Presentation on theme: "Chapter 15 Electric Circuits"— Presentation transcript:
1 Chapter 15 Electric Circuits Physics Beyond 2000Chapter 15Electric Circuits
2 Electric currentAn electric current is a flow of electric charges through a conductor.Current = rate of flow of chargesConstant currentChanging currentunit: ampere A
3 Definition of ampere and coulomb Definition of 1 Ampere of current:Two infinitely longparallel wires 1 m apart3. The forces onthe wires are2 × 10-7N m-1FF1m4. The current ineach wire is1 ampere2. The same amount ofcurrents are in the wireI = 1AI = 1A
4 Definition of ampere and coulomb Definition of 1 coulomb (C):Q = I.t1 C = The amount of charges flowing through in 1 second with a current of 1 A.I = 1At = 1 s
5 Charge-carriers The charge-carriers move and form the current. The type of charges carried by a charge-carrier may either be negative or positive depending on the material.
6 Charge-carriers Substance charge-carrier Type of charge Metal Free electronsNegative chargeElectrolytePositive ionsPositive chargesnegative ionsNegative chargesGas discharge tubeThermionic electrons
7 Charge-carriersCharge-carriers in electrolyte: positive and negative ions.+-positive ionnegative ionnegativeelectrodepositiveelectrodewater
8 Electric field in a circuit The switch is open.The p.d. of AB is zero.Not any current to flow.1. Cell with emf ξ2. Switch SAB3. Resistor R
9 Electric field in a circuit Close the switch S.An electric field is established inside the circuit.ξSRABE
10 Electric field in a circuit The electric field exerts force on the charge-carriers(electrons). Charge-carriers are accelerated.ξSRABE
11 Electric field in a circuit The charge-carriers are also retarded by the collisionwith atoms.ξSABRE
12 Electric field in a circuit As a result, the charge-carriers are moving atconstant speed.ξSRABE
13 Drift velocityCharge-carriers in a metal move with constant speed under an electric field.The constant speed is called the drift velocity vD.E = electric field strength-vD
14 Drift velocityConsider a conductor with n charge-carriers per unit volume.E = electric field strength-vD
15 Drift velocityIn time Δt , all the charge-carriers in the shaded region have passed the imaginary cross-section.cross-sectionalarea AvD. ΔtE = electric field strength-vD
16 Drift velocityWhat is the number of charge-carriers ΔN in the shaded area?cross-sectionalarea AΔN = nA.(vD.Δt)vD. ΔtE = electric field strength-vD
17 Drift velocitySuppose each charge-carrier has charge q. What is the amount of charges ΔQ passing the cross-sectional area in time Δt?cross-sectionalarea AΔQ = q.ΔN = q.nA.(vD. Δt)vD. ΔtE = electric field strength-vD
18 Drift velocity What is the current I in the conductor? cross-sectional area AvD. Δtelectric field strength-vD
19 Drift velocitySo the drift velocity of the charge-carriers in a conductor iscross-sectionalarea AvD. Δtelectric field strength-vD
20 Drift velocity In small cross-sectional area, charge-carriers So the drift velocity of the conductor isIn small cross-sectional area, charge-carrierswould move faster if other quantities are the same.In semi-conductor, n is small. The drift velocityis fast.
21 Drift velocity In metal, the charge-carriers are electrons So the drift velocity of the conductor isIn metal, the charge-carriers are electronswhich are in random motion at very fast speed.This random speed does not contribute tothe current.
22 Example 1Drift velocity of the charge-carriers in a copper wire. (10-5 ms-1)
23 Signal and drift velocity The drift velocity of electrons in a circuit is very small.The electric signal travels at a very fast speed in the circuit. It is close to the speed of light.
24 Electromotive force1. Suppose that the charge-carriers are positively charged.2. The charge-carriers gain energy from the batterywhen they flow from the low potential to high potentialinside the battery.4. The charge-carrierslose all energy whenthey return to thebattery.3. The charge-carrierslose energy when theymove from high potentialto low potential in thecircuit.
25 Electromotive force1. Suppose that the charge-carriers are positively charged.2. The battery does work on the charge-carriers in the battery.The energy transferred to one coulomb of charge is calledthe electromotive force of the battery.
26 Electromotive force The electromotive force ξof a source is the energy transferred into electricalenergy per unit charge within the source.unit: J C-1 or V, voltwhere Q is the amount of chargethrough the battery andU is the total energy provided bythe battery to the charge Q.
