswithanother resistorand a 4.5 V battery. Thecurrent in the circuit is0.60A.Calculatethevalueoftheunknown resistance. Resistors connected in Parallel I I1 I2 R1 R2 + + - - 1) TheP.dacrosseachresistoristhesameasthatof the battery 2) Thetotalcurrent,I,throughthebatteryisdivided between R1 & R2

3) The effective resistance in parallel is: Note:theparallelconnectionreducetheeffective resistor thus increase the current in the circuit I =I1 + I2

1 2 31 1 1 1tR R R R= + +Example An 18.0 , 9.00 , and 6.00 resistor are connected inparalleltoanemfsource.Acurrentof4.00Aisin the 9.00 resistor. a. Calculate the equivalent resistance of the circuit. b. What is the potential difference across the source? c. Calculate the current in the other resistors. Example Find the current flowing in each resistor ? Solution a) 6 & 3 are connected in parallel Rt = R1 + Rt1 = 4+2 = 6 Current in the battery: 2 312 32tR RRR R= = O+1226tVI AR= = =Current Iacross 4 & 2 : 2AAcross 4 : So ; V =R x I = 4 x 2= 8 V Across 2 : V = R x I = 2 x 2 = 4 V Voltage across 6 & 3 : 4 V Across 6 :Across 3 : 40.676VI AR= = =41.333VI AR= = =Example For this circuit(internal resistance is neglected), calculate: 1) The current in each resistor 2) The power dissipated in each resistor 3) The power developed by the battery a) The two resistors are (in parallel) so the p.d across each is the same as the battery, so; b) We can calculate the power dissipated: 111222632623VI ARVI AR= = == = =P1 = I2 R1=(3)2 x 2 = 18W P2 = I2 R2=(2)2 x 3 = 12W c) The power developed by the battery is given by: P=IV Where I = I1 +I2 = 5A P = 5x6 = 30W Power developed by the battery = power dissipated in the two resistors Redo the question if the internal resistance of the battery r=0.8 1 21 1 1 1 10.832 311.20.83tR R RR= + = + = O= = ORt = R + r =1.2+0.8 = 2 We can now calculate the current I: The potential difference across the resistors: 632tI ARc= = =6 (3 0.8) 3.6V IrV Vc = = =Hence: 1112223.61.823.61.23VI ARVI AR= = == = =P1 = I2 R1=(1.8)2 x 2 =6.5W P2 = I2 R2=(1.2)2 x 3 = 4.3W P = 6.5 + 4.3 = 10.8W The power developed by the battery: Theother7.2Wisthepowerwastedinthe battery(theworkdonepersecondbychargeto overcome the internal resistance) P=I2 r = (3)2 x 0.8 = 7.2W 3 6 18 P I W c = = =The Resistance of a Metal Wire , R i. R is directly proportional to the length , L. ii. R is inversely proportional to cross sectional area , A. iii. R depends on the resistivity of the substance, . The unit of resistivity is ohm m (m) LRA=R R RR RR LA radius 1A1radiusTemp0 C Example Usethedatafromthegraphbelowtocalculatethe resistivity of nichrome (diameter:0.2743mm) Answer:1.08x10-6 m Length(m) R() 18.3 1m I-V relationships Varying V & I in a circuit Variable Resistor(Rheostat): ismadeoflongwirecoilandaslidermovingon along a thick bar and touching the wire. The current enters from terminal A and leaves from terminal B MovingthesliderawayfromterminalA(where currententers)theresistanceincrease,andvise versa Rheostat can be used to control the currentOR give a continuously variable p.d (in this case its said to be potential divider) I-V characteristics for metallic conductor 1) Cut just over a meter length nichrome wire and tape it to a meter ruler as shown below. 2) Set up a potential divider to provide a variable p.d and connect a 1m length of the wire to the circuit by mean of crocodile clips. 