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 The figure shows a resistor, inductor, and capacitor connected in series. The same current i passes through all of the elements in the loop. From Kirchhoff’s loop law, E = v R + v L + v C . Because of the capacitive and inductive elements in the circuit, the current i will not in general be in phase with E , so we will have i = I cos( ϖ t- φ ) where φ is the phase angle between current i and drive voltage E . If v L >v C then the current 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 = = + - + - E E / 1/ C C X v I C = =
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10/06/09 Physics 2212 - Lecture 25 Analyzing an LRC Circuit Draw the current vector I at some arbitrary angle. All elements of the circuit will have this current. Draw the resistor voltage V R in phase with the current. Draw the inductor and capacitor voltages V L and V C 90 0 before and behind the current, respectively. Draw the emf E 0 as the vector sum of V R and V L -V C . The angle of this phasor is ϖ t , where the time-dependent emf is E 0 cos ϖ t . The phasors V R and V L - V C form the sides 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 We can define the impedance Z of the circuit as: 2 2 2 2 ( ) ( 1/ ) L C Z R X X R L C ϖ + - = + - Then / I Z = E From the phasor diagram ,we see that the phase angle f of the current 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 - - - - = = 0 cos R V = E
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10/06/09 Physics 2212 - Lecture 25 Resonance 0 2 2 ( 1/ ) I Z R L C ϖ = = + - E E The current I will be a maximum when ϖ L=1/ ϖ C . This defines the resonant frequency of the system ϖ 0 : 0 1 LC ϖ= ( 29 ( 29 0 2 2 2 2 0 1 I R L 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 voltage of 5.0 mV. The tuning circuit consists of a 60 μ H inductor in series with a variable capacitor. The inductor coil has a resistance of 0.25 , and the resistance of the rest of the circuit is negligible. (a) To what capacitance should the capacitor be tuned to listen to this radio station. (b) What is the peak current through the circuit at resonance? (c) A stronger station at 1050 kHz produces a 10 mV antenna signal. What is the
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Phys2212_34.1-34.5 - Physics 2212 Electricity and Magnetism...

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