Presentation on theme: "Physics Beyond 2000 Chapter 15 Electric Circuits."— Presentation transcript:
Physics Beyond 2000 Chapter 15 Electric Circuits
Electric current An electric current is a flow of electric charges through a conductor. Current = rate of flow of charges unit: ampere A Constant current Changing current
Definition of ampere and coulomb Definition of 1 Ampere of current: 1m I = 1A FF 1.Two infinitely long parallel wires 1 m apart 2. The same amount of currents are in the wire 3. The forces on the wires are 2 × 10 -7 N m -1 4. The current in each wire is 1 ampere
Definition of ampere and coulomb Definition of 1 coulomb (C): Q = I.t 1 C = The amount of charges flowing through in 1 second with a current of 1 A. I = 1A t = 1 s
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.
Charge-carriers Charge-carriers in electrolyte: positive and negative ions. +- water positive electrode negative electrode positive ion negative ion
Electric field in a circuit 1. Cell with emf ξ 2. Switch S 3. Resistor R AB The switch is open.The p.d. of AB is zero. Not any current to flow.
Electric field in a circuit Close the switch S. An electric field is established inside the circuit. ξ S R AB E E
Electric field in a circuit ξ The electric field exerts force on the charge-carriers (electrons). Charge-carriers are accelerated. S R AB E E
Electric field in a circuit ξ S R AB The charge-carriers are also retarded by the collision with atoms. E E
Electric field in a circuit As a result, the charge-carriers are moving at constant speed. ξ S R AB E E
Drift velocity Charge-carriers in a metal move with constant speed under an electric field. The constant speed is called the drift velocity v D. E = electric field strength - - - - - - - - - - - - - - - - - - - - - vDvD
Drift velocity Consider a conductor with n charge-carriers per unit volume. E = electric field strength - - - - - - - - - - - - - - - - - - - - - vDvD
Drift velocity In time Δt, all the charge-carriers in the shaded region have passed the imaginary cross-section. E = electric field strength v D. Δt - - - - - - - - - - - - - - - - - - - - - vDvD cross-sectional area A
Drift velocity What is the number of charge-carriers ΔN in the shaded area? E = electric field strength v D. Δt cross-sectional area A ΔN = nA.(v D.Δt) - - - - - - - - - - - - - - - - - - - - - vDvD
Drift velocity Suppose each charge-carrier has charge q. What is the amount of charges ΔQ passing the cross- sectional area in time Δt? E = electric field strength v D. Δt ΔQ = q.ΔN = q.nA.(v D. Δt) - - - - - - - - - - - - - - - - - - - - - vDvD cross-sectional area A
Drift velocity What is the current I in the conductor? electric field strength v D. Δt - - - - - - - - - - - - - - - - - - - - - vDvD cross-sectional area A
Drift velocity So the drift velocity of the charge-carriers in a conductor is electric field strength v D. Δt - - - - - - - - - - - - - - - - - - - - - vDvD cross-sectional area A
Drift velocity So the drift velocity of the conductor is In small cross-sectional area, charge-carriers would move faster if other quantities are the same. In semi-conductor, n is small. The drift velocity is fast.
Drift velocity So the drift velocity of the conductor is In metal, the charge-carriers are electrons which are in random motion at very fast speed. This random speed does not contribute to the current.
Example 1 Drift velocity of the charge-carriers in a copper wire. (10 -5 ms -1 )
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.
Electromotive force 1. Suppose that the charge-carriers are positively charged. 2. The charge-carriers gain energy from the battery when they flow from the low potential to high potential inside the battery. 3. The charge-carriers lose energy when they move from high potential to low potential in the circuit. 4. The charge-carriers lose all energy when they return to the battery.
Electromotive force 1. 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 called the electromotive force of the battery.
Electromotive force The electromotive force ξof a source is the energy transferred into electrical energy per unit charge within the source. where Q is the amount of charge through the battery and U is the total energy provided by the battery to the charge Q. unit: J C -1 or V, volt
Electromotive force The electromotive force ξof a source exists though there is not any current.