27 Electromotive force The electromotive force ξof a source exists though there is not any current.
28 Example 2 The e.m.f. (electromotive force) of an AA cell. Ah (ampere-hour) is a unit for charge.Q = I.t
29 Combination of cellsCells in seriesCells in parallel
30 Cells in seriesξ= ξ1 + ξ2 +ξ3Produce a high e.m.f.ξξ1ξ2ξ3
31 Cells in parallel Total e.m.f. = ξ It can supply a larger current. It lasts longer.It decreases the internal resistance of the source.ξξ
32 Potential difference (p.d.) The p.d. between two points in the circuit is the amount of energy change as one coulomb of charge passes from one point to another.Unit: V (volt)An amount of chargeQ passes from X to Y2. The charge loseselectrical energy UXY3.
33 Voltage Voltage is a general term. It may refer to the e.m.f. of a source or the p.d. between two points.Unit: V (volt)
34 Measurement of voltage Use voltmeterUse CROto voltmeter or CROto voltmeter or CROIXY
35 Example 3 First find the amount of charge of one mole of electrons. U = QV
36 Voltage and electric field strength A p.d. V betweenpoints X and Y.d2. The separationXY is d.XY4. The average electricfield strength is E = V/d.3. There is anelectric field Ebetween X and Y.
37 Example 4 Use E = V/d and F = e.E E d Note that the electrons move with uniform drift speed.
38 Resistance The p.d. between the ends of the conductor is V V I 2. A current I passesthrough the conductor3. The resistance ofthe conductor is R = V/I4. Unit of R is ohm, Ω
39 I-V graph Pure metal R = constant or V I (Ohm’s law) The slope gives V
40 I-V graph Bulb filament. The filament resistance R increases with temperature.Ohm’s law is not obeyed.IV
41 I-V graph Semi-conducting diode Current increases sharply in forward bias.Current is zero in reverse bias.IVcurrent startsto increase
42 Dynamic resistance RdFind the slope from the I-V graphUseIV
43 Dependence of resistance Depending on the physical dimension such as length ( ) and cross-sectional area (A).Depending on the material which is described by its resistivity (ρ).A
45 Effect of temperature (metals) Resistance increases uniformly with temperature for metals.θ/oCresistivity ρρoρ= ρo (1 + α.θ)αis called the temperaturecoefficient.
46 Effect of temperature (metals) An increase in temperature increases the thermal agitation of atoms.The interaction between atoms and electrons increases.The resistance increases.atomelectron
47 SuperconductivitySome metals or alloys lose the resistance at very low temperature.No dissipation of energy when current flows in a superconductor.
48 Internal resistance of battery When current flows inside a battery, energy is also lost.The battery has internal resistance.The e.m.f. of a battery > The terminal voltage of a battery.e.m.f. describesthe provision of power = e.m.f.r2. terminal voltagedescribes the lossof powerV= terminal voltage
49 To measure the internal resistance r of a battery 2. An ammeterto measure thecurrent I.3. A variable resistor Rto adjust the currentRAIIrA battery with emf andinternal resistance r
50 To measure the internal resistance r of a battery Express R in terms of , I and r.ArRI
51 To measure the internal resistance r of a battery Plot the graph of R-How to find r from the graph?ArRIR-r
52 The internal resistance r of a battery We assume that the internal resistance is a constant.ArRIR-r
53 Combination of resistors Resistors in seriesResistors in parallel
54 Combination of resistors Resistors in seriesA common current passing through all resistorsThe equivalent resistance R of the system isR = R1 + R2 + R3R1R2R3I
55 Combination of resistors Resistors in parallelSame voltage across each resistorThe equivalent resistor R givesR1R2R3V
56 Power and heating effect When a current I passes through a resistor of resistance R and the p.d. is V, some electric energy is converted into internal energy, the power output isIRV
57 Power of a cellIf the e.m.f. of a cell is ξand it provides a current I for an external circuit, then the power of the cell isrRI
58 Example 5Note that the internal resistor r is not counted in the power output.Vpower outputrRI
59 Power output and external resistance The power output Pout ispower outputpower outputPmaxRrrRI
60 Power output and external resistance The power output is maximum when R = r.power outputpower outputPmaxRrrRI
61 Power output and external resistance What is the efficiency when R = r?Hint: efficiency = Pout/Popower outputEfficiencyrRI100%50%Rr
63 Power transmissionHigh voltage is used for transmission of electric power.
64 Power transmissionWhat is the loss of power in the cable? (Use IL, the current in the cable, and R, the resistance of the cable.)
65 Domestic Electricity Live wire: with potential changing between positive andnegative3. Electrical devicesare connected inparallel.Same p.d. foreach device.mains2. Neutral wire: maintainingat zero potential by earthingin the power station.
