PHYS 1444 – Section 02 Lecture #20AC Circuits – the preambleAC Circuit w/ Resistance onlyAC Circuit w/ Inductance onlySlide 5LR CircuitsDischarge of LR CircuitsSlide 8Example 31 – 1AC Circuit w/ Capacitance onlySlide 11Slide 12Example 31 – 2AC Circuit w/ LRCSlide 15Maxwell’s EquationsModifying Ampere’s LawDisplacement CurrentGauss’ Law for MagnetismSlide 20Slide 21Example 32 – 1Slide 23Tuesday, April 26, 2011 1PHYS 1444-002 Dr. Andrew BrandtPHYS 1444 – Section 02Lecture #20 Tuesday April 26, 2011Dr. Andrew Brandt•AC Circuits•Maxwell EquationsTake home Quiz Ch 29 due nowHW9 isdue SundayMay 1 @ 10pm Note animation after slide 5 needs work!Tuesday, April 26, 2011AC 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 tv2PHYS 1444-002 Dr. Andrew BrandtTuesday, April 26, 2011AC 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 at its peak when voltage is at 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 R3PHYS 1444-002 Dr. Andrew BrandtAC 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 fieldthen 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 +o0LIvTuesday, April 26, 20114PHYS 1444-002 Dr. Andrew BrandtTuesday, April 26, 2011AC 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 by –Similarly, for an inductor we can write , where and XL is the inductive reactance of the inductor BUT since the voltage and current are out of phase this equation is not valid at any particular time. However Note the unit of XL is , while the unit of L is Henry (what is a Henry?)Stores 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, losing it to the environment0V =LX Lv=0I R0 0 LV I X=rms rms LV I X=is valid!5PHYS 1444-002 Dr. Andrew BrandtTuesday, April 26, 2011LR Circuits•This can be shown w/ Kirchhoff rule loop rules–The emfs in the circuit are the battery voltage V0 and the emf =-L(dI/dt) in the inductor opposing the current increase–The sum of the potential changes through the circuit is –Where I is the current at any instance–By rearranging the terms, we obtain a differential eq.– –We can integrate just as in RC circuit–So the solution is–Where =L/R•This is the time constant of the LR circuit and is the time required for the current I to reach 0.63 of the maximum 0V IRe+ - =L dI dt IR+ =00IIdIV IR==-�001lnV IRR V� �-- =� �� �( ) ( )0 max1 1t tI V e R I et t- -= - = -0V L dI dt IR- - =00V0ttdtL=�tL6PHYS 1444-002 Dr. Andrew BrandtDischarge of LR Circuits•If the switch is flipped away from the battery–The differential equation becomes– –So the integration is–Which results in the solution– –The current decays exponentially to zero with the time constant =L/R–So there always is a reaction time when a system with a coil, such as an electromagnet, is turned on or off.–The current in LR circuit behaves in a similar manner as for the RC circuit, except that in steady state RC current is 0, and the time constant is inversely proportional to R in LR circuit unlike the RC circuitL dI dt IR+ =0IIdIIR==�0lnI RtI L=-0 0RttLI I e I et--= =00ttdtL=�7Tuesday, April 26, 2011AC Circuit w/ Inductance only•From Kirchhoff’s loop rule, we obtain•Thus–Using the identity • –where –Current and voltage are “out of phase by /2 or 90o”. 0dIV Ldt- =V =dILdt=cosq =V =0V =( )0sind I tLdtv=0cosLI tv v( )sin 90q +o( )0sin 90LI tv v + =o( )0sin 90V tv +o0LIv–For an inductor we can write•Where XL is the inductive reactance of the inductor•The relationship is not generally valid since V0 and 0 do not occur at the same time, but is validLX Lv=0 0 LV I X=0 0 LV I X=rms rms LV I X=8PHYS 1444-002 Dr. Andrew BrandtTuesday, April 26, 2011Example 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 =09PHYS 1444-002 Dr. Andrew BrandtTuesday, April 26, 2011AC 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 direction10PHYS 1444-002 Dr. Andrew BrandtTuesday, April 26, 2011AC Circuit w/ Capacitance only•From Kirchhoff’s loop rule, we obtain•Current at any instant is•Thus, the charge Q on the plate at any instance is•The voltage across the capacitor is –Using the identity –Where– V =cosq =( )sin 90q- -oI =Q =V =V =0V =dQdt=0sinI
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