Example 2 The e.m.f. (electromotive force) of an AA cell. Ah (ampere-hour) is a unit for charge. Q = I.t
Combination of cells Cells in series Cells in parallel
Cells in series ξ= ξ 1 + ξ 2 +ξ 3 Produce a high e.m.f. ξ1ξ1 ξ2ξ2 ξ3ξ3 ξ
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. ξ ξ
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) XY 1.An amount of charge Q passes from X to Y 2. The charge loses electrical energy U 3.
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)
Measurement of voltage Use voltmeter Use CRO XY to voltmeter or CRO I
Example 3 First find the amount of charge of one mole of electrons. U = QV
Voltage and electric field strength XY 1.A p.d. V between points X and Y. d 2. The separation XY is d. 3. There is an electric field E between X and Y. 4. The average electric field strength is E = V/d.
Example 4 Use E = V/d and F = e.E E d Note that the electrons move with uniform drift speed.
Resistance 1.The p.d. between the ends of the conductor is V V I 2. A current I passes through the conductor 3. The resistance of the conductor is R = V/I 4. Unit of R is ohm, Ω
I-V graph Pure metal R = constant or V I (Ohms law) The slope gives I V0 http://www.fed.cuhk.edu.hk/sci_lab/Simulations/phe/ohmslaw.htm
I-V graph Bulb filament. The filament resistance R increases with temperature. Ohms law is not obeyed. I V0
I-V graph Semi-conducting diode Current increases sharply in forward bias. Current is zero in reverse bias. I V0 current starts to increase
Dynamic resistance R d Find the slope from the I-V graph Use I V0
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
Effect of temperature (metals) Resistance increases uniformly with temperature for metals. θ/ o C 0 resistivity ρ ρoρo ρ= ρ o (1 + α.θ) αis called the temperature coefficient.
Effect of temperature (metals) An increase in temperature increases the thermal agitation of atoms. The interaction between atoms and electrons increases. The resistance increases. atom electron
Superconductivity Some metals or alloys lose the resistance at very low temperature. No dissipation of energy when current flows in a superconductor.
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. r = e.m.f. V = terminal voltage 1.e.m.f. describes the provision of power 2. terminal voltage describes the loss of power
To measure the internal resistance r of a battery A r R II 1.A battery with emf and internal resistance r 2. An ammeter to measure the current I. 3. A variable resistor R to adjust the current
To measure the internal resistance r of a battery A r R II Express R in terms of, I and r.
To measure the internal resistance r of a battery A r R II Plot the graph of R- R How to find r from the graph? -r
The internal resistance r of a battery We assume that the internal resistance is a constant. A r R II R -r
Combination of resistors Resistors in series Resistors in parallel
Combination of resistors Resistors in series A common current passing through all resistors The equivalent resistance R of the system is R = R 1 + R 2 + R 3 R1R1 R2R2 R3R3 I http://www.lightlink.com/sergey/java/java/resist2/index.html
Combination of resistors Resistors in parallel Same voltage across each resistor The equivalent resistor R gives R1R1 R2R2 R3R3 V http://www.lightlink.com/sergey/java/java/resist4/index.html
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 is I R V
Power of a cell If the e.m.f. of a cell is ξand it provides a current I for an external circuit, then the power of the cell is r R II
Example 5 Note that the internal resistor r is not counted in the power output. r R II power output V
Power output and external resistance The power output P out is r R II power output P max R r 0 power output
Power output and external resistance The power output is maximum when R = r. r R II power output P max R r 0 power output
Power output and external resistance What is the efficiency when R = r? r R II Efficiency R r 0 power output 100% 50% Hint: efficiency = P out /P o
Example 6 Use a cell of r = 2 Ω
Power transmission High voltage is used for transmission of electric power.
Power transmission What is the loss of power in the cable? (Use I L, the current in the cable, and R, the resistance of the cable.)
Domestic Electricity mains 1.Live wire: with potential changing between positive and negative 2. Neutral wire: maintaining at zero potential by earthing in the power station. 3. Electrical devices are connected in parallel. Same p.d. for each device.
Ring circuit 1. Same p.d. for each device. liveneutral A B C
Ring circuit liveneutral A B C 1. Same p.d. for each device. 2. If a part is broken, can it be operated? YES, it can.
Combination of two cells in parallel with same e.m.f. 1.A cell with emf and internal resistance r 1 2. A cell with emf and internal resistance r 2 3. Connect them in parallel 4. Note that there is not any current flow between the cells because their emf are the same.