66 Ring circuitliveneutralABC1. Same p.d.for each device.
67 Ring circuit live neutral 1. Same p.d. for each device. A C 2. If a part isbroken, can itbe operated?BYES, it can.
68 Combination of two cells in parallel with same e.m.f. A cell with emf andinternal resistance r13. Connect themin parallel2. A cell with emf andinternal resistance r24. Note that there is not any currentflow between the cells because theiremf are the same.
69 Combination of two cells in parallel with different e.m.f. A cell with emf 1 andinternal resistance r15. There is current flowingfrom the cell of high emfto that of low emf2. A cell with emf 2 andinternal resistance r24. Connect themin parallel3. 1 > 2
70 Example 6 (p.327) A cell with emf 14V and internal resistance 5. 5. There is current flowingfrom the cell of high emfto that of low emf. Find the current.2. A cell with emf 12V andinternal resistance 15 .4. Connect themin parallel3. 1(14V) > 2(12V)
71 Combination of two cells in parallel with same emf r1Two cells of emf areconnected in parallel.2. Both cells providecurrent I for the externalresistor R.I1I2Ir23. Set up equations forthese quantities.R
72 Combination of two cells in parallel with same emf r1I1 + I2 = I (1)I1I2r2IIR
73 Combination of two cells in parallel with same emf r1I1 + I2 = I (1)I1 = I.R + I1.r (2)IIR
74 Combination of two cells in parallel with same emf I1 + I2 = I (1) = I.R + I1.r (2)I2 = I.R + I2.r (3)r2IIR
75 Example 6 Given: = 1.5V, r1 = 2 , r2 = 4 and R = 6 . r1 I1 I2 Find the currentsand the power output.IIR
76 I-V characteristics Linear: Obey Ohm’s law. Ohmic device. Non-linear: Not obey Ohm’s law. Non-ohmic device.INon-ohmicOhmicIVVNote that R =
77 I-V characteristicsIRIOhmicVV= 3 Vr = 1 Find R if I = 1 A.
78 I-V characteristics Junction diode I/mA I + - 20 VD VD/V 0.8 VD is 0.8 V for a currentof 20 mA.If the current is too large, thediode would be damaged.
79 I-V characteristics Junction diode I/mA I R 0.8V 20 VD/V 0.8 = 3V, r = 0Find the suitable value of R.What would happen if R is too big or too small?
80 I-V characteristics Thermionic diode I/mA I 2 VD 4 VD/V 24IVDThe maximum current is 2.0 mA.This current is called the saturation current.
81 I-V characteristics Thermionic diode I = 2 mA I/mA R 2 VD 4 VD/V 24RFind the minimum value for R.=12V, r = 0
82 Kirchhoff’s rules 1. Kirchhoff's Junction Rule, 2. Kirchhoff's Loop Rule and3. Ohm's Law (i.e. V/I = R, where I is the current, V is the voltage, and R is the resistance).
83 Kirchhoff’s rules 1. Kirchhoff's Junction Rule The algebraic sum of the currents at any branch point or junction in a circuit is zero. I1I2I3I4I5Note that it is necessaryto add a minus sign beforeI4 and I5 because they areleaving the junction.I1 + I2 + I3 - I4 - I5 = 0
84 Kirchhoff’s rules 2. Kirchhoff's Loop Rule The algebraic sum of the potential changes around any complete loop in the network is zero. R3CDNote the the potentialdrops by IR when currentpassing a resistor. It is necessary to add a minus before I.R following the current.I3I2BER22R4I1AFR11
85 Kirchhoff’s rules 2. Kirchhoff's Loop Rule The algebraic sum of the potential changes around any complete loop in the network is zero. R3CDConsider the loop ABEF:I3I21-I1R1-I2R2 - 2-I2R4 = 0BER22R4I1AFR11
91 Measuring devices Potentiometer To measure small p.d.To measure the internal resistance of a cellTo calibrate a voltmeterTo measure resistanceAdvantage: It can measure the voltage of a device without drawing any current from the device.
92 PotentiometerAn accumulator provides a steady current for a long period of time.A resistance wire with uniform cross-section.1. accumulator Vo (2V)IAB+-2. A uniform resistancewire of length
93 To measure an unknown voltage (p.d.) I5. Both +ends areconnected4. A slide tocontact wire AB.AB+-+-3. A protectiveresistorVDevicewith p.d.2. A galvanometerto detect current throughthe device
94 To measure an unknown voltage (p.d.) BI+-VPVoMove theslide on ABto find abalance point.2. The galvanometerdoes not deflect when Pis the balance point.3.Show that
95 Using the protective resistor 1.Before the balance pointis found, the current passingthe galvanometer is large.Use the protective resistor todecrease the current.ABI+-VPVo2. When the balance point isnear, the current passing thegalvanometer is small. Shortthe protective resistor toincrease the current.