3) Setthep.dtoitsminimumvalue(rheostatslider near the negative contact) and switch on 4) Graduallyincrease the p.d andrecordthevalues of p.d.; V, andcurrent, I,up tomaximumpotential difference possible 5) Tabulate your values of V and I. 6) Repeat the experiment but with terminals of the cell reversed(negativevaluesaredisplacedonthe meters) 7) Plot your data on (I-V) graph. Note(its conventional to plot I(on the y-axis) against V(on the x-axis) Results: With cell terminal reversed V/V00.511.522.53 I/mA0285483108138163 V/V0-0.5-1-1.5-2-2.5-3 I/mA0-26-56-81-110-136-165 I-V for a tungsten filament lamp 1) Set up the circuit as shown in the fig. below 2) Set the p.d to its minimum value(rheostat slider near the negative contact) and switch on 3) Graduallyincrease the p.d andrecordthevalues of p.d.; V, andcurrent, I,up tomaximumpotential difference (12V) 4) Tabulate your values of V and I and switch off. 5) Plot your data on (I-V) graph 6) After at least five minutes have elapsed, reverse the powersupplyconnectionsandrepeatthe experiment. Add this data to your graph. Diode Adiodeisanelectricaldeviceallowingcurrentto movethroughitinonedirectionwithfargreater ease than in the other Semiconductor diode is symbolized as shown Whenplacedinasimplebattery-lampcircuit, the diode will either allow or prevent current through thelamp,dependingonthepolarityoftheapplied voltage (a)Currentflowis permitted;thediodeis forward biased.(b) Current flow is prohibited; the diode is reversed biased. I-V characteristic semiconductor diode 1) Set up the circuit as shown 2) Becarfultoobservethepolarityofthediode(The diodeissaidtobeforwardbiasedwhen connected this way around) 3) Set a 10 resistor to limit thecurrent in the diode and prevent it being damaged 4) Switchonandveryslowlyincreasethep.d, observe what happens to the current when you do this. 5) Record the values of I and V 6) Reverse the power supply connections so that the diodeisreversebiasedandrepeatthe experiment 7) Plot your data on I-V graph Results: With cell terminal reversed V/V00.40.560.640.670.710.730.74 I/mA001510203040 V/V0-0.2-0.4-0.6-0.8-1-1.2-1.4 I/mA00000000 Example Explainwhythegraphshowsthatthecomponent obeysOhmslawandcalculatetheresistanceofthe component Answer: Asthegraphisastraightlinethroughtheorigin,the currentinthecomponentisproportionaltothe potential difference 3650120 10VRI= = = OExample Calculate the resistance of the component in the fig. below when the current is 80mA Answer: Thiscomponentdoesnotobeyohmslaw,andits resistanceisnotconstant,wecannotusethe gradient of the graph. Simply we use30.81080 10VRI= = = ODrift Velocity Formetallicconductors,thechargeiscarriedby delocalized or free electron . Theseelectronsarenotfixedandfreetomove randomlyathighspeedsaroundtheatomic lattice,(similar to gas particles in the atmosphere) Whenapotentialdifferenceisappliedacrossthe conductor,anelectricfieldissetupinsidethe conductor. Theforceduetothatfieldsetstheelectronsin motion, thereby creating a current. Theseelectronsdonotmoveinstraightlines Instead,theyundergoazigzagpatterndueto collision with metal atoms Drift velocity is the net(average) velocity of a charge carrier moving in an electric field Note: Thedriftvelocityismuchmuchlessthanthe electron's actual speed (106 ms-1) The current is given by the expression: I = n q v A Where: I: electric current in (A) n: number of charge carrier per m3(m-3) A: cross-sectional area in (m2)v: drift velocity (ms-1) q: charge carried by each charge carrier (C) Example Atungstenfilamentlampcarriesacurrentof 240mA.The diameter of the filament is 0.024 mm and that for copper connecting leads is 0.8mm Calculate the drift velocity of the electrons ina) The filamentb) The connecting lead Youmaytakethenumberoffreeelectronsperunit volume to be 8x1028 m-3 for copper and 4x1028 m-3 for tungsten 3128 19 3 235 128 19 3 2240 100.083(4 10 ) (1.6 10 ) (0.012 10 )240 103.7 10(8 10 ) (1.6 10 ) (0.4 10 )IvnqAv msv mstt == = = = Metals, Semiconductors and Insulators The value n determines the ability of a given material to conduct electric current 1) Metals have a value of n in the order of 1028 1029 m-3