Combination of two cells in parallel with different e.m.f. 1.A cell with emf 1 and internal resistance r 1 2. A cell with emf 2 and internal resistance r 2 4. Connect them in parallel 3. 1 > 2 5. There is current flowing from the cell of high emf to that of low emf
Example 6 (p.327) 1.A cell with emf 14V and internal resistance 5. 2. A cell with emf 12V and internal resistance 15. 4. Connect them in parallel 3. 1(14V) > 2(12V) 5. There is current flowing from the cell of high emf to that of low emf. Find the current.
Combination of two cells in parallel with same emf r1r1 r2r2 R 1.Two cells of emf are connected in parallel. 3. Set up equations for these quantities. 2. Both cells provide current I for the external resistor R. I1I1 I2I2 I I
Combination of two cells in parallel with same emf r1r1 r2r2 R I1I1 I2I2 I I I 1 + I 2 = I................. (1)
Combination of two cells in parallel with same emf r1r1 R I1I1 I I I 1 + I 2 = I................. (1) = I.R + I 1.r 1............ (2)
Combination of two cells in parallel with same emf r2r2 R I2I2 I I I 1 + I 2 = I................. (1) = I.R + I 1.r 1............ (2) = I.R + I 2.r 2............ (3)
Example 6 r1r1 r2r2 R I1I1 I2I2 I I Given: = 1.5V, r 1 = 2, r 2 = 4 and R = 6. Find the currents and the power output.
I-V characteristics Linear: Obey Ohms law. Ohmic device. Non-linear: Not obey Ohms law. Non- ohmic device. I V I V Ohmic Non-ohmic Note that R =
I-V characteristics I V Ohmic I V = 3 V r = 1 Find R if I = 1 A. R
I-V characteristics Junction diode I/mA V D /V 0 I VDVD 20 0.8 V D is 0.8 V for a current of 20 mA. If the current is too large, the diode would be damaged. +-
I-V characteristics Junction diode I/mA V D /V 0 20 0.8 I = 3V, r = 0 0.8V R Find the suitable value of R. What would happen if R is too big or too small?
I-V characteristics Thermionic diode I VDVD I/mA V D /V0 2 4 The maximum current is 2.0 mA. This current is called the saturation current.
I-V characteristics Thermionic diode I = 2 mA VDVD =12V, r = 0 R I/mA V D /V0 2 4 Find the minimum value for R.
Kirchhoffs rules 1. Kirchhoff's Junction Rule, 2. Kirchhoff's Loop Rule and 3. Ohm's Law (i.e. V/I = R, where I is the current, V is the voltage, and R is the resistance).
Kirchhoffs rules 1. Kirchhoff's Junction Rule The algebraic sum of the currents at any branch point or junction in a circuit is zero. I1I1 I2I2 I3I3 I4I4 I5I5 I 1 + I 2 + I 3 - I 4 - I 5 = 0 Note that it is necessary to add a minus sign before I 4 and I 5 because they are leaving the junction.
Kirchhoffs rules 2. Kirchhoff's Loop Rule The algebraic sum of the potential changes around any complete loop in the network is zero. I1I1 I2I2 I3I3 1 2 R1R1 R2R2 R3R3 R4R4 A B CD E F Note the the potential drops by IR when current passing a resistor. It is necessary to add a minus before I.R following the current.