96 Calibrating a potentiometer Use a standard cell (e.g. Weston cell with emf = V at 20oC) to calibrate a potentiometer.ABI+-PVostandardcell
97 Calibrating a potentiometer Find the balance point P and measure the length AP.Find the voltage per unit length of the resistance wire.ABI+-PVostandardcellWhat is the voltageacross AP?2. So what is ?3. =
98 Calibrating a potentiometer Use this calibrated potentiometer to measure the p.d. V of a device.ABI+-P’VodeviceV1. V = .AP’
99 No balance point is found The terminals of the device are reversed.The p.d. of the device is too large. V > Vo.ABI+-P’VodeviceV
100 Measure the emf of a cell Find the balance point Pemf V = AP where is the voltage per unit length.3. Why is the result the emf?Why not the p.d.?ABI+-PVocell withemf VIt is because there isnot any current throughthe cell. The pd is equalto the emf.
102 Measuring very small p.d. AP would be too short if the measured V is very small because V = .AP.The error of measurement is too big.ABI+-PVodevice with verysmall V
103 Measuring very small p.d. Add a variable resistor R in series with the resistance wire. Adjust R so that VAB is just greater than V.Must re-calibrate the potentiometer when R is changed.ABI+-PVodevice with verysmall VR
104 Examples 9-10 The emf of a thermocouple is very small. Adjust R so that the balance point P is near B.ABI+-PVoRcold junctionhotjunctionV
105 Measure the internal resistance r of a cell BI+-P1VoSRr‘1. The cell2. r is the internal resistance3. R is a known resistor4. S is a switch
106 Measure the internal resistance r of a cell With switch Sopen, locate thebalance point P1.2. Measure lengthAP1.3. The length AP1represents the emf’ of the cell.ABI+-P1VoSRr‘4. There is notany current flowingthrough the cell
107 Measure the internal resistance r of a cell With switch Sclosed, locate thebalance point P2.2. Measure lengthAP2.3. There is currentI2 passing the celland the resistor R.4. Length AP2represents thevoltage V acrossR.Measure the internal resistance r of a cellABI+-P2VoSRr‘I2V
108 Measure the internal resistance r of a cell 1. Note that the two currentsI and I2 do not mix up.ABI+-P2VoSRr‘I2V2. We have ’ = I2.R + I2.r
109 Measure the internal resistance r of a cell ‘ = AP (1)V = AP (2)’ = I2.R + I2.r (3)V = I2.R (4)‘rI2RVFind r in terms of AP1, AP2 and R.
110 Example 11 First find the voltage per unit length . The terminal voltage V is the voltage across the cell with current flowing.The e.m.f. is the voltage across the cell without current flowing.The above two are not equal if the cell has internal resistance r.
111 Calibrating a voltmeter To check that the reading of a voltmeter is correct.V+-
112 Calibrating a voltmeter 1. V = . APAPB+-VRR’E
113 Calibrating a voltmeter +-PVoVRR’E1. V = . AP2. Connect a voltmeterto measure the voltageacross R. Compare itsreading with the abovevalue3. Adjust R for other readings of the voltage.Each time locate the new balance point
114 Measure resistance RABI+-PVoRER’Measure V across R using the potentiometer. V= ξ.APRead the current I through the resistor.R =
115 Electrical metersIdeal measuring instruments: should not change the system being measured.Ammeter: its resistance should be as small as possible.Voltmeter: its resistance should be as large as possible.
116 Moving-coil galvanometer It can measure a small current.
119 Moving-coil galvanometer Can be changed into an ammeter or voltmeter.Typical data of a moving-coil galvanometer:f.s.d. current = 1.0 mAf.s.d. voltage = 0.15 Vresistance = 150 Ω+-
120 Ammeter It is connected in series in the circuit. R A A I Note that the total resistance of thecircuit is R + r where r is the resistanceof the ammeter.
121 Adapt a galvanometer to measure a large current (an ammeter) Shunt: add a resistor in parallel with the galvanometer.The adapted galvanometer has a smaller resistance.moving-coil galvanometerthe ammeter+-shuntR
122 Adapt a galvanometer to measure a large current (an ammeter) Example: A moving-coil galvanometer with f.s.d. current 1.0 mA is adapted to measure 1.0A at f.s.d.. What is the resistance of the shunt?moving-coil galvanometerI = 1.0AshuntI2I1=1.0mAVR = 0.15RNote that the p.d. across R = the p.d. across the galvanometer.