2) Insulators (glass) have no charge carriers at room temp. 3) Semiconductorssuchasgermanium(n=1019m-3) andsilicon(n=1017m-3)areabletoconductbut not as well as metals Effect of temp. on Metals 1) Chargeiscarriedthroughmetalsbyfree electrons in the lattice of positive ions 2) Heatingupametaldoesnotaffecthowmany electrons there are, but it does make it harder for them to move about. 3) Theionsvibratemorewhenheated,sothe electrons collide with themmore 4) Theresistanceofmetallicconductorsgoesup linearly with temp. In terms of drift velocity equation: I=nAvq A and q are constant for a given wire. Formetallicconductor.ndoesnotdependon temp.(n is also constant) Asthetemp.risesthedriftvelocity(v)willbe reduced, so the current will decrease So the resistance increases with temp. Effect of temp. an Semiconductors Insemiconductors,anincreaseintemp.can provideextraenergytoreleasemorecharge carriers. Thismeansthatnincreaseswithtemp.(rapid increase in n as temp. rises) A and q are constant Decrease in drift velocity The increase in (n) is much more than the decrease in (v) The over all effect is that current (I) increases Resistance will decrease QuantityMetallic conductor Semiconductor nconstantIncrease AConstantConstant vDecreaseDecrease qConstantConstant IDecreaseIncrease RIncreaseDecrease 1) Thermistors: As the current (and hence the temp) increases, the resistance decrease. Rdecreaseastemp.rises.Theriseintemp. makesmorefreeelectronsavailabletocarrythe current. 2) The light Dependent Resistor:(LDR) Its resistance depends on the brightness of the light, so its called light Dependent Resistor (day light) R decrease (Dark) R increase Note: Metalsaresaidtobe(PTC)positivetemperature coefficient Semiconductors:aresaidtobe(NTC)negative temperature coefficient Potential divider(Potentiometer) 12Vbatteryprovides12Vonly.However,weoften wishtouseonlyapartofthisvoltage,soweusea potentialdividercircuitasshowninthefollowing figures. B C A R1 =6 R2 =18 12V VAB VBC VAB = IR1 =0.5 x 6 = 3V VBC = IR2 =0.5 x 18 = 9V The network of resistors has divided the p.d of 12V to 3V and 9V across both resistors. Such an arrangement is called potential divider 120.524VI AR= = =Example Calculate the p.d across the 6 resistor in Fig 1. Calculate the p.d and currentacross the lamp in Fig 2. B C A R1 =12 R2 =6 7.5V B C A R1 =12 R2 =6 7.5V 6 V=2.5V I=0.5 A ;VBC = 3x0.5 V Thisexampleshowshowaddingaloadtothe output can affect the output voltage Wecannowformulateageneralexpressionfor potential divider by considering this circuit R1

R2

V V1 V2 If no current is taken by the output: 221 222 1 2221 2VIRVIR RV VR R RRV VR R==+=+= +Using thermistor /LDR to control voltage Applicationspotentiometersincludeheatorlight control switches. If the thermistor or LDR is connected in series with aresistor,theoutputvoltageacrosstheresistor can be controlled by temp of light variation. Example Theresistanceofthethermistoris2Kat200Cand 400Kat 500 C .Calculate the output voltage at each of the temperatures. Answer: 1.8V 5.0V