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Phys2212_34.1-34.5

# Phys2212_34.1-34.5 - Physics 2212 Electricity and Magnetism...

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Physics 2212 Electricity and Magnetism Lecture 25 (Knight: 34.1 to 34.5) Maxwell’s Equations

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10/06/09 Physics 2212 - Lecture 25 Thefigureshows a resistor, inductor, and capacitor connected in series. Thesamecurrent i passes through all of theelements in theloop. From Kirchhoff’s loop law, E = v R + v L + v C . Becauseof thecapacitiveand inductiveelements in thecircuit, thecurrent i will not in general bein phase with E , so wewill have i = I cos( ϖ t- φ ) where φ is thephaseanglebetween current i and drivevoltage E . If v L >v C then thecurrent i lags E and φ >0 . If v C >v L then i leads E and φ <0. The Series RLC Circuit / L L X v I L ϖ = = 2 2 2 2 2 2 0 ( ) ( ) R L C L C v v v R X X I = + - = + - E 0 0 2 2 2 2 ( ) ( 1/ ) L C I R X X R L C ϖ ϖ = = + - + - E E / 1/ C C X v I C ϖ = =
10/06/09 Physics 2212 - Lecture 25 Analyzing an LRC Circuit Draw thecurrent vector I at somearbitrary angle. All elements of thecircuit will havethis current. Draw theresistor voltage V R in phasewith the current. Draw the inductor and capacitor voltages V L and V C 90 0 beforeand behind the current, respectively. Draw theemf E 0 as the vector sum of V R and V L -V C . Theangleof this phasor is ϖ t , where thetime-dependent emf is E 0 cos ϖ t . Thephasors V R and V L - V C form thesides of a right triangle, with E 0 as the hypotenuse. Therefore, E 0 2 = V R 2 + (V L - V C ) 2 .

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10/06/09 Physics 2212 - Lecture 25 Impedance and Phase Angle Wecan definethe impedance Z of the circuit as: 2 2 2 2 ( ) ( 1/ ) L C Z R X X R L C ϖ ϖ + - = + - Then / I Z = E From thephasor diagram ,weseethat the phase angle f of thecurrent is given by: ( 29 tan L C L C R I X X V V V IR φ - - = = 1 1 1/ tan tan L C X X L C R R ϖ ϖ φ - - - - = = 0 cos R V φ = E
10/06/09 Physics 2212 - Lecture 25 Resonance 0 2 2 ( 1/ ) I Z R L C ϖ ϖ = = + - E E Thecurrent I will bea maximum when ϖ L=1/ ϖ C . This defines the resonant frequency of thesystem ϖ 0 : 0 1 LC ϖ = ( 29 ( 29 0 2 2 2 2 0 1 I R L ϖ ϖ ϖ = + - E C

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10/06/09 Physics 2212 - Lecture 25 Example : Designing a Radio Receiver An AM radio antenna picks up a 1000 kHz signal with a peak voltageof 5.0 mV. Thetuning circuit consists of a 60 μ H inductor in series with a variablecapacitor. Theinductor coil has a resistanceof 0.25 , and theresistanceof therest of the circuit is negligible. (a) To what capacitance should thecapacitor betuned to listen to this radio station. (b) What is thepeak current through thecircuit at resonance? (c) A stronger station at 1050 kHz produces a 10 mV antenna signal. What is the current in theradio at this frequency when thestation is tuned to 1000 kHz.
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Phys2212_34.1-34.5 - Physics 2212 Electricity and Magnetism...

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