123 voltmeter It is connected in parallel in the circuit. V R V I Note that the total resistance of the circuit isR.r/(R + r) where r is the resistance of thevoltmeter.
124 Adapt a galvanometer to measure a large voltage (a voltmeter) Multiplier: add a resistor in series with the galvanometer.The adapted galvanometer has a larger resistance.multiplierR+-voltmeter
125 Adapt a galvanometer to measure a large voltage (a voltmeter) Example: A moving-coil galvanometer with f.s.d. voltage 0.1 V is adapted to measure 15 V at f.s.d.. What is the resistance of the multiplier?R15V0.1VI=1.0mAIR = 15 kNote that the same current is passingthrough the multiplier and the galvanometer.
126 Load effect of a voltmeter: a voltmeter draws current 1. Without voltmeter2. With a voltmeter of resistance 15 k6V5 kI = 0.6 mA3V6V5 kI = mAV15k3.43V2.57V0.171 mA0.515 mA
127 Load effect of a voltmeter: a voltmeter draws current The best solution is to add a voltage follower.A voltage follower draws a current as small as 0.1 nA from the circuit.6V5 kV0.6mA< 0.1nA-+The voltmeterdraws currentfrom the voltagefollower.Not fromthe circuit.15k
128 Ohm-meter (digital)To read the resistance directly from the ohm-meter.
129 Ohm-meter (analog) cell with emf 2. Moving-coil galvanometer 3. Variable resistancefor setting zeroZ+-4. Connect resistor to these terminalsfor measurement
130 Ohm-meter (analog) 1. When measuring a resistor R, + - R I Z 2. The deflection of the galvanometeris proportional to I.3. Adjust the scale of thegalvanometer to read R fromthe deflection.4. The scale is non-linear.
131 Measuring resistance by voltmeter-ammeter method Set up either of the following circuits. Measure V and I.Calculate R from1. For small R2. For big RAVRAVRII
132 Measuring resistance by voltmeter-ammeter method Discuss why there are two different circuits.Assume values for R in your discussion.1. For small R2. For big RAVRAVRII
133 Multimeter A-V-O meter: ammeter, voltmeter and ohm-meter. (Analog) (Analog)(Digital)
134 Multimeter moving-coil galvanometer shunts To set zero Z multipliers E 100mA1VE10V100V+_unknownresistor
135 Shunts and currents Given: f.s.d. current of the galvanometer = 1.0 mA resistance of the galvanometer = 150 resistance of each shunt = 0.15 Find: the f.s.d current in each of the following adapted galvanometer.I1I2 1A 0.5ARRRResistance of the shuntis smaller.Resistance of the shuntis larger.
136 Multimeter To measure large current shunts E Z + _ To set zero I 100mA+_To setzeroTo measure largecurrentI1. external currentflows into the multimeter2. externalcurrent throughthe shunt (oneresistor, biggercurrent).3. external current throughthe galvanometer4. external currentflows out of themultimeter
137 Multimeter To measure small current 3. external current through the galvanometerTo measure smallcurrentshunts2. externalcurrent throughthe shunt (threeresistors, lesscurrent).To setzeroZ1A10A100mAE+_4. external currentflows out of themultimeter1. external currentflows into the multimeterII
138 Multiplier and voltages Given: f.s.d. current of the galvanometer = 1.0mAf.s.d. voltage of the galvanometer = 0.15Vresistance of the galvanometer = 150 resistance of each multiplier = 15 kFind: the f.s.d voltage in each of the following adapted galvanometer.I1Resistance of the multiplieris smaller.RV1RV2I2Resistance of the multiplieris larger.15V30V
139 Multimeter To measure a high voltage 3. external current through the galvanometer2. externalcurrent throughthe multiplier(three resistors,higher voltage).To setzeroZmultipliers1VE10V100V4. external currentflows out of themultimeter_+1. external currentflows into the multimeterV
140 Multimeter To measure a low voltage 3. external current through the galvanometer2. externalcurrent throughthe multiplier(one resistor,lower voltage).To setzeroZmultipliers1VE10V100V4. external currentflows out of themultimeter_+1. external currentflows into the multimeterV
141 Multimeter 4. The galvanometer deflects to show the resistance R. To measure Rshort theinputs and setzero with Z.shuntsTo setzeroZmultipliersIE3. The internalcell providesthe current.+_2. connect an unknownresistor R to the inputs5. The currentdepends on R.R