Wednesday, Oct. 5, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu1PHYS 1444 – Section 003Lecture #11Wednesday, Oct. 5, 2005Dr. Jaehoon Yu• Alternating Current (AC)•Power in AC• Microscopic view of current• Superconductivity• Electric shock hazards• EMF and Terminal VoltageWednesday, Oct. 5, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu2Announcements• Homework due has been changed to noon Tuesdays starting #6• First term exam next Wednesday, Oct. 12– Time: 1 – 2:20 pm– Location: SH103– Coverage: CH. 21 – 25• Reading Assignments– CH25 – 9– CH25 – 10• There will be a workshop 1 – 5pm this Saturday in SH103 for construction of the World’s Largest Cloud Chamber– Food from 12:30pmWednesday, Oct. 5, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu3Alternating Current• Does the direction of the flow of current change when a battery is connected to a circuit?–No. Why?• Because its source of potential difference stays put.– This kind of current is called the Direct Current (DC), and it does not change its direction of flow.• How would DC look as a function of time?– A straight line• Electric generators at electric power plant produce alternating current (AC)– AC reverses direction many times a second– AC is sinusoidal as a function of time• Most the currents supplied to homes and business are AC.Wednesday, Oct. 5, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu4Alternating Current• The voltage produced by an AC electric generator is sinusoidal– This is why the current is sinusoidal• Voltage produced can be written as • What are the maximum and minimum voltages?–V0and –V0– The potential oscillates between +V0and –V0, the peak voltages or amplitude– What is f?• The frequency, the number of complete oscillations made per second. What is the unit of f? What is the normal size of f in the US?–f=60Hz in the US and Canada. – Many European countries have f=50Hz.– ω=2πfV =0sin 2Vftπ=0sinVtϖWednesday, Oct. 5, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu5Alternating Current• Since V=IR, if a voltage V exists across a resistance R, the current I is• What are the maximum and minimum currents?–I0and –I0– The current oscillates between +I0and –I0, the peak currents or amplitude. The current is positive when electron flows to one direction and negative when they flow opposite.– AC is as many times positive as negative. What’s the average current?• Zero. So there is no power and no heat is produced in a heater?– Yes, the electrons actually flow back and forth and power is delivered.VIR==0sin 2VftRπ=0sinItϖWhat is this?Wednesday, Oct. 5, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu6Power Delivered by Alternating Current• AC power delivered to a resistance is: – Since the current is squared, the power is always positive• The average power delivered is• Since the power is also P=V2/R, we can obtain• The average of the square of current and voltage are important in calculating power:2220sinPIRIR tϖ==2012PIR=()220sinPVR tϖ=2012VPR⎛⎞=⎜⎟⎜⎟⎝⎠Average power22012II=22012VV=Wednesday, Oct. 5, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu7Power Delivered by Alternating Current• The square root of each of these are called root-mean-square, or rms:• rms values are sometimes called effective values– These are useful quantities since they can substitute current and voltage directly in power, as if they are in DC – In other words, an AC of peak voltage V0or peak current I0produces as much power as DC voltage of Vrmsor DC current Irms.– So normally, rms values are specified in AC are specified or measured.• US uses 115V rms voltage. What is the peak voltage?•• Europe uses 240V•2000.7072rm sIIII===2000.7072rmsVVV V===22012rm sPIRIR==22012rm sVVPRR==0V =2rm sV=2 115 162.6VV⋅=0V =2rm sV=2 240 340VV⋅=rm s rm sPIV=Wednesday, Oct. 5, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu8Example 25 – 11 Hair Dryer. (a) Calculate the resistance and the peak current in a 1000-W hair dryer connected to a 120-V line. (b) What happens if it is connected to a 240-V line in Britain? The rms current is:(b) If connected to 240V in Britain … rmsI=rmsPV=10008.33120WAV=Thus the resistance is:R=2rmsPI=()2100014.48.33WA=ΩThe peak current is:0I=2rmsI=2 8.33 11.8AA⋅=The average power provide by the AC in UK is P =So? The heating coils in the dryer will melt! 2rmsVR=()2240400014.4VW=ΩWednesday, Oct. 5, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu9• When a potential difference is applied to the two ends of a wire of uniform cross-section, the direction of electric field is parallel to the walls of the wire, this is possible since the charges are moving, electrodynamics• Let’s define a microscopic vector quantity, the current density,j, the electric current per unit cross-sectional area– j=I/A or I = jA if the current density is uniform– If not uniform – The direction of j is the direction a positive charge would move when placed at that position, generally the same as E• The current density exists on any point in space while the current I refers to a conductor as a whole so a macroscopic Microscopic View of Electric CurrentIjdA=⋅∫GGWednesday, Oct. 5, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu10• The direction of j is the direction of a positive charge. So in a conductor, since negatively charged electrons move, the direction is –j.• Let’s think about the current in a microscopic view again:– When voltage is applied to the end of a wire– Electric field is generated by the potential difference– Electrons feel force and get accelerated – Electrons soon reach to a steady average speed due to collisions with atoms in the wire, called drift velocity, vd– The drift velocity is normally much smaller than electrons’ average random speed. Microscopic View of Electric CurrentWednesday, Oct. 5, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu11• How do we relate vdwith the macroscopic current I?–In time ∆t, the electrons travel l=vd∆t on average– If wire’s x-sectional area is A, in time ∆t electrons in a volume V=lA=Avd∆t will pass through the area A– If there are n free electrons ( of charge –e) per unit volume, the total charge ∆Q that pass through A in time ∆tis–– The current I in the wire is– The density in vector form is– For any types of charge: Microscopic View of Electric CurrentQ∆=QIt∆==+IjA==GiidiiInqv
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