Kirchhoffs rules 2. Kirchhoff's Loop Rule The algebraic sum of the potential changes around any complete loop in the network is zero. I1I1 I2I2 I3I3 1 2 R1R1 R2R2 R3R3 R4R4 A B CD E F Consider the loop ABEF: 1 -I 1 R 1 -I 2 R 2 - 2 -I 2 R 4 = 0
Bridge circuits 4 resistors are connected as a bridge circuit. Two currents in the bridge circuit. R1R1 R2R2 R3R3 R4R4 I I I1I1 I2I2 I = I 1 + I 2 A B D C
Bridge circuits If the potentials at B and C are equal, Connect BD with a galvanometer. Is there any current passing the galvanometer? I = I 1 + I 2 R1R1 R2R2 R3R3 R4R4 I I I1I1 I2I2 A B D C
Example 7 The galvanometer shows no reflection No current flows between B and D The potentials at B and D are equal
Measuring devices Potentiometer
Measuring devices Potentiometer –To measure small p.d. –To measure the internal resistance of a cell – To calibrate a voltmeter – To measure resistance Advantage: It can measure the voltage of a device without drawing any current from the device. http://www.plus2physics.com/current_electricity/study_material.asp?chapter=5
Potentiometer An accumulator provides a steady current for a long period of time. A resistance wire with uniform cross-section. 1. accumulator V o (2V) AB 2. A uniform resistance wire of length I + -
To measure an unknown voltage (p.d.) AB I +- V 1.Device with p.d. 2. A galvanometer to detect current through the device 3. A protective resistor 4. A slide to contact wire AB. + - 5. Both + ends are connected VoVo
To measure an unknown voltage (p.d.) 1.Move the slide on AB to find a balance point. 2. The galvanometer does not deflect when P is the balance point. AB I +- V + - P VoVo 3.Show that
Using the protective resistor AB I +- V + - P VoVo 1.Before the balance point is found, the current passing the galvanometer is large. Use the protective resistor to decrease the current. 2. When the balance point is near, the current passing the galvanometer is small. Short the protective resistor to increase the current.
Calibrating a potentiometer Use a standard cell (e.g. Weston cell with emf = 1.0186 V at 20 o C) to calibrate a potentiometer. AB I + - P VoVo standard cell
Calibrating a potentiometer Find the balance point P and measure the length AP. Find the voltage per unit length of the resistance wire. AB I + - P VoVo standard cell 1.What is the voltage across AP? 2. So what is ? 3. =
Calibrating a potentiometer Use this calibrated potentiometer to measure the p.d. V of a device. AB I + - P VoVo device +- V 1. V =.AP
No balance point is found The terminals of the device are reversed. The p.d. of the device is too large. V > V o. AB I + - P VoVo device +- V
Measure the emf of a cell AB I + - P VoVo cell with emf V 1.Find the balance point P 2.emf V = AP where is the voltage per unit length. 3. Why is the result the emf? Why not the p.d.? It is because there is not any current through the cell. The pd is equal to the emf.
Example 8 The SI unit of is V m -1.
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. AB I + - P VoVo device with very small V + -
Measuring very small p.d. Add a variable resistor R in series with the resistance wire. Adjust R so that V AB is just greater than V. Must re-calibrate the potentiometer when R is changed. AB I + - P VoVo device with very small V + - R
Examples 9-10 The emf of a thermocouple is very small. Adjust R so that the balance point P is near B. AB I + - P VoVo R cold junction hot junction V
Measure the internal resistance r of a cell AB I + - P1P1 VoVo S R r 3. R is a known resistor 4. S is a switch 2. r is the internal resistance 1. The cell
Measure the internal resistance r of a cell AB I + - P1P1 VoVo S R r 1.With switch S open, locate the balance point P 1. 2. Measure length AP 1. 3. The length AP 1 represents the emf of the cell. 4. There is not any current flowing through the cell
Measure the internal resistance r of a cell 1.With switch S closed, locate the balance point P 2. 2. Measure length AP 2. 3. There is current I 2 passing the cell and the resistor R. 4. Length AP 2 represents the voltage V across R. AB I + - P2P2 VoVo S R r I2I2 V
Measure the internal resistance r of a cell AB I + - P2P2 VoVo S R r I2I2 V 2. We have = I 2.R + I 2.r 1. Note that the two currents I and I 2 do not mix up.
Measure the internal resistance r of a cell r R I2I2 V = AP 1.............(1) V = AP 2.............(2) = I 2.R + I 2.r.........(3) V = I 2.R................ (4) Find r in terms of AP 1, AP 2 and R.
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.
Calibrating a voltmeter To check that the reading of a voltmeter is correct. V + -
Calibrating a voltmeter AB I + - P VoVo V RR E 1. V =. AP
Calibrating a voltmeter 1. V =. AP AB I + - P VoVo V RR E 2. Connect a voltmeter to measure the voltage across R. Compare its reading with the above value 3. Adjust R for other readings of the voltage. Each time locate the new balance point
Measure resistance R Measure V across R using the potentiometer. V= ξ.AP Read the current I through the resistor. R = AB I + - P VoVo R E A R I
Electrical meters Ideal 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.
