PHYS 1444 – Section 003 Lecture #21AnnouncementsWhy do we care about circuits on AC?AC Circuits – the preambleAC Circuit w/ Resistance onlyAC Circuit w/ Inductance onlySlide 7Example 31 – 1AC Circuit w/ Capacitance onlySlide 10Slide 11Example 31 – 2AC Circuit w/ LRCSlide 14Phasor DiagramsSlide 16Slide 17Monday, Nov. 21, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu1PHYS 1444 – Section 003Lecture #21Monday, Nov. 21, 2005Dr. Jaehoon Yu•AC Circuit w/ Resistance only•AC Circuit w/ Inductance only•AC Circuit w/ Capacitance only•AC Circuit w/ LRCToday’s homework is homework #11, due noon, next Tuesday!!Monday, Nov. 21, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu2Announcements•Reading assignment–CH. 31 – 6, 31 – 7 and 31 – 8 •There is class this Wednesday!!!•Final term exam –Time: 11am – 12:30pm, Monday Dec. 5–Location: SH103–Covers: 29.3 – which ever chapter we finish next, Wednesday, Nov. 30–Please do not miss the exam–Two best of the three exams will be used for your gradesMonday, Nov. 21, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu3Why do we care about circuits on AC?•The circuits we’ve learned so far contain resistors, capacitors and inductors and have been connected to a DC source or a fully charged capacitor–What? This does not make sense. –The inductor does not work as an impedance unless the current is changing. So an inductor in a circuit with DC source does not make sense.–Well, actually it does. When does it impede?•Immediately after the circuit is connected to the source so the current is still changing. So?–It causes the change of magnetic flux.–Now does it make sense?•Anyhow, learning the responses of resistors, capacitors and inductors in a circuit connected to an AC emf source is important. Why is this?–Since most the generators produce sinusoidal current–Any voltage that varies over time can be expressed in the superposition of sine and cosine functionsMonday, Nov. 21, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu4AC Circuits – the preamble•Do you remember how the rms and peak values for current and voltage are related?•The symbol for an AC power source is •We assume that the voltage gives rise to current –wherermsV =rmsI =I =2 fv p=0V 20I 20sin 2I ftp =0sinI tvMonday, Nov. 21, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu5AC Circuit w/ Resistance only•What do you think will happen when an ac source is connected to a resistor?•From Kirchhoff’s loop rule, we obtain•Thus–where•What does this mean?–Current is 0 when voltage is 0 and current is in its peak when voltage is in its peak.–Current and voltage are “in phase”•Energy is lost via the transformation into heat at an average rate• 0V IR- =V =0 0V I R=P =0sinI R tv =0sinV tvIV =2rmsI R =2rmsV RMonday, Nov. 21, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu6AC Circuit w/ Inductance only•From Kirchhoff’s loop rule, we obtain•Thus–Using the identity • –where •What does this mean?–Current and voltage are “out of phase by /2 or 90o” in other words the current reaches its peak ¼ cycle after the voltage•What happens to the energy?–No energy is dissipated –The average power is 0 at all times–The energy is stored temporarily in the magnetic field–Then released back to the source0dIV Ldt- =V =dILdt=cosq =V =0V =( )0sind I tLdtv=0cosLI tv v( )sin 90q +o( )0sin 90LI tv v + =o( )0sin 90V tv +o0LIvMonday, Nov. 21, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu7AC Circuit w/ Inductance only•How are the resistor and inductor different in terms of energy?–Inductor–Resistor•How are they the same?–They both impede the flow of charge–For a resistance R, the peak voltage and current are related to –Similarly, for an inductor we can write•Where XL is the inductive reactance of the inductor•What do you think is the unit of the reactance? •The relationship is not valid at a particular instance. Why not?–Since V0 and 0 do not occur at the same timeStores the energy temporarily in the magnetic field and then releases it back to the emf source Does not store energy but transforms it to thermal energy, getting it lost to the environment0V =LX Lv=0 0 LV I X=0I R0 0 LV I X=rms rms LV I X=is valid!0 when =0.Monday, Nov. 21, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu8Example 31 – 1 Reactance of a coil. A coil has a resistance R=1.00 and an inductance of 0.300H. Determine the current in the coil if (a) 120 V dc is applied to it; (b) 120 V ac (rms) at 60.0Hz is applied.Is there a reactance for dc? So for dc power, the current is from Kirchhoff’s rule0V IR- =For an ac power with f=60Hz, the reactance is Nope. Why not? Since =0, LX =0I =LX =rmsI �0VR=1201201.00VA=WLv =2 fLp =( )12 60.0 0.300 113s Hp-� � = WrmsLVX=1201.06113VA=WSince the resistance can be ignored compared to the reactance, the rms current is Lv =0Monday, Nov. 21, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu9AC Circuit w/ Capacitance only•What happens when a capacitor is connected to a dc power source?–The capacitor quickly charges up.–There is no steady current flow in the circuit•Since a capacitor prevents the flow of a dc current•What do you think will happen if it is connected to an ac power source?–The current flows continuously. Why?–When the ac power turns on, charge begins to flow one direction, charging up the plates–When the direction of the power reverses, the charge flows in the opposite directionMonday, Nov. 21, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu10AC Circuit w/ Capacitance only•From Kirchhoff’s loop rule, we obtain•Current at any instance is•This the charge Q on the plate at any instance is•Thus the voltage across the capacitor is –Using the identity –Where– V =cosq =( )sin 90q- -oI =Q =V =V =0V =dQdt=0sinI tv0QQdQ==�00sinttI tdtv==�0cosItvv-QC=01cosI tCvv-( )01sin 90I tCvv- =o( )0sin 90V tv -o0ICvQCMonday, Nov. 21, 2005 PHYS 1444-003, Fall 2005Dr. Jaehoon Yu11AC Circuit w/ Capacitance only•So the voltage is •What does this mean?–Current and voltage are “out of phase by /2 or 90o” but in this case, the voltage reaches its peak ¼ cycle after the current•What happens to the energy?–No energy is dissipated –The average power is 0 at all times–The energy is stored temporarily in the electric field–Then released back to the source•Applied voltage and the current in the capacitor can be written
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