Moving-coil galvanometer It can measure a small current. http://hyperphysics.phy- astr.gsu.edu/hbase/magnetic/galvan.html#c1http://hyperphysics.phy- astr.gsu.edu/hbase/magnetic/galvan.html#c1
Can be changed into an ammeter or voltmeter. Typical data of a moving-coil galvanometer: f.s.d. current = 1.0 mA f.s.d. voltage = 0.15 V resistance = 150 Ω Moving-coil galvanometer +-
Ammeter It is connected in series in the circuit. I R A A Note that the total resistance of the circuit is R + r where r is the resistance of the ammeter.
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 galvanometer R shunt the ammeter +-
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 galvanometer I = 1.0AI 1 =1.0mA shunt I2I2 R Note that the p.d. across R = the p.d. across the galvanometer. R = 0.15 V
voltmeter It is connected in parallel in the circuit. I R V Note that the total resistance of the circuit is R.r/(R + r) where r is the resistance of the voltmeter. V
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. R multiplier +- voltmeter
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? R 15V 0.1V I=1.0mAI Note that the same current is passing through the multiplier and the galvanometer. R = 15 k
Load effect of a voltmeter: a voltmeter draws current 1. Without voltmeter 6V 5 k I = 0.6 mA 3V 2. With a voltmeter of resistance 15 k 6V 5 k I = 0.686 mA V 15k 3.43V 2.57V 0.171 mA 0.515 mA
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. 6V 5 k V 0.6mA < 0.1nA 0.6mA - + 15k The voltmeter draws current from the voltage follower. Not from the circuit.
Ohm-meter (digital) To read the resistance directly from the ohm-meter.
Ohm-meter (analog) 2. Moving-coil galvanometer 3. Variable resistance for setting zero 1.cell with emf 4. Connect resistor to these terminals for measurement +- Z
Ohm-meter (analog) +- R I Z 1. When measuring a resistor R, 2. The deflection of the galvanometer is proportional to I. 3. Adjust the scale of the galvanometer to read R from the deflection. 4. The scale is non-linear.
Measuring resistance by voltmeter-ammeter method A V R A V R 1. For small R2. For big R I I Set up either of the following circuits. Measure V and I. Calculate R from
Measuring resistance by voltmeter-ammeter method A V R A V R 1. For small R2. For big R I I Discuss why there are two different circuits. Assume values for R in your discussion.
Multimeter moving-coil galvanometer shunts E Z multipliers unknown resistor 10A 1A 100mA 1V 10V 100V +_ To set zero
Shunts and currents I1I1 I2I2 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. RR R Resistance of the shunt is smaller. Resistance of the shunt is larger. 1A 0.5A
Multimeter shunts E Z 10A 1A 100mA +_ To set zero To measure large current I I 1. external current flows into the multimeter 2. external current through the shunt (one resistor, bigger current). 3. external current through the galvanometer 4. external current flows out of the multimeter
Multimeter shunts E Z 10A 1A 100mA +_ To set zero To measure small current I I 1. external current flows into the multimeter 2. external current through the shunt (three resistors, less current). 3. external current through the galvanometer 4. external current flows out of the multimeter
Multiplier and voltages Given: f.s.d. current of the galvanometer = 1.0mA f.s.d. voltage of the galvanometer = 0.15V resistance of the galvanometer = 150 resistance of each multiplier = 15 k Find: the f.s.d voltage in each of the following adapted galvanometer. I1I1 Resistance of the multiplier is smaller. R V1V1 R R V2V2 I2I2 Resistance of the multiplier is larger. 15V 30V
Multimeter To measure a high voltage E Z multipliers 1V 10V 100V + _ To set zero V 1. external current flows into the multimeter 2. external current through the multiplier (three resistors, higher voltage). 3. external current through the galvanometer 4. external current flows out of the multimeter
Multimeter To measure a low voltage E Z multipliers 1V 10V 100V + _ To set zero V 1. external current flows into the multimeter 2. external current through the multiplier (one resistor, lower voltage). 3. external current through the galvanometer 4. external current flows out of the multimeter
Multimeter shunts E Z multipliers 2. connect an unknown resistor R to the inputs +_ To set zero R I 1.short the inputs and set zero with Z. 3. The internal cell provides the current. 4. The galvanometer deflects to show the resistance R. 5. The current depends on R. To measure R